Question

When shunted the , the current sensitivity of a galvanometer reduces by a factor of 100 . The resistance of shunt is -
A) 22/23 B) 1/3 C) 2/3 D ) 3/2

Answer 1

Answer:

The resistance of the shunt will becomes 1/33 times the resistance of the galvanometer

Explanation:

as we know that current sensitivity is defined as the number of deflection per unit current

so here when the galvanometer is shunted with some resistance R

then its current sensitivity is decreased by factor of 100

so here we can say

$i_g R_g = (i - i_g) R_s$

here we know that

$i = 100 i_g$

$i_g R_g = 99 i_g R_s$

so we have

$R_s = \frac{R_g}{99}$

So it will becomes 1/33 times the resistance of the galvanometer

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Topic : Galvanometer

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Answer 2

The resistance of shunt is $\dfrac{1}{3}\ \Omega$

(B) is correct option.

Explanation:

Given that,

Factor = 100

Suppose, the resistance of galvanometer is 33 ohm.

We know that the current sensitivity is defined as the number of deflection per unit current.

The the current sensitivity of a galvanometer reduces by a factor of 100

We need to calculate the resistance of shunt

Using formula of shunt resistance of galvanometer

$i_{g}R_{g}=(i-i_{g})R_{s}$

Put the value into the formula

$i_{g}R_{g}=(100i_{g}-i_{g})R_{s}$

$R_{s}=\dfrac{33}{99}$

$R_{s}=\dfrac{1}{3}\ \Omega$

Hence, The resistance of shunt is $\dfrac{1}{3}\ \Omega$

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Topic : shunt resistance

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