Example 01:

A steady P.D. is maintained between the ends of the potentiometer wire. A cell of e.m.f. 1.02 V is on an open circuit when its terminals are in contact with two points on the wire distant 150 cm. When the cell is shunted by a resistance of 4 ohms, this distance reduces to 120 cm. Find the internal resistance of the cell.

Given: e.m.f of a cell = e = 1.02 V, Balancing length when circuit is open = l = 150 cm = 1.5 m, Balancing length when cell is shunted l₁ = 120 cm = 1.2 m, Value of shunt = R = 4 ohm.

To Find: Internal resistance of cell = r =?

Solution:

r = 4 x ((1.5/1.2) – 1) = 4 x (1.25 – 1)

r = 4 x 0.25 = 1 ohm

Ans: Internal resistance of the cell is 1 ohm

Example 02:

A steady P.D. is maintained between the ends of the potentiometer wire. A Daniel cell when in an open circuit is balanced by a length of 108 cm. When the cell is shunted by a resistance of 10 ohms, the balancing length reduces to 90 cm. Find the internal resistance of the cell.

Given: Balancing length when circuit is open = l = 108 cm = 1.08 m, Balancing length when cell is shunted l₁ = 90 cm = 0.9 m, Value of shunt = R = 10 ohm.

To Find: Internal resistance of cell = r =?

Solution:

r = 10 x ((1.08/0.9) – 1) = 10 x (1.2 – 1)

r = 10 x 0.2 = 2 ohm

Ans: Internal resistance of the cell is 2 ohm

Example 03:

A steady P.D. is maintained between the ends of the potentiometer wire. A cell when in an open circuit is balanced by a length of 1.812 m. When the cell is shunted by resistance of 5 ohms, the balancing length reduces to 1.51 m. Find the internal resistance of the cell.

Given: Balancing length when circuit is open = l = 1.812 m, Balancing length when cell is shunted l₁ = 1.51 m, Value of shunt = R = 5 ohm.

To Find: Internal resistance of cell = r =?

Solution:

r = 5 x ((1.812/1.51) – 1) = 5 x (1.2 – 1)

r = 5 x 0.2 = 1 ohm

Ans: Internal resistance of the cell is 1 ohm

Example 04:

A potentiometer wire of length 4 m and resistance 8 ohms is connected in series with a battery of e.m.f. 2 V and negligible internal resistance. If the e.m.f. of cell balances against the length of 217 cm of the wire, find the e.m.f. of the cell. When a cell is shunted by a resistance of 15 ohms, the balancing length is reduced by 17 cm. Find the internal resistance of a cell.

Part 1:

Given: Length of potentiometer wire = l_AB = 4m, Resistance of potentiometer wire = R_AB = 8 ohm, Applied e.m.f = E = 2 V, Internal resistance = r = 0, Balancing length by cell = 217 cm = 2.17 m.

To Find: e.m.f. of cell = e =?

I = E/R_AB = 2/8 = 0.25 A

V_AB = I x R_AB = 0.25 x 8 = 2 V

Potential Drop = V_AB/l_AB = 2/4 = 0.5 V m^-1

E.m.f. of cell = Potential drop x Balancing length

E.m.f. of cell = 0.5 x 2.17 = 1.085 V

Part 2:

Given: Balancing length when circuit is open = l = 2.17 m, Balancing length when cell is shunted l₁ = 2.17 m – 17 cm = 2.17 m – 0.17 m = 2 m, Value of shunt = R = 15 ohm.

To Find: Internal resistance of cell = r =?

r = 15 x ((2.17/2) – 1) = 15 x (1.085 – 1)

r = 15 x 0.085 = 1.275 ohm

Ans: E.m.f. of the cell is 1.085 V and internal resistance of the cell is 1.275 ohm

Example 05:

A cell balances against a length of 150 cm on potentiometer wire when it is shunted by resistance of 5 ohms. The balancing length reduces to 175 cm when it is shunted by a resistance of 10 ohms. Find the balancing length when the ell is in an open circuit.

To Find: Balancing length of the cell when in open circuit = l =?

Solution:

Let ‘l’ be the balancing length when the cell is in open circuit

Case 1: Balancing length when cell is shunted l₁ = 150 cm = 1.5 m, Value of shunt = R = 5 ohm.

Case 2: Balancing length when cell is shunted l₁ = 175 cm = 1.75 m, Value of shunt = R = 10 ohm.

From equations (1) and (2)

Ans: Balancing length of the cell when in open circuit is open is 2.1 m

Example 06:

A cell balances against a length of 250 cm on potentiometer wire when it is shunted by a resistance of 10 ohms. The balancing length reduces to 200 cm when it is shunted by resistance of 5 ohms. Find internal resistance of the cell.

To Find: Internal resistance of cell = r =?

Solution:

Let ‘l’ be the balancing length when cell is in open circuit

Case 1: Balancing length when cell is shunted l₁ = 250 cm = 2.5 m, Value of shunt = R = 10 ohm.

Case 2: Balancing length when cell is shunted l₁ = 200 cm = 2 m, Value of shunt = R = 5 ohm.

From equations (1) and (2)

Substituting in equation (1)

Ans: Internal resistance of the cell is 10/3 ohm

Example 07:

A cell balances against a length of 250 cm on potentiometer wire when it is shunted by resistance of 5 ohms. The balancing length reduces to 400 cm when it is shunted by a resistance of 20 ohm. Find internal resistance of the cell.

To Find: Internal resistance of cell = r =?

Solution:

Let ‘l’ be the balancing length when the cell is in open circuit

Case 1: Balancing length when cell is shunted l₁ = 250 cm = 2.5 m, Value of shunt = R = 5 ohm.

Case 2: Balancing length when cell is shunted l₁ = 400 cm = 4 m, Value of shunt = R = 20 ohm.

From equations (1) and (2)

Substituting in equation (1)

Ans: Internal resistance of the cell is 5 ohm

Previous Topic: Numerical Problems on Potential Drop

Science > Physics > Current Electricity > Numerical Problems on Internal Resistance of Cell

The post Numerical Problems: (Internal Resistance of Cell) appeared first on The Fact Factor.

In this article, we shall study to solve numerical problems on the calculation of potential drop (potential gradient) on potentiometer wire.

Example 01:

A potentiometer wire is 10 m long and a P.D. of 6 V is maintained between its ends. Find the potential drop per centimeter of the wire. Also, find the e.m.f. of a cell which balances against a length of 180 cm of the wire.

Given: Length of potentiometer wire = l_AB = 10 m = 1000 cm, P.D. across potentiometer wire = V_AB = 6 V, Balancing length = l = 180 cm

To Find: Potential drop per cm =? e.m.f. of cell =?

Solution:

Potential drop per cm = V_AB/l_AB in cm = 6/1000 = 0.006 V cm^-1

E.m.f. of cell = Potential drop per cm x Balancing length in cm

E.m.f. of cell = 0.006 x 180 = 1.08 V

Ans: Potential drop = 0.006 V cm^-1 and e.m.f. of cell = 1.08 V

Example 02:

A potentiometer wire is 8 m long and a P.D. of 10 V is maintained between its ends. Find the potential drop of the wire. Also find the e.m.f. of a cell which balances against a length of 208 cm of the wire.

Given: Length of potentiometer wire =  l_AB = 8 m, P.D. across potentiometer wire = V_AB = 10 V, Balancing length = l = 208 cm = 2.08 m

To Find: Potential drop =? e.m.f. of cell =?

Solution:

Potential drop = V_AB/l_AB = 10/8 = 1.25 V m^-1

E.m.f. of cell = Potential drop x Balancing length

E.m.f. of cell = 1.25 x 2.08 = 2.6 V

Ans: Potential drop = 1.25 V m^-1 and e.m.f. of cell = 2.6 V

Example 03:

A potentiometer wire has a length of 2 m and a resistance of 10 ohm. It is connected in series with a cell of e.m.f. 4 V and internal resistance 6 ohm. Find the potential gradient on the wire. Find also where a cell of e.m.f. 1 volt will balance on the wire.

Given: Length of potentiometer wire = l_AB = 2 m, Resistance of potentiometer wire = R_AB = 10 ohm, e.m.f. of cell = E = 4 V, Internal resistance of cell = r = 6 ohm, e.m.f. of cell = e =  1 V

To Find: Potential gradient =? Balancing length = l =?

