Physics  Chemistry  Biology  Mathematics 
Science > Physics > Current Electricity > You are Here 
Example – 14:
 In the 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 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 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
Let L_{L} = l cm and L_{R} = (100 – l) cm
∴ (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 – 15:
 Two coils are connected in series in one gap of Wheatstone’s metre bridge and 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 R_{1} and R_{2} be the resistances of the two coils
 Case – I: Given R_{1} and R_{2} be in series, resistance in left gap = R_{1 }+ R_{2}, Resistance in right gap = 50 Ω
l = 50 cm and, 100 – l = 100 – 50 = 50 cm
For balanced metre bridge
∴ (R_{1} + R_{2})/50 = 50/50
∴ (R_{1} + R_{2}) = 50 ………. (1)
 Case – II:
Now the two coils are connected in parallel. Hence their effective resistance is decreasing
Two keep 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
∴ 50R_{1} – R_{1}^{2} = 1350
∴ R_{1}^{2} – 50 R_{1} + 600 = 0
∴ (R_{1} – 30)(R_{1} – 20) = 0
∴ R_{1} = 30 Ω or R1_{1} = 20 Ω
Hence R_{2} = 20 Ω or R_{2} = 30 Ω
Ans: The restances of two coils are 30 Ω and 20 Ω.
Example – 16:
 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 – 17:
 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 – 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 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 – 19:
 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 R_{1} and R_{2} 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 ρ_{1 }= ρ_{2}
Thus resistance in left gap = R_{1} Ω, Resistance in right gap = R_{2} Ω
Let l be the position of null point from the left end
For balanced metre bridge
∴ R_{1}/R_{2} = 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 – 20:
 With resistance R_{1} in the left gap and R_{2} in a right gap of a Wheatstone’s metre bridge, the null point is obtained at 30 cm from the right end. When R_{1} is reduced by 2 Ω and R_{2} is increased by 2 Ω, the null point is obtained at 30 cm from the left end. Find R_{1} and R_{2}.
 Solution:
 Case – I: Resistance in left gap = R_{1}, Resistance in right gap = R_{2}
Null point from right end = 100 – l = 30 cm
Null point from left end = l = 100 – 30 = 70 cm
For balanced metre bridge
∴ R_{1}/R_{2} = 70/30= 7/3
∴ R_{1} = (7/3)R_{2 } ………. (1)
 Case – II: Resistance in left gap = R_{1 }– 2 Ω Resistance in right gap = R_{2 }+ 2 Ω
Null point from left end = l = 30 cm and 100 – l = 100 – 30 = 70 cm
For balanced metre bridge
∴ (R_{1 }– 2)/(R_{2 }+ 2) = 30/70= 3/7
∴ 7R_{1} – 14 = 3R_{2 } + 6
∴ 7R_{1} – 3R_{2} = 20
∴ 7 (7/3)R_{2} – 3R_{2} = 20
∴ (49/3)R_{2} – 3R_{2} = 20
∴ ((49 – 9)/3)R_{2 } = 20
∴ (40/3)R_{2 } = 20
∴ R_{2 } = 20 x (3/40) = 1.5 Ω
R_{1} = (7/3)R_{2 } = (7/3) x 1.5 = 3.5 Ω
Ans: R_{1} = 3.5 Ω and R_{2} = 1.5 Ω
Example – 21:
 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: 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: 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 – 22:
 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. When the value of X is changed by 20 Ω, the null point divides the wire in the ratio 1:4. Calculate each resistance.
 Solution:
 Case – I: 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: Resistance in left gap = X_{ }± 20 Ω Resistance in right gap = Y_{ } Ω
l /(100 – l )= 1/4
∴ 4 l = 100 – l
∴ 5 l = 100
∴ l = 20 cm from left end
Now in l second case is less than l in the first case
Thus value of X is reducing, Hence resistance in left gap = x – 20
For balanced metre bridge
∴ (X_{ }– 20)/Y = 1/4
∴ 4X – 80 = Y
∴ 4X – Y = 80
∴ 4 x (2/3)Y – Y = 80
∴ (8/3)Y – Y = 80
∴ (5/3)Y = 80
∴ Y = 48 Ω
X = (2/3)Y = (2/3) x 48 = 32 Ω
Ans: X = 32 Ω and Y = 48 Ω
Example – 23:
 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 left end. Find the ratio of diameters of wirtes 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 – 24:
 Equal lengths of wires of manganin(ρ_{1}) and nichrome (ρ_{2}) 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 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 R_{1} and R_{2} be the resistances of the two wires connected in left and right gaps.
Resistance in left gap = R_{1} Resistance in right gap = R_{2}
Null point from left end = l = 40 cm and 100 – l = 100 – 40 = 60 cm
For balanced metre bridge
∴ R_{1}/R_{2} = 40/60= 2/3 …………. (1)
ρ_{1 }= 4.8 x 10^{8} Ωm, ρ_{2 }= 10^{6} Ωm
Let r_{1} and r_{2} be the radii of the two wires
Ans: The ratio of diameters of two wires is 0.2683:1
Example – 25:
 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 R_{1} and R_{2} be the resistances of the two cut wires
R = R_{1 }+ R_{2}
Ratio of their lengths l_{1}/l_{2} = 1/2
The material of wires is the same, hence resistivity is the same ρ_{1 }= ρ_{2}
The wire is uniform, hence radii are the same r_{1 }= r_{2}
Resistance in left gap = (2/3)R_{1}, 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)R_{1}/25 = 40/60= 2/3
∴ R_{1}/25 = 1
∴ R_{1} = 25 Ω
R_{2} = 2R_{1} = 2 x 25 = 50 Ω
Resistance of uncut wire = R = R_{1 }+ R_{2 }= 25 + 50 = 75 Ω
Ans: The resistance of the wire before it was cut into two pieces is 75 Ω.
Example – 26:
 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 R_{1} and R_{2} be the resistances of the two cut wires
R = R_{1 }+ R_{2}
Ratio of their lengths l_{1}/l_{2} = 1/2
The material of wires is the same, hence resistivity is the same ρ_{1 }= ρ_{2}
The wire is uniform, hence radii are the same r_{1 }= r_{2}
Resistance in left gap = (2/3)R_{1}, 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)R_{1}/20 = 40/60= 2/3
∴ R_{1}/20 = 1
∴ R_{1} = 20 Ω
R_{2} = 2R_{1} = 2 x 20 = 40 Ω
Resistance of uncut wire = R = R_{1 }+ R_{2 }= 20 + 40 = 60 Ω
Ans: The resistance of the wire before it was cut into two pieces is 60 Ω.
Science > Physics > Current Electricity > You are Here 
Physics  Chemistry  Biology  Mathematics 