Question

A galvanometer of resistance $G$ is converted into an ammeter using a shunt of resistance $R$. If the ratio of the heat dissipated through the galvanometer and shunt is $3:4$ then $R$ equals

• A. $\frac{4}{3}G$
• B. $\frac{3}{4}G$
• C. $\frac{16}{9}G$
• D. $\frac{9}{16}G$
• E. G

Answer 1

Answer: (B)

Heat produced in a resistor
$H=l^{2}Rt=\frac{V^2}{R}t$
$\therefore H\propto \frac{1}{R}$
Ratio of heat dissipated,
$\frac{H_1}{H_2}=\frac{3}{4}$ (given)
$\frac{H_1}{H_2}=\frac{R_2}{R_1}\Rightarrow \frac{R_2}{R_1}=\frac{3}{4}...(i)$
$\because R_1=G$
Therefore, Eq. (i) becomes,
$R_2=\frac{3}{4}G$