Question 566: Current Electricity

Question

If only $$1 \%$$ of total current is passed through a galvanometer of resistance '$$G$$' then the resistance of the shunt is

Accepted Answer

Given: $$I_g=\frac{1}{100} I$$
$$\begin{aligned}
& 	herefore \quad \frac{\mathrm{I}}{\mathrm{I}_{\mathrm{g}}}=100 \quad 	ext{..... (i)} \
& \mathrm{~S}=\frac{\mathrm{GI}_{\mathrm{g}}}{\mathrm{I}-\mathrm{I}_{\mathrm{g}}}=\frac{\mathrm{G}}{\frac{\mathrm{I}}{\mathrm{I}_{\mathrm{g}}}-1} \quad 	ext{..... (ii)}
\end{aligned}$$
Substituting, (i) into (ii),
$$\mathrm{S}=\frac{\mathrm{G}}{100-1}=\frac{\mathrm{G}}{99}$$

Answer

$$\frac{G}{25} \Omega$$

Answer

$$\frac{\mathrm{G}}{49} \Omega$$

Answer

$$\frac{\mathrm{G}}{2} \Omega$$

Answer

$$\frac{\mathrm{G}}{99} \Omega$$

Current Electricity

If only $1 \% of total current is passed through a galvanometer of resistance 'G$' then the resistance of the shunt is

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