In the circuit,the galvanometer G shows zero | vedclass

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(B) Since the galvanometer $G$ shows zero deflection,no current flows through the branch containing the galvanometer. Let the current flowing through

Vedclass · Accepted Answer

$100$

Vedclass · Answer

(B) Since the galvanometer $G$ shows zero deflection,no current flows through the branch containing the galvanometer. Let the current flowing through the left loop be $i$. Applying Kirchhoff's Voltage Law $(KVL)$ to the loop containing the $12 	ext{ V}$ battery and the $500 \, \Omega$ resistor: $12 - 500i - iR = 0$ $12 = i(500 + R)$ Applying $KVL$ to the loop containing the $2 	ext{ V}$ battery and the resistor $R$: Since no current flows through the galvanometer,the potential difference across $R$ must be equal to the $EMF$ of battery $A$. $iR = 2 	ext{ V}$ Substituting $i = \frac{2}{R}$ into the first equation: $12 = \left(\frac{2}{R}ight)(500 + R)$ $12R = 1000 + 2R$ $10R = 1000$ $R = 100 \, \Omega$

In the circuit,the galvanometer $G$ shows zero deflection. If the internal resistance of batteries $A$ and $B$ is negligible,then the value of resistance $R$ is ....... $\Omega$.

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