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Example – 01:
 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 left gap = 20 Ω, Resistance in right gap = 40 Ω
Let l 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
 Case – II: Given 40 Ωresistance 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 be the distance of the null point from the left end.
For balanced metre bridge
Let L_{L} = l cm and L_{R} = (100 – l) cm
∴ 20/20 = l / (100 – l)
∴ 1= l / (100 – l)
∴ 100 – l = l
∴ 2 l = 100
∴ l = 50 cm
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 – 02:
 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.
 Solution:
 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
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 – 03:
 In a metre bridge experiment, with the resistance R_{1} 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 R_{2} 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 R_{1} and R_{2} are put parallel in the left gap and the right gap still containing X?
 Solution:
 Case – I: Given resistance in left gap = R_{1}, Resistance in right gap = X

l = 40 cm and, 100 – l = 100 – 40 = 60 cm
 For balanced metre bridge

∴ R_{1}/X = 40/60
∴ ∴ R_{1} = (2/3) X ………. (1)
 Case – II: Given resistance in left gap = R_{2}, Resistance in right gap = X

l = 50 cm and, 100 – l = 100 – 50 = 50 cm
 For balanced metre bridge

∴ R_{2}/X = 50/50
∴ R_{2} = 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 – 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 R_{1} and R_{2} be the resistances of the two coils
 Case – I: Given R_{1} and R_{2} be resistance in left gap = R_{1 }+ R_{2}, Resistance in right gap = 75 Ω

l = 50 cm and, 100 – l = 100 – 50 = 50 cm
 For balanced metre bridge

∴ (R_{1} + R_{2})/75 = 50/50
∴ (R_{1} + R_{2}) = 75 ………. (1)

 Case – II:

∴ R1 (75 – R1) = 1350
∴ 75R_{1} – R_{1}^{2} = 1350
∴ R_{1}^{2} 75 R_{1} = 1350 = 0
∴ (R_{1} – 30)(R_{1} – 45) = 0
∴ R_{1} = 30 Ω or R1_{1} = 45 Ω
Hence R_{2} = 45 Ω or R_{2} = 30 Ω

Ans: The restances of two coils are 30 Ω and 45 Ω.
Example – 05:
 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 – 06:
 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: 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 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 – 07:
 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: 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: 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 – 08:
 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: 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: The value of R is increased by 2 Ω. The balance point should shift towards rig

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 – 09:
 In a metre bridge experiment, with the resistance R_{1} 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 R_{2} 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 R_{1} and R_{2} are put series in the left gap and the right gap still containing X?
 Solution:
 Case – I: Given resistance in left gap = R_{1}, Resistance in right gap = X

l = 40 cm and, 100 – l = 100 – 40 = 60 cm
 For balanced metre bridge

∴ R_{1}/X = 40/60
∴ ∴ R_{1} = (2/3) X ………. (1)
 Case – II: Given resistance in left gap = R_{2}, Resistance in right gap = X

l = 50 cm and, 100 – l = 100 – 50 = 50 cm
 For balanced metre bridge

∴ R_{2}/X = 50/50
∴ R_{2} = X ………. (2)
 Case – III:
Resistance in left gap = R_{1} + R_{2} , Resistance in right gap = X
Let l be the distance of the null point from the left end.
For balanced metre bridge

∴ (R_{1} + R_{2})/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 – 10:
 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 a 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: 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.
Science > Physics > Current Electricity > You are Here 
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