Solution:

I = E/(R_AB + r) = 4/(10 + 6) = 4/16 = 0.25 A

V_AB = I R_AB = 0.25 x 10 = 2.5 V

Potential gradient = V_AB/l_AB = 2.5/2 = 1.25 V m^-1

E.m.f. of cell = Potential drop x Balancing length

1 = 1.25 x l_AB

l_AB  = 1/1.25 = 0.8 m

Ans: Potential gradient = 1.25 V m^-1 and balancing length = 0.8 m

Example 04:

A potentiometer wire has a length of 4 m and a resistance of 10 ohm. It is connected in series with a cell of e.m.f. 4 V and internal resistance 2 ohm. Find the potential gradient on the wire. Find also where a cell of e.m.f. 1.5 volt will balance on the wire.

Given: Length of potentiometer wire = l_AB = 4m, Resistance of potentiometer wire = R_AB = 10 ohm, e.m.f. of cell = E = 4 V, Internal resistance of cell = r = 2 ohm, e.m.f. of cell = e = 1.5 V

To Find: Potential gradient =? Balancing length = l =?

Solution:

I = E/(R_AB + r) = 4/(10 + 2) = 4/12 = (1/3) A

V_AB = I R_AB = (1/3)  x 10 = (10/3) V

Potential gradient = V_AB/l_AB = (10/3)/4 = 10/12 = 5/6 =  0.8333V m^-1

E.m.f. of cell = Potential drop x Balancing length

1.5 = (5/6) x l_AB

l_AB  = (1.5 x 6)/5 = 1.8 m

Ans: Potential gradient = 0.8333 V m^-1 and balancing length = 1.8 m

Example 05:

A potentiometer wire has a length of 2 m and a resistance of 10 ohm. It is connected in series with a cell of e.m.f. 2 V and a resistance 990 ohm. Find the potential gradient on the wire.

Given: Length of potentiometer wire = l_AB = 2m, Resistance of potentiometer wire = R_AB = 10 ohm, e.m.f. of cell = E = 2 V, Resistance in series = R = 990 ohm.

To Find: Potential gradient =?

Solution:

I = E/(R_AB + R) = 2/(10 + 990) = 2/1000 = 0.002 A

V_AB = I R_AB = 0.002 x 10 = 0.02 V

Potential gradient = V_AB/l_AB = 0.02/2 = 0.01 V m^-1

Ans: Potential gradient = 0.01 V m^-1

Example 06:

A potentiometer wire has a length of 2 m and a resistance of 5 ohm. It is connected in series with a cell of e.m.f. 2 V and internal resistance 2 ohm and resistance of 993 ohm. Find the potential gradient on the wire. Find also where a cell of e.m.f. 4 mV will balance on the wire.

Given: Length of potentiometer wire = l_AB = 2 m, Resistance of potentiometer wire = R_AB = 5 ohm, e.m.f. of cell = E = 2 V, Internal resistance of cell = r = 2 ohm, Resistance in series = R = 993 ohm, e.m.f. of cell = e = 4mV = 4 x 10^-3 V

To Find: Potential gradient =? Balancing length = l =?

Solution:

I = E/(R_AB + r + R) = 2/(5 + 2 + 993) = 2/1000 = 2 x 10^-3 A

V_AB = I R_AB = 2 x 10^-3 x 5 = 0.01 V

Potential gradient = V_AB/l_AB = 0.01 /2 = 5 x 10^-3 V m^-1

E.m.f. of cell = Potential drop x Balancing length

4 x 10^-3 = 5 x 10^-3  x l_AB

l_AB = 4 x 10^-3 /5 x 10^-3  = 0.8 m

Ans: Potential gradient = 5 x 10^-3 V m^-1 and balancing length = 0.8 m

Example 08:

A potentiometer wire of length 5 m has a resistance of 5 ohm. What resistance must be connected in series with this wire and an accumulator of e.m.f. 4 volt so as to get a potential drop of 10^-2 Vm^-1 on the wire?

Given: Length of potentiometer wire = l_AB = 5 m, Resistance of potentiometer wire = R_AB = 5 ohm, e.m.f. of cell = E = 4 V, Resistance in series = R ohm, Potential drop = 10^-2 V m^-1.

To Find: Resistance R =?

Solution:

Ans: A resistance of 395 ohm to be connected in series with the wire

Example 09:

A potentiometer wire of length 8 m has a resistance of 8 ohm. A resistance box is connected in series with it. An accumulator of e.m.f. 2 volt so as to get a potential drop of 1µ V mm^-1. What resistance must be the value of the resistance in the resistance box?

Given: Length of potentiometer wire = l_AB = 5 m, Resistance of potentiometer wire = R_AB = 5 ohm, e.m.f. of cell = E = 4 V, Resistance in series = R ohm, Potential drop = 1µ V mm^-1= 10^-3 V m^-1.

To Find: Resistance R =?

Solution:

Ans: A resistance of 1992 ohm to be connected in series with the wire

Example 10:

A potentiometer wire of length 4 m has a resistance of 4 ohm. What resistance must be connected in series with this wire and an accumulator of e.m.f. 2 volt so as to get a potential drop of 10^-3 V cm^-1 on the wire?

Given: Length of potentiometer wire = l_AB = 4 m = 400 cm, Resistance of potentiometer wire = R_AB = 4 ohm, e.m.f. of cell = E = 2 V, Resistance in series = R ohm, Potential drop = 10^-3 V cm^-1.

To Find: Resistance R =?

Solution:

Ans: A resistance of 16 ohm to be connected in series with the wire

Example 11:

A potentiometer wire of length 4 m has a resistance of 4 ohm. What resistance must be connected in series with this wire and an accumulator of e.m.f. 2 volt and internal resistance 2 ohm so as to get a potential drop of 10^-3 V cm^-1 on the wire?

Given: Length of potentiometer wire = l_AB = 4 m = 400 cm, Resistance of potentiometer wire = R_AB = 4 ohm, e.m.f. of cell = E = 2 V, Internal resistance = r = 2 ohm, Resistance in series = R ohm, Potential drop = 10^-3 V cm^-1.

To Find: Resistance R =?

Solution:

Ans: A resistance of 14 ohm to be connected in series with the wire

Example 12:

A potentiometer wire of length 5 m has a resistance of 10 ohm. It is connected in series with a cell of e.m.f. 2 V and resistance R. If a P.D. of 3 mV is balanced against a length of 3 m. Find R.

Given: Length of potentiometer wire = l_AB = 5 m, Resistance of potentiometer wire = R_AB = 10 ohm, e.m.f. of cell = E = 2 V, Resistance in series = R ohm, Banaced P.D. = 3 mV = 3 x 10^-3 V, Balancing length = 3m.

To Find: Resistance R =?

Solution:

Balanced P. D. = Potential drop x Balancing length

3 x 10^-3 = Potential drop x 3

Potential drop = 3 x 10^-3 / 3 = 10^-3 Vm^-1

Ans: A resistance of 3990 ohm to be connected in series with the wire

Example 13:

A potentiometer wire of length 10 m has a resistance of 10 ohm. It is connected in series with a cell of e.m.f. 4 V and resistance R. If a source of 100 mV is balanced against a length of 4 m of potentiometer wire. Find R.

Given: Length of potentiometer wire = l_AB = 10 m, Resistance of potentiometer wire = R_AB = 10 ohm, e.m.f. of cell = E = 4 V, Resistance in series = R ohm, Balanced e.m.f. = 100 mV = 100 x 10^-3 V = 0.1 V, Balancing length = 4m.

To Find: Resistance R =?

Solution:

Balanced P. D. = Potential drop x Balancing length

0.1 = Potential drop x 4

Potential drop = 0.1 / 4 = 0.025Vm^-1

Ans: A resistance of 150 ohm to be connected in series with the wire

Previous Topic: Theory of Potentiometer

Next Topic: Numerical Problems on to Find Internal Resistance of a cell Using Potentiometer

Science > Physics > Current Electricity > Numerical Problems on Potential Drop

The post Numerical Problems on Potential Drop appeared first on The Fact Factor.

In this article, we shall study the principle, construction, and working of a potentiometer and its uses.

Principle of Potentiometer:

When a steady current flows through a wire of uniform cross-section the potential difference per unit length of the wire is constant throughout the length of the wire (or p.d. across any two points of the wire is directly proportional to the length of the wire. It can be explained as below.

Let us consider a uniform wire AB of length l_AB and uniform cross-sectional area A. Let R_AB be its resistance. Let ‘I’ be the steady current flowing through the wire. Let V_AB be the p.d. across the ends of the wire. Let ‘ρ be the specific resistance of the material of the wire. Let there be a uniform potential drop across the length of wire.

Let us consider point P on the wire and the length of wire between A and P be ‘ l_AP’. Thus, the resistance of a wire of length ‘R_AP’ is given by

When a constant current flows through a wire, then the potential difference between any two points of the wire is directly proportional to the length of wire between these two points. In such a case, the p.d. per unit length of the wire is constant and called the potential gradient of the wire or voltage drop across the wire.

