Question

The value of \(R\) in the given circuit, so that there is no current flow in the \(10\,\Omega\) resistor, will be

• A. \(2\,\Omega\)
• B. \(4\,\Omega\)
• C. \(8\,\Omega\)
• D. \(12\,\Omega\)

Hint:

If no current flows through the central branch of a bridge network, immediately apply the Wheatstone bridge balance condition: \[ \frac{R_1}{R_2}=\frac{R_3}{R_4} \] This avoids lengthy circuit calculations.

Answer 1

The Correct Option is D

Concept: For no current to flow through the central \(10\,\Omega\) resistor, the potential difference across its ends must be zero. Hence, the circuit behaves as a balanced Wheatstone bridge. For a balanced Wheatstone bridge, \[ \frac{R_1}{R_2} = \frac{R_3}{R_4} \]

Step 1: Identify the four arms of the bridge. Upper left arm: \[ 12\,\Omega \] Upper right arm: \[ 15\,\Omega \] Lower right arm: \[ 5\,\Omega \] Lower left arm: \[ 6\,\Omega \parallel R \]

Step 2: Apply the bridge balance condition. \[ \frac{12}{15} = \frac{6\parallel R}{5} \] \[ 6\parallel R = \frac{12}{15}\times 5 = 4\,\Omega \]

Step 3: Calculate \(R\). For parallel combination, \[ \frac{6R}{6+R}=4 \] \[ 6R=24+4R \] \[ 2R=24 \] \[ R=12\,\Omega \]

Step 4: State the answer. \[ \boxed{R=12\,\Omega} \] Hence, the correct option is \[ \boxed{(D)} \]