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(B) Let $i_1$ be the current flowing through the $500 \Omega$ resistor and the $12 	ext{ V}$ battery.Since the ammeter reading is zero,no current flo

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$100$

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(B) Let $i_1$ be the current flowing through the $500 \Omega$ resistor and the $12 	ext{ V}$ battery.Since the ammeter reading is zero,no current flows through the right branch containing the $2 	ext{ V}$ battery and the ammeter.Therefore,the current $i_1$ flows through the $500 \Omega$ resistor and the resistor $R$ in series.Applying Kirchhoff's voltage law $(KVL)$ to the left loop:$12 - 500 i_1 - R i_1 = 0$$12 = i_1 (500 + R) \quad \dots (i)$Now,consider the potential difference across the resistor $R$. Since no current flows through the right branch,the potential difference across $R$ must be equal to the electromotive force $(EMF)$ of the battery in the right branch to maintain zero current in that loop.The potential difference across $R$ is $V_R = i_1 R$.For the ammeter to read zero,the potential difference across $R$ must balance the $2 	ext{ V}$ battery.Thus,$i_1 R = 2 	ext{ V} \Rightarrow i_1 = \frac{2}{R}$.Substituting $i_1$ into equation $(i)$:$12 = \frac{2}{R} (500 + R)$$6 R = 500 + R$$5 R = 500$$R = 100 \Omega$

In the circuit shown below,the ammeter reading is zero. Then the value of the resistance $R$ is (in $\Omega$)

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