In the adjoining circuit,the galvanometer G s | vedclass

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(D) The galvanometer $G$ shows zero deflection,which means no current flows through it. This implies that the potential at the node between the $500 \

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$100$

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(D) The galvanometer $G$ shows zero deflection,which means no current flows through it. This implies that the potential at the node between the $500 \, \Omega$ resistor and resistor $R$ must be equal to the potential of the battery $A$ $(2 \, V)$.Let the potential at the bottom wire be $0 \, V$. The potential at the top node is $V_x = 2 \, V$ (due to battery $A$ and zero current in $G$).Now,consider the left loop. The current flowing through the $500 \, \Omega$ resistor and resistor $R$ is $I = \frac{12 \, V}{500 \, \Omega + R}$.The voltage drop across resistor $R$ is $V_x = I 	imes R = 2 \, V$.Substituting the expression for $I$:$\frac{12}{500 + R} 	imes R = 2$$12R = 2(500 + R)$$12R = 1000 + 2R$$10R = 1000$$R = 100 \, \Omega$.

In the adjoining circuit,the galvanometer $G$ shows zero deflection. If the batteries $A$ and $B$ have negligible internal resistance,the value of the resistor $R$ will be .............. $\Omega$.

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