Conditions for Zero Current in Resistance R2 in Circuit

To determine the condition under which the current flowing through resistor $R_2$ vanishes, we can perform a nodal analysis on the circuit.

1. Nodal Potential Assignment

Let us choose the wire connecting the negative terminal of $V_1$, the left terminal of $R_2$, and the left side of $R_3$ as the reference node with potential: $V_L = 0\text{ V}$

Let $V_R$ be the potential at the junction to the right of $R_2$ (between $R_1$, $R_2$, and $V_2$).

• Top Branch: The battery $V_1$ has its negative terminal connected to the reference node ($0\text{ V}$) and its positive terminal connected to the top wire. Therefore, the potential at the top of resistor $R_1$ is $+V_1$.
• Middle Branch: Resistor $R_2$ is connected between potential $V_R$ and $0\text{ V}$.
• Bottom Branch: The battery $V_2$ has its negative terminal connected to junction $R$ (potential $V_R$) and its positive terminal connected towards resistor $R_3$. Thus, the potential at the junction between $V_2$ and $R_3$ is $V_R + V_2$.

2. Current Expressions and Kirchhoff's Current Law (KCL)

The current flowing from the top branch into junction $R$ through resistor $R_1$ is: $I_1 = \frac{V_1 - V_R}{R_1}$

The current flowing through resistor $R_2$ from junction $R$ to junction $L$ is: $I_2 = \frac{V_R - 0}{R_2} = \frac{V_R}{R_2}$

The current flowing downwards from junction $R$ through battery $V_2$ and resistor $R_3$ to the reference node is: $I_3 = \frac{(V_R + V_2) - 0}{R_3}$

For the current through $R_2$ to be zero ($I_2 = 0$), the potential difference across $R_2$ must be zero: $V_R = 0\text{ V}$

Applying KCL at junction $R$ with $V_R = 0\text{ V}$: $I_1 = I_2 + I_3 \implies \frac{V_1}{R_1} = 0 + \frac{V_2}{R_3}$

Thus, the condition for zero current in $R_2$ is independent of $R_2$ and is given by: $\frac{V_1}{R_1} = \frac{V_2}{R_3}$

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3. Verification of Options

• Option A: $V_1 = V_2 \text{ and } R_1 = R_2 = R_3$ $\frac{V_1}{R_1} = \frac{V_2}{R_3} \implies \frac{V_1}{R_1} = \frac{V_1}{R_1} \quad \text{(Satisfied)}$

• Option B: $V_1 = V_2 \text{ and } R_1 = 2R_2 = R_3$ Here $R_1 = R_3$, so: $\frac{V_1}{R_1} = \frac{V_2}{R_3} \implies \frac{V_1}{R_1} = \frac{V_1}{R_1} \quad \text{(Satisfied)}$

• Option C: $V_1 = 2V_2 \text{ and } 2R_1 = 2R_2 = R_3$ Here $V_1 = 2V_2$ and $R_3 = 2R_1$: $\frac{V_1}{R_1} = \frac{2V_2}{R_1}, \quad \text{while } \frac{V_2}{R_3} = \frac{V_2}{2R_1}$ $\frac{2V_2}{R_1} \neq \frac{V_2}{2R_1} \quad \text{(Not Satisfied)}$

• Option D: $2V_1 = V_2 \text{ and } 2R_1 = R_2 = R_3$ Here $V_2 = 2V_1$ and $R_3 = 2R_1$: $\frac{V_2}{R_3} = \frac{2V_1}{2R_1} = \frac{V_1}{R_1} \quad \text{(Satisfied)}$

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Conclusion

requirements are satisfied by options A, B, and D.