Question 9: Current Electricity

Question

In the circuit, shown the galvanometer $$G$$ of resistance $$60 \Omega$$ is shunted by a resistance $$r=0.02 \Omega$$. The current through $$R$$ is nearly $$1 \mathrm{~A}$$. The value of resistance $$R$$ (in ohm) is nearly

Accepted Answer

Here, resistance of the galvanometer
$$R_G=60\Omega$$
When the galvanometer is shunted by a resistance r, its effective resistance
$$R_P=\frac{R_G r}{R_G+r}=\frac{60 	imes 0.02}{60+0.02} \approx 0.02 \Omega$$
Total resistance of the circuit $$=R+R_P=R+0.02$$
$$\begin{aligned}
	ext { Current, } I & =\frac{5}{R+0.02} \Rightarrow 1=\frac{5}{R+0.02} \
R+0.02 & =5 \Rightarrow R=5-0.02=4.98 \approx 5 \Omega
\end{aligned}$$

Answer

1.00 $$\Omega$$

Answer

5.00 $$\Omega$$

Answer

11.0 $$\Omega$$

Answer

6.0 $$\Omega$$

Current Electricity

In the circuit, shown the galvanometer $G of resistance 60 \Omega is shunted by a resistance r=0.02 \Omega. The current through R is nearly 1 \mathrm{~A}. The value of resistance R$ (in ohm) is nearly

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