Question

The scale of galvanometer is divided into 150 equal divisions. The galvanometer has a current sensitivity of 10 divisions per $mA$ and the voltage sensitivity of 2 divisions per $mV$. How can the galvanometer be designed to read:

I. 6 A per division
The value of shunt is $\times {10}^{-5}\Omega$

II. 1 V per division
The value of resistance to be connected in series is $\Omega$

Answer 1

Let $I_G$ be the full scale current through galvanometer.
$\therefore I_G=\frac{150}{10}=15mA$
Similarly $V_G=\frac{150}{2}=75mV$
$⟹G=\frac{V_G}{I_G}=\frac{75mV}{15mA}=5\Omega$

Case 1: Convert galvanometer to ammeter
To do this we need to add a stunt resistance $S$ in parallel to galvanometer.
Total current measured by this ammeter will be
$I=150\times 6A$
$I=900A$
Now we can say
$(I-I_G)S=I_{G}G$
$⟹S=\frac{I_{G}G}{(I-I_G)}=\frac{15\times {10}^{-3}\times 5}{(900-15\times {10}^{-3}}$)
$S=8.3\times {10}^{-5}\Omega$

Case 2: Convert galvanometer to voltameter
For this we need to add a resistance $R$ in series to a galvanometer.
$V=i_G(G+R)$
$⟹\frac{V}{i_G}-G=R$
$⟹\frac{150}{15\times {10}^{-3}}-5=R$
$⟹10000-5=R$
$⟹R=9995\Omega$