In the circuit shown,the galvanometer G of re | vedclass

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(B) The galvanometer $G$ has a resistance $R_G = 60 \, \Omega$ and is shunted by a resistance $r = 0.02 \, \Omega$.The effective resistance $R_P$ of t

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$5$

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(B) The galvanometer $G$ has a resistance $R_G = 60 \, \Omega$ and is shunted by a resistance $r = 0.02 \, \Omega$.The effective resistance $R_P$ of the parallel combination is given by:$R_P = \frac{R_G 	imes r}{R_G + r} = \frac{60 	imes 0.02}{60 + 0.02} = \frac{1.2}{60.02} \approx 0.02 \, \Omega$.The total resistance of the circuit is $R_{total} = R + R_P = R + 0.02 \, \Omega$.Given that the current $I$ through the circuit is $1 \, A$ and the voltage source is $5.0 \, V$,we use Ohm's law:$I = \frac{V}{R_{total}}$$1 = \frac{5}{R + 0.02}$$R + 0.02 = 5$$R = 5 - 0.02 = 4.98 \, \Omega$.Thus,the value of $R$ is nearly $5 \, \Omega$.

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 $1 \, A$. The value of resistance $R$ (in $\Omega$) is nearly:

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