Question

The scale of a galvanometer is divided into $150$ equal divisions. The galvanometer has current sensitivity of $10$ divisions per $\text{mA}$ and a voltage sensitivity of $2$ divisions per $\text{mV}$. How can the galvanometer be designed to read
$\left(i\right)6\text{A}$ per division and
$\left(ii\right)1\text{V}$ per division
Choose the correct answer in terms of the resistance that should be added with the galvanometer.

• A. $\left(i\right)8.3\times {10}^{-5}\Omega$ in parallel $\left(ii\right)9995\Omega$ in series
• B. $\left(i\right)4.3\times {10}^{-5}\Omega$ in parallel $\left(ii\right)999\Omega$ in series
• C. $\left(i\right)2.15\times {10}^{-5}\Omega$ in parallel $\left(ii\right)99\Omega$ in series
• D. $\left(i\right)4.15\times {10}^{-5}\Omega$ in parallel $\left(ii\right)999\Omega$ in series

Answer 1

The correct option is A $\left(i\right)8.3\times {10}^{-5}\Omega$ in parallel $\left(ii\right)9995\Omega$ in series

Given,
Total number of divisions, $\theta =150$
Current sensitivity, $\text{CS}=10\text{divisions/mA}$
Voltage sensitivity, $\text{VS}=2\text{divisions/mV}$

So, the full scale current of the galvanometer,

$I_G=\frac{\theta }{CS}=\frac{150}{10}=15\text{mA}$,

Full scale voltage of the galvanometer,

$V_G=\frac{\theta }{VS}=\frac{150}{2}=75\text{mV}$

So, resistance of galvanometer,

$G=\frac{V_G}{I_G}=\frac{75\times {10}^{-3}}{15\times {10}^{-3}}=5\Omega$

$\left(i\right)$ In order to design a galvanometer to read current $6\text{A}$ per division:

Maximum current of galvanometer, $I=\theta \times 6=150\times 6=900\text{A}$.

So, shunt resistance,

$S=\frac{I_{G}G}{I-I_G}=\frac{15\times {10}^{-3}\times 5}{(900-15\times {10}^{-3})}=8.3\times {10}^{-5}\Omega$

$\left(ii\right)$ In order to design a galvanometer to read voltage $1\text{V}$ per division.

Maximum voltage reading, $V=150\times 1=150V$

So, the resistance to be added in series $R=\frac{V}{I_G}-G=\frac{150}{15\times {10}^{-3}}-5=9995\Omega$

Hence, option (a) is the correct answer.