Question

If the potential difference between points B and D is zero in the given circuit, the value of x is \(\frac{1}{n} \Omega\). The value n is

• A. \(2\)
• B. \(1\)
• C. \(3\)
• D. \(4\)

Hint:

Zero potential difference between B and D means the bridge ratio holds.

Answer 1

The Correct Option is A

Step 1: Simplify each arm
Upper left: \(6\Omega\parallel3\Omega = \frac{18}{9} = 2\Omega\). Lower left: \(1+2 = 3\Omega\). Upper right: \(x\parallel1 = \frac{x}{x+1}\). Lower right: \(x\Omega\).

Step 2: Balance condition
For \(V_B = V_D\), \(\frac{2}{3} = \frac{x/(x+1)}{x} = \frac{1}{x+1}\).

Step 3: Solve
\(x + 1 = \frac32\), so \(x = \frac12\Omega = \frac1n\Omega\), giving \(n = 2\). Option (A).

Final Answer:
The value of n is 2. \[ \boxed{\text{(A)}\ 2} \]