Precautions to be Taken While Using a Potentiometer:

The e.m.f. of the cell connected across the potentiometer wire should b greater than the e.m.f. to be compared.

The positive terminal of the cells whose e.m.f. is to be compared must be connected to that end of potentiometer wire where positive terminal of the battery (driving cell) is connected.

The potentiometer wire must be uniform.

The resistance of potentiometer wire should be high.

Advantages of a Potentiometer Over a Voltmeter:

• A potentiometer can be used to measure the internal resistance of cell which cannot be measured by the voltmeter.
• A Potentiometer can be to measure e.m.f of a cell which cannot be measured by a voltmeter. When a voltmeter is connected in a circuit it draws current through the circuit and thus can measure the potential difference across the cell terminals. When the potentiometer is connected in a circuit it draws no current when the null point is obtained. Thus it measures the e.m.f. of the cell.
• A potentiometer can be used to measure extremely small p.d. accurately which cannot be measured by a voltmeter. It can be done by using very long wire and adjusting a very small potential gradient.
• Potentiometer is more sensitive compared to voltmeter.
• The accuracy of the potentiometer can be increased by increasing the length of the wire. The accuracy of the voltmeter cannot be increased beyond the limit.

Disadvantages of a Potentiometer:

• A voltmeter is a direct reading instrument while potentiometer is not so. We have to perform calculations to find the result.
• A voltmeter is portable while potentiometer is non-portable

Construction of Potentiometer:

A potentiometer consists of a uniform wire AB several meters long.  It is stretched between two points A and B on the wooden board. A battery having a sufficiently large e.m.f. E is connected between A and B of the wire.  On closing, the key current will flow through the wire. The current in the wire can be adjusted by adjusting rheostat connected in series with the battery.  The battery maintains a uniform potential gradient along the length of wire.

Uses of Potentiometer:

To Measure e.m.f. of a Cell or to Compare e.m.f.s of Two Cells by Individual Method

Let E₁ and E₂ be the e.m.f.’s of the two cells to be compared by using the potentiometer. The positive terminal of the cell of e.m.f. E₁ is connected to end A and a negative terminal is connected to jockey through the galvanometer. By closing the key the jockey is moved along wire AB and null point P is determined such that galvanometer shows no deflection. The length of wire AP = l ₁ is measured.  The p.d. across this length balances e.m.f. E₁

e.m.f. of the cell =  potential difference across AP

E₁ = K l₁ ………..  (1)

where K is the Potential gradient of the wire

Then cell of e.m.f. E₁ is disconnected and cell of e.m.f. E₂ is connected in circuit and procedure is repeated

E₂ = K l₂ ………..   (2)

Dividing equation  (1) by (2), we get,

Thus knowing the values of l₁ and l₂ we can compare e.m.f.s of two cells.

To Measure e.m.f. of a Cell or to Compare e.m.f.s of Two Cells by Sum and Difference Method:

Let E₁ and E₂ be the e.m.f.’s of the two cells to be compared by using the potentiometer. In this method both the cells whose e.m.f.s are to be compared are connected together.

When the two cells are connected in series such that the negative terminal of one cell is connected to positive terminal of the other, then the two cells are said to assist each other and their resultant e.m.f. is given by the sum of the e.m.f.s of the two cells. (E₁ + E₂)

When the two cells are connected in series such that the negative terminal of one cell is connected to the negative terminal of the other, then the two cells are said to oppose each other and their resultant e.m.f. is given by the difference of the e.m.f.s of the two cells. ( E₁ – E₂)

In the first step, the cells are connected to assist each other. The positive terminal of the combination of cells is connected to end A and another terminal is connected to jockey through the galvanometer. By closing the key the jockey is moved along wire AB and null point P is determined such that galvanometer shows no deflection.  The length of wire AP = l₁ is measured.  The p.d. across this length balances e.m.f. (E₁ + E₂)

∴  e.m.f. of the cell  = potential difference across AP.

E₁ + E₂  = K l₁ . . . (1)

where K is the Potential gradient of the wire

In the second step, the cells are connected to oppose each other and the procedure is repeated.

E₁ – E₂ = K l₂  . . . (2)

Dividing equation  (1) by (2), we get,

Thus knowing the values of l₁ and l₂ we can compare e.m.f.s of two cells.

To Find Internal Resistance of a Cell:

A battery B having an e.m.f. greater than the e.m.f. (E) of the cell whose internal resistance (r) is to be measured, is connected in series with the potentiometer wire AB, a key K₁, and a rheostat. The positive terminal of the cell of e.m.f. E is connected to the end A of the potentiometer wire. The negative terminal of E is connected to a jockey through the galvanometer G. A resistance box and a keyK₂ are connected across the cell E.

Initially, the key K₂ is kept open.  By closing the key K₁ current is passed through the potentiometer wire so that uniform potential gradient is produced along the wire. By sliding the Jockey along the wire, a point of contact P₁ for which the galvanometer shows zero deflection is found.  The length of the wire AP₁ = l is measured. As the cell is in an open circuit, e.m.f. of the cell is equal to the p.d. across the length l, of the potentiometer wire.

E = K l          . . . (1)

Where K is the potential gradient of the wire.

Now a suitable resistance (R) is connected from the resistance box and the key K₂ is closed and once again null point P₂ is found on the potentiometer wire.  The length AP₂ = l₁  is measured.  Let V be the terminal p.d. of the cell

Thus knowing R, l and l₁ we can calculate the value of r i.e. the internal resistance of the cell using this formula.

Previous Topic: Numerical Problems on Metre Bridge

Next Topic: Numerical Problems on Potential Drop (Potentiometer)

Science > Physics > Current Electricity > Potentiometer

The post Potentiometer appeared first on The Fact Factor.

Example – 15:

In Wheatstone’s metre bridge experiment with unknown resistance X in the left gap and 60 Ω resistance in a right gap null point is obtained at l cm from the left end. If the unknown resistance is shunted by equal resistance, what should be the value of resistance in the right gap in order to get the null point at the same point?

Solution:

Case – I:

Given: resistance in the left gap = X Ω, Resistance in right gap = 60 Ω

l is the distance of the null point from the left end.

For balanced metre bridge

∴ X/60 = l / (100 – l) …………. (1)

Case – II: 

Given: X Ωresistance is shunted by X Ω, Resistance in the left gap = (X x X)/(X + X) = X²/2X = X/2 Ω,

Let the resistance in right gap be R Ω

Again l is the distance of the null point from the left end.

For balanced metre bridge

∴  (X/2)/R = l / (100 – l)

∴  X/2R = l / (100 – l)  ………….. (2)

From equations (1) and (2)

∴   X/60 = X/2R

∴   2R = 60

∴   R = 30 Ω

Ans: The resistance in the right gap should be 30 Ω

Example – 16:

With resistance R₁ in the left gap and R₂ in the right gap of a Wheatstone’s metre bridge, the null point is obtained at 30 cm from the right end. When R₁ is reduced by 2 Ω and R₂ is increased by 2 Ω, the null point is obtained at 30 cm from the left end. Find R₁ and R₂.

Solution:

Case – I: 

Given: Resistance in left gap = R₁, Resistance in right gap = R₂

Null point from right end = 100 – l = 30 cm

Null point from left end = l = 100 – 30 = 70 cm

For balanced metre bridge

∴ R₁/R₂  = 70/30= 7/3

∴ R₁  = (7/3)R₂  ………. (1)

Case – II: 

Given: Resistance in left gap = R₁ – 2 Ω  Resistance in right gap = R₂ + 2 Ω

Null point from left end = l = 30 cm and 100 – l = 100 – 30 = 70 cm

For balanced metre bridge

∴ (R₁ – 2)/(R₂ + 2)  = 30/70= 3/7

∴ 7R₁  – 14 = 3R₂  + 6

∴ 7R₁  – 3R₂ = 20

∴ 7 (7/3)R₂ – 3R₂ = 20

∴  (49/3)R₂ – 3R₂ = 20

∴  ((49 – 9)/3)R₂  = 20

∴  (40/3)R₂  = 20

∴  R₂  = 20 x (3/40) = 1.5 Ω

R₁  = (7/3)R₂  =  (7/3) x 1.5 = 3.5 Ω

Ans:  R₁  = 3.5 Ω and R₂  = 1.5 Ω

Example – 17:

Two coils are connected in series in one gap of Wheatstone’s metre bridge and the null point is obtained at the midpoint of wire when a 50 Ω resistance is connected in the right gap. The two coils are then connected in parallel and it is found that the resistance in other gap is to be changed by 38 Ω to get the null point at the same point as before. Find the resistance of the coils.

