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Standard XII

Physics

Series and Parallel Combination of Resistors

An unknown re...

Question

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

6 Ω, 4 Ω

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6 Ω, 3 Ω

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4 Ω, 6 Ω

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3 Ω, 4 Ω

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Solution

The correct option is A $6 Ω, 4 Ω$

using wheatstone bridge connection:
$\frac{X}{R} = \frac{60}{100 - 60} = \frac{60}{40}$
$\frac{X}{R} = \frac{3}{2}$ ....(1)
and after shifting,
$\frac{X}{R + 2} = 1$ ....(2)

on solving both eq. (1) and (2)
$R = 4 Ω$
$X = 6 Ω$

Hence, option A is correct.

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

The correct option is A 6 Ω, 4 Ω using wheatstone bridge connection: XR=60100−60=6040 XR=32 ....(1) and after shifting, XR+2=1 ....(2) on solving both eq. (1) and (2) R=4 Ω X=6 Ω Hence, option A is correct.

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

The correct option is A $6 Ω, 4 Ω$

using wheatstone bridge connection:
$\frac{X}{R} = \frac{60}{100 - 60} = \frac{60}{40}$
$\frac{X}{R} = \frac{3}{2}$ ....(1)
and after shifting,
$\frac{X}{R + 2} = 1$ ....(2)

on solving both eq. (1) and (2)
$R = 4 Ω$
$X = 6 Ω$

Hence, option A is correct.