Physics - Current Electricity Question with Solution | TestHub

Given $I_g=16\text{ mA}=16\times 10^{-3}\text{ A}$ and $V_g=80\text{ mV}=80\times 10^{-3}\text{ V}$.

The resistance of the galvanometer is $G=\frac{V_g}{I_g}=\frac{80\times 10^{-3}}{16\times 10^{-3}}=5\Omega$.

To convert the galvanometer into a voltmeter of range $V=160\text{ V}$, a high resistance $R$ must be connected in series.

The required series resistance is given by $R=\frac{V}{I_g}-G$.

Substituting the values, we get $R=\frac{160}{16\times 10^{-3}}-5=10000-5=9995\Omega$.

Thus, a resistance of $9995\Omega$ must be connected in series.

Answer: $9995\Omega$ in series.