Question

In balanced meter bridge, the resistance of bridge wire is \(0.1\ \Omega /\)cm. Unknown resistance \(X\) is connected in the left gap and \(6\ \Omega\) in right gap. Null point divides the wire in the ratio \(2:3\). The current drawn from the battery of \(5\)V having negligible resistance is

• A. \(1\) A
• B. \(1.6\) A
• C. \(2\) A
• D. \(5\) A

Hint:

Find X from the balance ratio, then treat the two resistance arms in parallel with the bridge wire across the battery.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
At balance \(\dfrac{X}{6}=\dfrac{l}{100-l}=\dfrac{2}{3}\), so \(X=4\ \Omega\).

Step 2: Resistance of Each Branch
Branch with X and 6 \(\Omega\): \(4+6=10\ \Omega\).
Bridge wire: \(100\text{ cm}\times0.1=10\ \Omega\).

Step 3: Equivalent Resistance
The two branches are in parallel across the battery: \(R_{eq}=\dfrac{10\times10}{20}=5\ \Omega\).

Step 4: Current
\(I=\dfrac{5}{5}=1\) A.

Final Answer:
The current drawn from the battery is 1 A. \[ \boxed{\text{(A) }1\text{ A}} \]