Solution:

Let R₁ and R₂ be the resistances of the two coils

Case – I:

Given: R₁ and R₂ be in series, resistance in left gap = R₁  +  R₂, Resistance in right gap = 50 Ω, l = 50 cm and, 100 – l = 100 – 50 = 50 cm

For balanced metre bridge

∴ (R₁ + R₂)/50  = 50/50

∴ (R₁ + R₂) = 50  ………. (1)

Case – II:

Now the two coils are connected in parallel. Hence their effective resistance is decreasing. To keep the null point at the same point, the resistance in the right gap should also decrease.

Hence resistance in the right gap = 50 – 38 = 12  Ω

∴ R₁ (50 – R₁ ) = 600

∴ 50R₁ – R₁² = 1350

∴ R₁² – 50 R₁ + 600 = 0

∴ (R₁ – 30)(R₁ – 20) = 0

∴ R₁ = 30 Ω or R1₁ = 20 Ω

Hence R₂ = 20 Ω or R₂ = 30 Ω

Ans: The restances of two coils are 30 Ω and 20 Ω.

Example – 18:

Two diametrically opposite points of a metal ring are connected by two wires in the left gap of a Wheatstone’s metre bridge. A resistance of 11 Ω. is connected in a right gap and the null point is obtained at a distance of 45 cm from the left end. Find the resistance of the metal ring.

Solution:

Let 2R be the resistances of the metal ring. As the diametrically opposite points are connected in the left gap. The resistance of each half is R and these two halves are connected in parallel.

Thus resistance in left gap = (R x R)/(R + R) = R²/2R = R/2 Ω, Resistance in right gap = 11 Ω

l = 45 cm and, 100 – l = 100 – 45 = 55 cm

For balanced metre bridge

∴ (R/2)/11  = 45/55

∴ (R/2)  = 45/5 = 9

∴ R = 18 Ω

Now resistance of ring = 2R = 2 x 18 = 36 Ω

Ans: The resitance of the ring is 36 Ω.

Example – 19:

Two diametrically opposite points of a metal ring are connected by two wires in the left gap of a Wheatstone’s metre bridge. A resistance of 15 Ω. is connected in a right gap and the null point is obtained at a distance of 40 cm from the left end. Find the resistance of the wire bent in the shape of a ring.

Solution:

Let 2R be the resistances of the metal ring. As the diametrically opposite points are connected in left gap. The resistance of each half is R and these two halves are connected in parallel.

Thus resistance in left gap = (R x R)/(R + R) = R²/2R = R/2 Ω,

Resistance in right gap = 15 Ω

l = 40 cm and, 100 – l = 100 – 40 = 60 cm

For balanced metre bridge

∴ (R/2)/15  = 40/60

∴ (R/2)  = 40/4 = 10

∴ R = 20 Ω

Now resistance of ring = 2R = 2 x 20 = 40 Ω

Ans: The resitance of the ring is 40 Ω.

Example – 20:

Two diametrically opposite points of a metal ring are connected by two wires in the left gap of a Wheatstone’s metre bridge. A resistance of 25 Ω. is connected in a right gap and the null point is obtained at a distance of 33.3 cm from the left end. Find the resistance of the metal ring.

Solution:

Let 2R be the resistances of the metal ring. As the diametrically opposite points are connected in left gap. The resistance of each half is R and these two halves are connected in parallel.

Thus resistance in left gap = (R x R)/(R + R) = R²/2R = R/2 Ω,

Resistance in right gap = 25 Ω

l = 33.3 cm and, 100 – l = 100 – 33.3 = 66.7 cm

For balanced metre bridge

∴ (R/2)/25  = 33.3/66.7

∴ R  = (33.3/66.7) x 25 x 2

∴ R = 24.96 Ω

Now resistance of ring = 2R = 2 x 24.96 = 49.92 Ω

Ans: The resitance of the ring is 49.92 Ω.

Example – 21:

Two resistance wires of the same material have diameters in the ratio 2:1 and the lengths in the ratio 4:1 are connected in left and the right gaps of Wheatstone’s metre bridge. Find the position of the null point from the left end.

Solution:

Let R₁ and R₂ be the resistances of the two wires connected in left and right gaps. The material of wires is the same, hence resistivity is the same ρ₁ = ρ₂

Thus resistance in the left gap = R₁ Ω, Resistance in right gap = R₂ Ω

Let l be the position of the null point from the left end

For balanced metre bridge

∴ R₁/R₂ = l / (100 – l)

∴ 1 = l / (100 – l)

∴ 100 – l = l

∴ 2l = 100

∴ l =  50 cm

Ans: The position of the null point from the left end is 50 cm

Example – 22:

Equal lengths of wires of material A and B are connected in the left gap and the right gap of a Wheatstone’s metre bridge to get a null point at 30 cm from the left end. Find the ratio of diameters of wires A and B. Specific resistance of A is 5 x 10^-8 Ωm and of B is 2 x 10^-6 Ωm.

Solution:

Let R_A and R_B be the resistances of the two wires connected in left and right gaps.

Resistance in left gap = R_A Resistance in right gap = R_B

Null point from left end = l = 30 cm and 100 – l = 100 – 30 = 70 cm

For balanced metre bridge

∴ R_A/R_B  = 30/70= 3/7 …………. (1)

ρ_A = 5 x 10^-8 Ωm, ρ_B = 2 x 10^-6 Ωm

Let r_A and r_B be the radii of the two wires

Ans: The ratio of diameters of two wires is 0.242:1

Example – 23:

Equal lengths of wires of manganin(ρ₁) and nichrome (ρ₂) are connected in the left gap and the right gap of Wheatstone’s a metre bridge to get a null point at 40 cm from the left end. Find the ratio of diameters of wires A and B. Specific resistance of manganin is 4.8 x 10^-8 Ωm and of nichrome is 10^-6 Ωm.

Solution:

Let R₁ and R₂ be the resistances of the two wires connected in left and right gaps.

Resistance in left gap = R₁ Resistance in right gap = R₂

Null point from left end = l = 40 cm and 100 – l = 100 – 40 = 60 cm

For balanced metre bridge

∴ R₁/R₂  = 40/60= 2/3 …………. (1)

ρ₁ = 4.8 x 10^-8 Ωm, ρ₂ = 10^-6 Ωm

Let r₁ and r₂ be the radii of the two wires

Ans: The ratio of diameters of two wires is 0.2683:1

Example – 24:

A uniform wire is cut into two pieces are so that their lengths are in the ratio 1:2. When these pieces are connected in parallel in the left gap of the Wheatstone’s metre bridge, with a resistance 25 Ω in a right gap, the null point is obtained at a distance of 40 cm from the left end of the wire. Find the resistance of the wire before it was cut into two pieces.

Solution:

Let R be the resistance of the uncut wire. Let R₁ and R₂ be the resistances of the two cut wires

R = R₁ + R₂

Ratio of their lengths l₁/l₂ = 1/2

The material of wires is the same, hence resistivity is the same ρ₁ = ρ₂

The wire is uniform, hence radii are the same  r₁ = r₂

Resistance in left gap = (2/3)R₁, Resistance in right gap = 25

Null point from left end = l = 40 cm and 100 – l = 100 – 40 = 60 cm

For balanced Wheatstone’s metre bridge

∴ (2/3)R₁/25  = 40/60= 2/3

∴ R₁/25  = 1

∴ R₁  = 25 Ω

R₂  = 2R₁  = 2 x 25 = 50 Ω

Resistance of uncut wire = R = R₁ + R₂ = 25 + 50 = 75 Ω

Ans: The resistance of the wire before it was cut into two pieces is 75 Ω.

Example – 25:

A uniform wire is cut into two pieces are so that one piece is twice as long as the other. When these pieces are connected in parallel in the left gap of the Wheatstone’s metre bridge, with a resistance 20 Ω in a right gap, the null point is obtained at a distance of 60 cm from the right end of the wire. Find the resistance of the wire before it was cut into two pieces.

Solution:

Let R be the resistance of the uncut wire. Let R₁ and R₂ be the resistances of the two cut wires

R = R₁ + R₂

Ratio of their lengths l₁/l₂ = 1/2

The material of wires is the same, hence resistivity is the same ρ₁ = ρ₂

The wire is uniform, hence radii are the same  r₁ = r₂

Resistance in left gap = (2/3)R₁, Resistance in right gap = 25

Null point from left end = l = 100 – 60 cm = 40 cm and 100 – l = 100 – 40 = 60 cm

For balanced Wheatstone’s metre bridge

∴ (2/3)R₁/20  = 40/60= 2/3

∴ R₁/20  = 1

∴ R₁  = 20 Ω

R₂  = 2R₁  = 2 x 20 = 40 Ω

Resistance of uncut wire = R = R₁ + R₂ = 20 + 40 = 60 Ω

Ans: The resistance of the wire before it was cut into two pieces is 60 Ω.

