Question 267: Magnetism and Matter

Question

A moving coil of galvanometer when shunted with $2 \Omega$ resistance gives a full scale deflection for a current of 500 mA . When a resistance of $470 \Omega$ is connected in series it gives a full scale deflection for 10 V potential applied on it. The value of resistance of galvanometer coil is $\_\_\_\_$ $\Omega$.

Accepted Answer

When a galvanometer of resistance $G$ is shunted with resistance $S$, the total current I splits. The current through the galvanometer is $\mathrm{I}_{\mathrm{g}}$.
The potential across the galvanometer and shunt is equal:
$$ \mathrm{I}_{\mathrm{g}} \mathrm{G}=\left(\mathrm{I}-\mathrm{I}_{\mathrm{g}}ight) \mathrm{S} $$
The resistance of shunt is $\mathrm{S}=2 \Omega$ and the current to measure with full deflection is $\mathrm{I}=500 \mathrm{~mA}=0.5 \mathrm{~A}$
$$ I_g G=\left(0.5-I_gight) 	imes 2 $$
$$\Rightarrow $$ $\mathrm{I}_{\mathrm{g}} \mathrm{G}=1-2 \mathrm{I}_{\mathrm{g}}$
$$ I_g(G+2)=1 \Rightarrow I_g=\frac{1}{G+2} \ldots (i)$$
When a resistance R is connected in series, the total potential V is :
$$ V=I_g(G+R) $$
The resistance added in series is given as $\mathrm{R}=470 \Omega$ and $\mathrm{V}=10 \mathrm{~V}$ :
$$ 10=I_g(G+470) \Rightarrow I_g=\frac{10}{G+470} \ldots (ii)$$
Since the full-scale deflection current $\mathrm{I}_{\mathrm{g}}$ is constant for the galvanometer,
$$ \frac{1}{G+2}=\frac{10}{G+470} $$
$$\Rightarrow $$ $\mathrm{G}+470=10(\mathrm{G}+2)$
$$\Rightarrow $$ $\mathrm{G}+470=10 \mathrm{G}+20$
$$\Rightarrow $$ $470-20=10 \mathrm{G}-\mathrm{G}$
$$\Rightarrow $$ $450=9 \mathrm{G}$
$$\Rightarrow $$ $$ \mathrm{G}=\frac{450}{9}=50 \Omega $$
Therefore, the value of resistance of the galvanometer coil is $50 \Omega$.

Magnetism and Matter

Integer / Numeric Answer Type