Question 532: Current Electricity

Question

In an ammeter, $4 \%$ of the main current is passing through the galvanometer, If shunt resistance is $5 \Omega$, then resistance of galvanometer will be

Accepted Answer

Shunt $\mathrm{S=\frac{I_g G}{I-I_g}}$
On substituting the given values, we get,
$$\begin{aligned}
	herefore \quad 5 & =\frac{\left(\frac{4}{100} I 	imes Gight)}{I-\frac{4}{100} I}=\frac{4 G}{96} \
& \Rightarrow G=\frac{96 	imes 5}{4}=120 \Omega
\end{aligned}$$

Answer

$60 \Omega$

Answer

$120 \Omega$

Answer

$240 \Omega$

Answer

$480 \Omega$

Current Electricity

In an ammeter, 4 \% of the main current is passing through the galvanometer, If shunt resistance is 5 \Omega, then resistance of galvanometer will be

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