Physics - Current Electricity Question with Solution | TestHub

Given $I_g=10\text{ mA}=0.01\text{ A}$ and $V_g=0.75\text{ V}$.

Resistance of the galvanometer is $G=\frac{V_g}{I_g}=\frac{0.75}{0.01}=75\Omega$.

To convert the galvanometer into an ammeter of range $I=10\text{ A}$, a shunt resistance $S$ is connected in parallel.

The value of the shunt resistance is given by $S=\frac{I_{g}G}{I-I_g}=\frac{V_g}{I-I_g}$.

Substituting the values, we get $S=\frac{0.75}{10-0.01}=\frac{0.75}{9.99}=\frac{75}{999}\Omega$.

Answer: $\frac{75}{999}\Omega$