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Science > Physics > Current Electricity > Numerical Problems on Wheatstone’s Metre Bridge 02

The post Numerical Problems on Wheatstone’s Metre Bridge 02 appeared first on The Fact Factor.

In this article and the next article, we shall study numerical problems based on Wheatstone’s metre bridge.

Example – 01:

A resistance of 10 Ω is connected in a left gap of a metre bridge. Two resistors of 20 Ω and 16 Ω are connected in parallel in the right gap. Find the position of a null point on the bridge wire.

Given: resistance in left gap = X = 10 Ω, in right gap 20 Ω and 16 Ω are in parallel, Resistance in right gap R = (20 x 16)/(20 + 16) = 320/36 = 80/9 Ω

To Find: Position of null point = l = ?

Solution:

Let l be the distance of the null point from the left end.

For balanced metre bridge

∴  10/(80/9) = l / (100 – l)

∴  9/8 = l / (100 – l)

∴  9(100 – l )=  8 l

∴  900 – 9l  =  8 l

∴  17 l = 900

∴   l = 900/17 = 52.94  cm

Ans: The null point is at 52.94 cm on the wire from the left end.

Example – 02:

A metre bridge is balanced with a 20 Ω resistance in the left gap and a 40 Ω resistance in the right gap. If 40 Ω resistance is now shunted with another of like resistance, find the shift in the null point.

Solution:

Case – I:

Given: resistance in the left gap = 20 Ω, Resistance in the right gap = 40 Ω

Let l cm be the distance of the null point from the left end.

For balanced metre bridge

∴  20/40 = l / (100 – l)

∴  1/2 = l / (100 – l)

∴  100 – l = 2 l

∴  3 l = 100

∴   l = 100/3  cm from left end

Case – II: 

Given: 40 Ω resistance is shunted by 40 Ω,  resistance in left gap = 20 Ω, Resistance in right gap = (40 x 40)/(40 + 40) = 1600/80 = 20 Ω

Let l cm be the distance of the null point from the left end.

For balanced metre bridge

∴  20/20 = l / (100 – l)

∴  1= l / (100 – l)

∴  100 – l =  l

∴  2 l = 100

∴   l = 50  cm from left end

Shift in null point = 50 – 100/3 = 50/3 = 16.67 cm towards right

Ans: Shift in null point is 16.67 cm towards the right

Example – 03:

A metre bridge is balanced with a 20 Ω resistance in the left gap and a 30 Ω resistance in the right gap. If 20 Ω resistance is now shunted with another 20 Ω resistance, find the shift in the null point.

Solution:

Case – I:

Given: resistance in the left gap = 20 Ω, Resistance in the right gap = 30 Ω

Let l cm be the distance of the null point from the left end.

For balanced metre bridge

∴  20/30 = l / (100 – l)

∴  2/3 = l / (100 – l)

∴  200 – 2l = 3 l

∴  5 l = 200

∴   l = 40  cm from left end

Case – II: 

Given: 20 Ωresistance is shunted by 20 Ω,  Resistance in left gap = (20 x 20)/(20 + 20) = 400/40 = 10 Ω, resistance in left gap = 30 Ω,

Let l cm be the distance of the null point from the left end.

For balanced metre bridge

∴  10/30 = l / (100 – l)

∴  1/3= l / (100 – l)

∴  100 – l =  3l

∴  4 l = 100

∴   l = 25  cm from left end

Shift in null point = 40 – 25 = 15 cm towards left

Ans: Shift in null point is 15 cm towards the left.

Example – 04:

When two resistance coils are connected in series in one gap of metre bridge with a resistance of 75 Ω in the right gap, the null point is obtained at the midpoint of the bridge wire. The coils are then connected parallel and inserted in one gap with resistance 18 Ω in the other gap to obtain the null point at the same point as before. Find the resistance of each coil.

Solution:

Let R₁ and R₂ be the resistances of the two coils

Case – I:

Given: R₁ and R₂ be resistance in left gap = R₁  +  R₂, Resistance in right gap = 75 Ω, l = 50 cm and, 100 – l = 100 – 50 = 50 cm

For balanced metre bridge

∴ (R₁ + R₂)/75  = 50/50

∴ (R₁ + R₂) = 75  ………. (1)

Case – II:

Given: R₁ and R₂ parallel in left gap, Resistance in right gap = 18 Ω

∴ R₁ (75 – R₁ ) = 1350

∴ 75R₁ – R₁² = 1350

∴ R₁² -75 R₁ + 1350 = 0

∴ (R₁ – 30)(R₁ – 45) = 0

∴ R₁ = 30 Ω or R1₁ = 45 Ω

Hence R₂ = 45 Ω or R₂ = 30 Ω

Ans: The restances of two coils are 30 Ω and 45 Ω.

Example – 05:

When two resistance coils are connected in series in one gap of metre bridge with a resistance of 75 Ω in the right gap, the null point is obtained at the midpoint of the bridge wire. The coils are then connected parallel and inserted in one gap with resistance in the other gap changed by 57 Ω to obtain the null point at the same point as before. Find the resistance of each coil.

Solution:

Let R₁ and R₂ be the resistances of the two coils

Case – I:

Given: R₁ and R₂ be resistance in left gap = R₁  +  R₂, Resistance in right gap = 75 Ω, l = 50 cm and, 100 – l = 100 – 50 = 50 cm

For balanced metre bridge

∴ (R₁ + R₂)/75  = 50/50

∴ (R₁ + R₂) = 75  ………. (1)

Case – II:

When the coils are connected in parallel, their effective resistance decreases. Hence the resistance in the left gap decreases but the null point is remaining the same. It means resistance in the right gap also decreases.

Resistance in the right gap = 75 – 57 = 18 Ω

∴ R1 (75 – R1) = 1350

∴ 75R₁ – R₁² = 1350

∴ R₁² -75 R₁ + 1350 = 0

∴ (R₁ – 30)(R₁ – 45) = 0

∴ R₁ = 30 Ω or R1₁ = 45 Ω

Hence R₂ = 45 Ω or R₂ = 30 Ω

Ans: The resistances of two coils are 30 Ω and 45 Ω

Example – 06:

With resistances X and Y ohm in left and right gaps respectively of a metre bridge, the point is obtained at 30 cm from the left of the wire. When Y is shunted with 15 Ω resistance, the shift in null point is 10 cm.

Solution:

Case – I:

Given: resistance in left gap = X Ω, Resistance in right gap = Y Ω, l = 30 cm and, 100 – l = 100 – 30 = 70 cm

For balanced metre bridge

∴ X/Y  = 30/70

∴ X =  (3/7)Y  ………. (1)

Case – II: 

Given: Y is shunted by 15 Ω,  resistance in left gap = X Ω, Resistance in right gap = (15 x Y)/(15 + Y)

As Y is shunted, the resultant resistance reduces thus the null point shifts towards the right by 10 cm

l = 40 cm and, 100 – l = 100 – 40 = 60 cm

For balanced metre bridge

∴ 45 + 3Y = 70

∴  3Y = 25

∴  Y = 25/3 Ω

X =  (3/7)Y = (3/7) x (25/3) = 25/7 Ω

Ans: X = 25/7 Ω and Y = 25/3 Ω

Example – 07:

With two resistances in two gaps of a metre bridge, the null point is at 0.4 m from zero end. When 10 Ω resistance coil is put in series with smaller resistance the null point is at 0.6 m from the same end. Find the resistances.

Solution:

Case – I: 

Given: Resistance in left gap = X Ω, Resistance in right gap = Y Ω, l = 0.4 m = 40 cm and, 100 – l = 100 – 40 = 60 cm

For balanced metre bridge

∴ X/Y  = 40/60

∴ Y =  (3/2)X  ………. (1)

Case – II:

Y =  (3/2)X Thus X < Y. Thus 10 Ω resistance is connected in series with X.

X is shunted by 10 Ω,  resistance in the left gap = (X + 10) Ω, Resistance in right gap = Y, l = 0.6 m = 60 cm and, 100 – l = 100 – 60 = 40 cm

For balanced metre bridge

∴ 4X + 40 = 9X

∴  5X = 40

∴  X =  8 Ω

Y =  (3/2)X = (3/2) x 8 = 12 Ω

Ans: The two resistances are 8 Ω and 12 Ω

Example – 08:

Two resistances X and Y are connected in the left and right gap of metre bridge. The null point is measured from the left end and the ratio of balancing lengths is found to be 2:3. If the value of X is changed by 20 Ω, the ratio of balancing lengths measure from the left end of the wire is found to be 1:4. Find X and Y.

