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.

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Next Topic: Numerical Problems on Potential Drop (Potentiometer)

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