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Q.
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
Heat produced in a resistor
$H=l^{2} R t=\frac{V^{2}}{R} t$
$\therefore H \propto \frac{1}{R}$
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,
$R_{2}=\frac{3}{4} G$