Wheatstone bridge balance condition

Steps:

• Identify the bridge arms: 2 Ω and 4 Ω in one branch, 7 Ω and R in the other.
• For zero voltmeter reading, the bridge is balanced: \frac{2}{4} = \frac{7}{R}.
• Simplify the left ratio: \frac{2}{4} = 0.5.
• Solve for R: R = \frac{7}{0.5} = 14 \Omega.

Why D is correct:

• 14 Ω makes the resistance ratios equal, satisfying the Wheatstone balance condition for zero potential difference.

Why the others are wrong:

• A: 2.0 Ω gives \frac{7}{2} = 3.5 \neq 0.5, causing imbalance.
• B: 1.8 Ω gives \frac{7}{1.8} \approx 3.89 \neq 0.5, causing imbalance.
• C: 7.0 Ω gives \frac{7}{7} = 1 \neq 0.5, causing imbalance.

Final answer: D