Solution:

Case – I: 

Given: Resistance in left gap = X Ω, Resistance in right gap = Y Ω, l  : (100 – l ) = 2:3

For balanced metre bridge

∴ X/Y  = 2/3

∴ X =  (2/3)Y  ………. (1)

Case – II: 

Given: The value of X is changed by 20 Ω. now new ratio of length (1:4) is less than the ratio of lengths in case – I (2:3). It means value of X is reduced.  resistance in left gap = (X – 20) Ω, Resistance in right gap = Y Ω.

l  : (100 – l ) = 1:4

For balanced metre bridge

∴ (X – 20)/Y  = 1/4

∴ X – 20 = (1/4)Y

∴ (2/3)Y – 20 = (1/4)Y

∴ (2/3)Y – (1/4)Y = 20

∴ (5/12)Y = 20

∴ Y = 48 Ω

∴ X =  (2/3)Y = (2/3) x 48 = 32 Ω

Ans: X = 32 Ω and Y = 48 Ω

Example – 09:

Two resistances X and Y in the two gaps of a Wheatstone’s metre bridge give a null point dividing the wire in the ratio 2: 3. If each resistance is increased by 30 Ω, the null point divides the wire in the ratio 5:6. Calculate each resistance.

Solution:

Case – I: 

Given: Resistance in left gap = X, Resistance in right gap = Y

l /(100 – l )= 2/3

For balanced metre bridge

∴ X/Y  = 2/3

∴ X  = (2/3)Y ………. (1)

Case – II: 

Given: Resistance in left gap = X+ 30 Ω  Resistance in right gap = Y+ 30 Ω

l /(100 – l )= 5/6

For balanced metre bridge

∴ (X+ 30)/(Y + 30)  = 5/6

∴ 6X + 180 = 5Y + 150

∴ 5Y – 6 X = 30

∴ 5Y – 6 x (2/3)Y = 30

∴ 5Y – 4Y = 30

∴ Y = 30 Ω

X  = (2/3)Y = (2/3) x 30 = 20 Ω

Ans:  X  = 20 Ω and Y  = 30 Ω

Example – 10:

A unknown resistance X is connected in a left gap and known resistance R is connected in the right gap of metre bridge. The balance point is obtained at 60 cm from the left end of wire. When R is increased by 2 Ω, the balance point shifts 10 cm. Find X and R

Solution:

Case – I: 

Given: Resistance in left gap = X Ω, Resistance in right gap = R Ω, l = 60 cm and, 100 – l = 100 – 60 = 40 cm

For balanced metre bridge

∴ X/R  = 60/40

∴ X =  (3/2)R  ………. (1)

Case – II: 

Given: The value of R is increased by 2 Ω. The balance point should shift towards right. l = 50 cm and, 100 – l = 100 – 50 = 50 cm

For balanced metre bridge

∴ X/(R + 2)  = 50/50

∴ X/(R + 2)  = 1

∴ X =  R + 2

∴  (3/2)R =  R + 2

∴  (3/2)R –  R = 2

∴  (1/2)R  = 2

∴  R = 4 Ω

X =  (3/2)R = (3/2) x 4 = 6 Ω

Ans: X = 6 Ω and R = 4 Ω

Example – 11:

In a metre bridge experiment, with the resistance R₁ in the left gap and a resistance X in the right gap, the null point is obtained at 40 cm from the left end.  With the resistance R₂ in the left gap and the same resistance X in the right gap, the null point is obtained at 50 cm from the left end. Where will be the null point if R₁ and R₂ are put series in the left gap and the right gap still containing X?

Solution:

Case – I:

Given: Resistance in left gap = R₁, Resistance in right gap = X, l = 40 cm and, 100 – l = 100 – 40 = 60 cm

For balanced metre bridge

∴ R₁/X  = 40/60

∴  ∴ R₁ = (2/3) X  ………. (1)

Case – II:

Given: Resistance in left gap = R₂, Resistance in right gap = X, l = 50 cm and, 100 – l = 100 – 50 = 50 cm

For balanced metre bridge

∴ R₂/X  = 50/50

∴ R₂ =  X  ………. (2)

Case – III:

Given: Resistance in left gap = R₁ + R₂ , Resistance in right gap = X

Let l be the distance of the null point from the left end.

For balanced metre bridge

∴  (R₁ + R₂)/X = l / (100 – l)

∴  (2/3 X + X)/X = l / (100 – l)

∴  (5/3 X) /X = l / (100 – l)

∴  5/3 = l / (100 – l)

∴ 500 – 5l= 3l 

∴ 8l = 500

∴ l = 500/8 = 62.5 cm

Ans: The null point is at 62.5 cm on the wire from the left end.

Example – 12:

In a metre bridge experiment, with the resistance R₁ in the left gap and a resistance X in the right gap, the null point is obtained at 40 cm from the left end.  With the resistance R₂ in the left gap and the same resistance X in the right gap, the null point is obtained at 50 cm from the left end. Where will be the null point if R₁ and R₂ are put parallel in the left gap and the right gap still containing X?

Solution:

Case – I:

Given: resistance in left gap = R₁, Resistance in right gap = X. l = 40 cm and, 100 – l = 100 – 40 = 60 cm

For balanced metre bridge

∴ R₁/X  = 40/60

∴  ∴ R₁ = (2/3) X  ………. (1)

Case – II:

Given: resistance in left gap = R₂, Resistance in right gap = X, l = 50 cm and, 100 – l = 100 – 50 = 50 cm

For balanced metre bridge

∴ R₂/X  = 50/50

∴ R₂ =  X  ………. (2)

Case – III:

Resistance in right gap = X

Let l be the distance of the null point from the left end.

For balanced metre bridge

∴  (2/5)X/X = l / (100 – l)

∴  2/5= l / (100 – l)

∴  2(100 – l)= 5l 

∴ 200 – 2l= 5l 

∴ 7l = 200

∴ l = 200/7 = 28.57 cm

Ans: The null point is at 28.57 cm on the wire from the left end.

Example – 13:

With a coil of unknown resistance X in the left gap and resistance R in the right gap of a metre bridge, the null point is obtained at 40 cm from the left end. When resistance of 10 Ω is put in series with X, the null point is at the centre of the wire with R still in the right gap. Find X and R. Where would the null point be if the 10 Ω resistance is put in parallel with X and R still in the right gap.

Solution:

Case – I:

Given: resistance in left gap = X, Resistance in right gap = R, l = 40 cm and, 100 – l = 100 – 40 = 60 cm

For balanced metre bridge

∴ X/R  = 40/60

∴ X = (2/3) R  ………. (1)

Case – II:

Given: a resistance of 10 Ω is put in series with X. resistance in left gap = X + 10, Resistance in right gap = R, l = 50 cm and, 100 – l = 100 – 50 = 50 cm

For balanced metre bridge

∴ (X + 10)/R  = 50/50 = 1

∴ X + 10 = R

∴ (2/3) R + 10 =  R

∴  10 =  R  –  (2/3) R

∴  10 =  (1/3) R

∴  R = 30 Ω

X = (2/3) R = (2/3) x 30 = 20 Ω

Case – III: 

Given: 10 Ω resistance is connected in parallel with X.

resistance in left gap = 10X/(X + 10) = (10 x 20)/(10 + 20) = 200/30 = (20/3) Ω, Resistance in right gap = R

Let   l cm be the distance of null point from left end.

For balanced metre bridge

∴ (20/3)/30  = l / (100 – l)

∴ 20/90 = l / (100 – l)

∴ 2/9 = l / (100 – l)

∴ 200 – 2l = 9l

∴  11l = 200

∴  l =  200/11 = 18.2 cm

Ans: X = 20 Ω, R = 30Ω and the required balance point at 18.2 cm from left end of wire.

Example – 14:

With an unknown resistance X in the left gap and a resistance of 30 Ω in the right gap of the metre bridge the null point is obtained at 40 cm from the left end of the wire. Find (i) the unknown resistance and (ii) the shift in the position of the null point a) when the resistance in both the gaps are increased by 15 Ω and b) when the resistances in each gap is shunted by 8 Ω.

Solution:

Part (i)

resistance in left gap = X Ω, Resistance in right gap = 30 Ω

l = 40 cm and, 100 – l = 100 – 40 = 60 cm

For balanced metre bridge

∴ X/30  = 40 / 60 = 2/3

∴ X = 20 Ω

Case a) when the resistance in both the gaps are increased by 15 Ω

resistance in left gap = X + 15 = 20 + 15 = 35 Ω, Resistance in right gap = 30 + 15 = 45 Ω

Let   l cm be the distance of null point from left end.

