A millivoltmeter of 25 , mV range is to be co | vedclass

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Question

(A) The shunt resistance $S$ required to convert a galvanometer (or millivoltmeter) into an ammeter is given by the formula: $S = \frac{V_g}{I - I_g}$

Vedclass · Accepted Answer

$0.001 \, \Omega$

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(A) The shunt resistance $S$ required to convert a galvanometer (or millivoltmeter) into an ammeter is given by the formula: $S = \frac{V_g}{I - I_g}$.Here,$V_g$ is the full-scale voltage of the millivoltmeter,$I$ is the desired range of the ammeter,and $I_g$ is the full-scale current of the millivoltmeter.Given $V_g = 25 \, mV = 25 	imes 10^{-3} \, V$ and $I = 25 \, A$.Since $I \gg I_g$,we can approximate $I - I_g \approx I$.Therefore,$S = \frac{V_g}{I} = \frac{25 	imes 10^{-3} \, V}{25 \, A} = 10^{-3} \, \Omega = 0.001 \, \Omega$.

$A$ millivoltmeter of $25 \, mV$ range is to be converted into an ammeter of $25 \, A$ range. The value (in $\Omega$) of the necessary shunt will be:

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