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.

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Science > Physics > Current Electricity > Kirchhoff’s Laws

The post Kirchhoff’s Laws appeared first on The Fact Factor.