For balanced metre bridge

∴ (20/3)/30  = l / (100 – l)

∴ 35/45 = l / (100 – l)

∴ 7/9 = l / (100 – l)

∴ 700 – 7l = 9l

∴  16l = 700

∴  l =  700/16 = 43.75 cm

Shift in null point = 43.75 – 40 = 3.75 cm towards right

Case b) when the resistances in each gap is shunted by 8 Ω.

Resistance in left gap = (20 x 8)/(20 + 8) = 160/28 = 40/7 Ω,

Resistance in right gap = (30 x 8)/(30 + 8) = 240/38 = 120/19

Let   l cm be the distance of null point from left end.

For balanced metre bridge

∴ (40/7)/(120/19)  = l / (100 – l)

∴ (1/7)/(3/19)  = l / (100 – l)

∴ ∴ (19/21)  = l / (100 – l)

∴ 1900 – 19l = 21l

∴  40l = 1900

∴  l =  1900/40 = 47.5 cm

Shift in null point = 47.5 – 40 = 7.5 cm towards right

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Science > Physics > Current Electricity > Numerical Problems on Wheatstone’s Metre Bridge 01

The post Numerical Problems on Wheatstone’s Metre Bridge 01 appeared first on The Fact Factor.

In this article, we shall study Kirchhoff’s laws, construction, working and uses of Wheatstone’s metre bridge. There are two Kirchhoff’s laws viz: current law or junction law and voltage law or mesh law.

Kirchhoff’s Current Law or Kirchhoff’s Junction Law:

Statement: 

In an electric network, the algebraic sum of currents at any junction is zero.

Explanation:

Any point in the circuit where the current split is called a junction. Currents approaching junction are taken positive and currents going away from junction are taken as negative. Let us consider a junction of a circuit as shown in the figure.

Applying Kirchhoff’s junction law at point O.

I₁ + I₂ – I₃ – I₄ – I₅ = 0

Since there is no loss or gain of the charge at the junctions, this law is in accordance with the law of conservation of charge.

Kirchhoff’s Voltage Law OR Kirchhoff’s Mesh Law:

Statement: 

The algebraic sum of the products of the current and resistance of each part of a closed circuit (mesh or loop) is equal to the algebraic sum of the e.m.f.s in that closed circuit.

Sign Convention:

• When passing through the circuit in the direction of current there is drop in the potential hence the potential difference should be taken as negative.
• When passing through the circuit in the opposite direction of current there is an increase in the potential hence the potential difference should be taken as positive.
• When passing through a cell from the negative terminal to the positive terminal the e.m.f. of a cell is taken as positive.
• When passing through a cell from the positive terminal to the negative terminal the e.m.f. of a cell is taken as negative.

Explanation:

Consider the following circuit

Let us consider a closed-loop E₁ARBr₁

E₁ – (I₁ + I₂) R – I₁r₁ = 0

Let us consider a closed-loop E2ARBr2

E2 – (I₁ + I2) R – I₂r₂ = 0

Let us consider a closed-loop E₁AE₂BE₁

E₁ – E₂ + I₂r₂ – I₁r₁ = 0

Wheatstone’s Network (Application of Kirchhoff’s Laws):

Circuit Arrangement:

A Wheatstone’s network consists of four resistances connected such that they form a quadrilateral of resistances. A battery is connected (or p.d. is applied) between one pair of opposite corners of the quadrilateral. A galvanometer is connected between another pair of opposite corners.

When the values of the resistance are such that, the galvanometer shows no deflection or null deflection then the network is said to be balanced Wheatstone’s network.

Condition for Balanced Wheatstone’s Network:

Let R₁, R₂, R₃, and R₄ be the four resistances connected as shown in the diagram to form Wheatstone’s network as shown in the diagram. Let I₁, I₂ be the currents through the resistances R₁ and R₃ respectively. Let I_G be the current through the galvanometer whose resistance is G. In balanced condition the current through galvanometer is zero hence I_G = 0

Applying Kirchhoff’s voltage law to the loop ABDA,

I₁R₁ –I_GG + I₂R₃ = 0

∴ – I₁R₁ –(0) G +I₂R₃ = 0

∴ – I₁R₁ + I₂R₃ = 0

∴ I₁R₁ = I₂R₃ ……….. (1)

Applying Kirchhoff’s voltage law to the loop BCDB

– (I₁ – I_G) R₂ + (I₂ + I_G) R₄   + I_GG = 0

∴  – (I₁ – 0)R₂ + (I₂ +0)R4   + (0)G = 0

∴ – I₁ R₂ + I₂R4  = 0

I₁ R₂ = I₂R4 ……..   (2)

Dividing equation (1) by (2)

This is the required condition for balanced Wheatstone’s network.

Condition for Balanced Wheatstone’s network using Ohm’s law:

When the values of the resistance are such that, the galvanometer shows no deflection or null deflection then the network is said to be balanced Wheatstone’s network. Thus potential at point B is equal to the potential at D. Thus (V_B = V_D)

∴ V_A – V_B = V_A – V_D

∴ By Ohm’s law I₁R₁ = I₂R₃ ……… (1)

Similarly V_B – V_C = V_D – V_C

∴ By Ohm’s law (I₁ – I_G)R₂ = (I₂ + I_G)R₄ ……… (2)

But no current flows through the galvanometer (I_G = 0)

I₁R₂ = I₂R₄ ……… (2)

Dividing equation (1) by (2)

This is the required condition for balanced Wheatstone’s network.

Wheatstone’s Metre Bridge:

The value of an unknown resistance can be determined by using Wheatstone’s meter bridge.

Construction:

It consists of a uniform wire AC, one meter long, stretched on a wooden board.  The two ends of the wire are fixed to two thick L shaped copper strips.  The third thick and straight copper strip is so put that it forms two gaps with the two copper strips.

The unknown resistance X is connected in one gap and a resistance box R (known resistance) is connected in the other gap. The junction of X and R is connected to one terminal of a galvanometer G.  The other terminal of the galvanometer is connected to a pencil Jockey which can slide along the wire AC. A cell, key, and rheostat are connected in series with the wire.

Working:

A suitable resistance R is taken in the resistance box and the current is sent around the circuit by closing the key K. The jockey is touched to different points of the wire AC and a point of contact D for which the galvanometer shows zero deflection is found. As at point D the galvanometer is showing null deflection, point D is called the null point.

Let the distance of the point from A be l₁ and that from C is l₂. These two distances are measured. Let σ be the resistance per unit length of wire. As the network is balanced

Using this formula unknown resistance X can be calculated.

Possible Errors in Metre Bridge Experiment:

• If the wire is not uniform, the resistance per unit length of wire is not the same throughout the wire and thus there is a possible error in the determination of unknown resistance.
• There may be the error due to contact resistances of the points where the wires are connected to copper strips.

Minimization of Errors:

• The resistance X and R are interchanged and experiment is repeated.
• Value of R is so selected that the deflection in the galvanometer is at the centre of the scale.
• The average of the readings of a number of readings gives most probable value.

Kelvin’s Method to Find Resistance of a Galvanometer:

By this method, the resistance of galvanometer itself can be found.

Circuit Arrangement:

The galvanometer whose resistance to be found is connected in one gap and the known resistance R is connected in another gap. The galvanometer itself is used to determine the condition of equilibrium without locating null point. The junction of galvanometer G and known resistance R is connected to the jockey.

The circuit is closed and adjusting the rheostat a suitable current is passed through the circuit. A suitable resistance is taken out from the resistance of the box and deflection in the galvanometer is noted. A jockey is moved over wire AC. A point D is noted for which the galvanometer shows the same deflection as before. Thus D point is such that at this point the deflection is the same with or without the jockey. Thus D is equal deflection point.

Distances of point D from two ends are measured.  Let AD = l₁ and CD =l₂. If G is the resistance of the galvanometer, then using the following formula the value of G can be calculated.

Previous Topic: Temperature Dependence of Resistance

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Science > Physics > Current Electricity > Kirchhoff’s Laws

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In this article, we shall study the effect of temperature on the value of resistance and thermistors and their uses.

A metallic conductor consists of a large number of free electrons. These electrons are always in a state of random motion. When a potential difference is applied across the ends of the conductor. These free electrons start moving in the definite direction i.e. towards the positive end of the conductor. During this process, the electrons flow through the crowd of vibrating atoms. These electrons collide with the atoms. Thus vibrating atoms offer obstruction to the flow of electrons. This obstruction to the flow of electrons is called the resistance of the conductor.

If the temperature of the conductor is increased, the kinetic energy of vibrating atoms is increased, due to which the atoms start vibrating with higher amplitude. Thus the obstruction to the flow of electrons increases and hence the resistance of the conductor also increases.

