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:

\frac{4}{3} G

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\frac{3}{4} G

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\frac{16}{9} G

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\frac{9}{16} G

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Solution

The correct option is B $\frac{3}{4} G$
Heat produced in a resistor
$H = l^{2} R t = \frac{V^{2}}{R} t$
$∴ H \propto \frac{1}{R}$
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)
$∵ R_{1} = G$
Therefore, Eq.(i) becomes,
$R_{2} = \frac{3}{4}$ G.

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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:

The correct option is B 34GHeat produced in a resistorH=l2Rt=V2Rt∴H∝1RRatio of heat dissipated,H1H2=34 (given)H1H2=R2R1⇒R2R1=34 ..........(i)∵R1=GTherefore, Eq.(i) becomes,R2=34G.

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: