Question

A balanced meter bridge is shown in the circuit diagram. The meter bridge wire has resistance \(2\frac{\Omega }{cm}\). The current drawn from the battery is (Internal resistance of the battery is negligible)

• A. \(0.6\) A
• B. \(0.8\) A
• C. \(0.95\) A
• D. \(1.05\) A

Hint:

Use the balance condition to find x, then find the equivalent resistance of the two parallel branches.

Answer 1

The Correct Option is D

Step 1: Balance condition
The balance point is at 30 cm, so \(R_{AP}=30\times2=60\,\Omega\) and \(R_{PB}=70\times2=140\,\Omega\). Then \(\dfrac{3}{x}=\dfrac{60}{140}\), so \(x=7\,\Omega\).

Step 2: Branches
The resistor branch has \(3+7=10\,\Omega\). The full wire has \(100\times2=200\,\Omega\). They are in parallel across the 10 V battery.

Step 3: Calculation
\(R_{eq}=\dfrac{10\times200}{210}=9.52\,\Omega\), so \(I=\dfrac{10}{9.52}\approx1.05\) A.

Final Answer:
The current drawn from the battery is about 1.05 A. \[ \boxed{\text{(D) }1.05\ \text{A}} \]