Expression for Temperature Coefficient of Resistance:

Let R_o be the initial resistance at 0° C. Let R be the resistance at t° C.

∴ Change in resistance             =    R  –  R_o

∴ Change in temperature (Δt)   =    t₂  – t₁

Experimentally it is found that the change in resistance is directly proportional to

• the original resistance.
• to change in temperature.

R  –  R_o   ∝    R_o       ——— (1)

R  –  R_o   ∝    t₂  – t₁  ——— (2)

From (1) & (2)

R  –  R_o    ∝   R_o  (t₂  – t₁)

R –  R_o    =      α R_o (t₂  – t₁)

Where α is constant called temperature coefficient of resistance.

But    t₂ – t₁ = Δ t

R – R_o    =      α R_o Δ t    ……….. (3)

R   =    R_o + α R_o Δ t

R   =    R_o (1 + α Δ t)

This is an expression which gives the value of resistance at the new temperature.

From equation (3), we have

This is an expression for the temperature coefficient of the resistance of a material of a conductor.

Temperature Coefficient of Resistance:

Temperature coefficient of resistance is defined as the change in resistance per unit resistance at 0° C per degree rise in temperature

Notes:

For good conductors value of temperature coefficient of resistance is positive hence the value of resistance increases as temperature increases and the value of resistance decreases if its temperature decreases

For semiconductors value of temperature coefficient of resistance has a negative value. Hence the value of resistance decreases as temperature increases and the value of resistance decreases if its temperature increases.

Thermistors: 

A thermistor is a special case of a semiconductor having a large negative temperature coefficient of resistance. Thermistors are also called as temperature-sensitive resistance. As they have a large negative value of alpha the value of resistance decreases very fast, as the temperature increases. Thermistors are very sensitive.

Thermistors are made up of oxides of copper, manganese, nickel, cobalt, iron, lithium, etc. These oxides are mixed and are powdered. After this, they are given the desired shape and are heated to very high temperatures. Thus ceramic thermistors are formed. Thermistors are used in the temperature controlling devices or as temperature sensors.

Previous Topic: Introduction to Current Electricity

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Science > Physics > Current Electricity > Temperature Dependence of Resistance

The post Temperature Dependence of Resistance appeared first on The Fact Factor.

In this article, we shall study the concept of current electricity, the resistance, ohm’s law, and the conductance of a conductor.

Electric Current Through Conductor:

A conductor is made up of very minute particles called atoms. Atoms consist of a positively charged nucleus and negatively charged electrons which move around the nucleus in different orbits. The electrons in the last orbit are loosely attached to the atoms. They can be removed by applying an external force, hence such electrons are called free electrons.

If one end of the conductor is connected to positive terminal of a battery and another end is connected to negative terminal of a battery, negatively charged free electrons start moving towards the positive terminal of battery Thus there is a flow of electron through the conductor and we can say that electric current is flowing through the conductor.

Actually, electrons flow from the negative terminal of a battery to the positive terminal of a battery through the external circuit. But conventionally it is assumed that electric current flows from positive terminal to the negative terminal of a battery.

If ‘e’ is a charge on one electron and ‘Q’ is a total charge flowing through the conductor, then t6he number of electrons (n) flowing through the conductor can be found by using the relation

Q = n e.

Types of Material:

Conductors:

The substances which allow an electric current to flow through them easily are called good conductors of electricity. In conductor electric current flows due to free electrons. e.g. All metals. Silver, Aluminium, Copper, etc.

Insulators:

The substances which do not allow the electric current to flow through them are called bad conductors or insulators of electricity. e.g. Plastic, Rubber, Glass, etc.

Rubber plastic, wood don’t have the free electrons in them, hence they do not allow the electric current to pass through them. Such materials are called as insulators or bad conductors.

Electric Current:

The rate of flow of charge with respect to time through a given cross-section of the conductor is called an electric current. The symbol of the current is ‘I’. The unit of current is ampere (A)

I = q/t

Where I    =   electric current

t    =   time, q   =   electric charge

Potential Difference:

A conductor contains free electrons, which are in random motion i.e. they move in any possible direction with any possible velocity. Due to which the number of electrons passing in unit time through any section of the conductor in one direction is equal to the number of electrons passing in unit time through that section in the opposite direction. Therefore, the net flow of change is equal to zero. Therefore, no current flows through the conductor.

When a potential difference is applied across the conductor the negatively charged electrons start moving towards the positive end of the conductor. Thus electrons start moving in a definite direction. Thus current flows through the conductor. Hence we can conclude that for a flow of electrons in a conductor, the potential difference across the end of the conductor is required.

Cause of a Resistance of a Conductor:

A metallic conductor consists of a large number of free electrons. These electrons are always in a state of random motion. When a potential difference is applied across the ends of the conductor. These free electrons start moving in the definite direction i.e. towards the positive end of the conductor. During this process, the electrons flow through the crowd of vibrating atoms. These electrons collide with the atoms. Thus vibrating atoms offer obstruction to the flow of electrons. This obstruction to the flow of electrons is called the resistance of the conductor.

If the temperature of the conductor is increased, the kinetic energy of vibrating atoms is increased, due to which the atoms start vibrating with higher amplitude. Thus the obstruction to the flow of electrons increases and hence the resistance of the conductor also increases.

Resistance of Wire:

Experimentally it is found that the value of resistance (R) depends on length (L) of a conductor, area of cross-section (A) of conductor and nature of a conductor as follows :

The resistance is directly proportional to the length of a conductor.

R ∝  L   …………..  (1)

The resistance is inversely proportional to the area of a cross-section.

R ∝  1/A   …………..  (2)

The resistance depends on the nature of the conductor.

From equation (1) & (2)

ρ is a constant called specific resistance or resistivity.

This is an expression for the resistance of a conducting wire

This is an expression for the specific resistance or the resistivity of a material of a conductor.

Resistivity or Specific Resistance:

We have,

Let A = 1 unit  and L = 1 unit

∴ R    = ρ

Thus specific resistance or resistivity of a material of a conductor is defined as that resistance of a conductor whose area of cross-section and its length is unity.

Unit of Resistivity:

Therefore, the unit of resistivity or coefficient of resistance is  ohm metre (Ωm)

Conductance:

Reciprocal of resistance is called conductance (K). Its unit is mho or siemens (S).

K = 1 / R

Conductivity:

Reciprocal of resistivity is called conductivity (k). Its S.I. unit is siemens per metre (S/m)

k = 1/ρ

Notes:

• Resistivity is a measure of opposition to the flow of electric current, while conductivity is a measure of assistance to the flow of electric current i.e. easiness of flow of electric current.
• A material exhibiting high resistivity has low conductivity, while material exhibiting high resistivity has low conductivity.

Copper wires are generally used as connecting leads in an electrical circuit.

Copper has an extremely small specific resistance, hence the resistance of a conductor made up of copper is very less. Thus there is a very small loss of electrical energy when a current flows through a copper wire. Due to its high conductivity, a thin copper wire can be used. Hence, copper wires are used to save energy and cost.

Coils of electric iron are made up of nichrome.

Electric-iron works on the principle of the heating effect of electric current. Hence in electric iron more heat is to be produced. When the value of resistance of the coil is more, the more heat is generated. Nichrome is an alloy whose specific resistivity is very high. Hence resistance or coil made from nichrome possesses higher resistance. Hence coils of electric iron are made up of nichrome.

The resistance coils in high-quality resistance boxes are made of manganin.

A resistance box is used in electrical experiments. The resistance box contains a number of resistances of different values. These values should remain constant even if there is a change in room temperature. Manganin has a very small temperature coefficient of resistance and therefore for a small change in temperature, the change in the resistance of a manganin coil is negligible. Hence the value of resistance made from manganin remains constant. Hence the resistance coils in high-quality resistance boxes are made of manganin.

Ohm’s Law:

Statement: 

Physical conditions of the conductor (i.e. the length, area of cross-section, the material, and the temperature) remain the same, the potential difference across the terminal of the conductor is directly proportional to the electric current flowing through the conductor.

Explanation:

Let ‘V’ be the potential difference across the conductor and ‘I’ be the current through it, then by ohm’s law

V ∝   I

V   =   R I

Where R = constant called resistance of the conductor.

Graphical Representation of Ohm’s Law

For conductors obeying ohm’s law, we get a straight line. The resistances obeying ohm’s law are called ohmic resistances or linear resistances.

Note:

For some conductors, we don’t get straight lines but we get curves as shown below.

From the above graphs, we can conclude that these resistances are not obeying ohm’s law that’s why they are called a non- ohmic or non-linear resistance.

Science > Physics > Current Electricity > Introduction

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