Question

Resistances are joint as shown in the figure. In the balanced condition, the current \(I\) drawn from the battery is

• A. \(0.1\) A
• B. \(0.15\) A
• C. \(0.2\) A
• D. \(0.25\) A

Hint:

The bridge is balanced, so ignore the middle arm. Combine 20 ohm and 60 ohm in parallel and use Ohm's law.

Answer 1

The Correct Option is C

Step 1: Branch currents
With no current in BD, the two arms act separately. Upper branch: $I_1=\frac{3}{20}=0.15$ A.

Step 2: Lower branch
$I_2=\frac{3}{60}=0.05$ A.

Step 3: Total
The battery supplies both: $I=I_1+I_2=0.15+0.05=0.2$ A, option (C).

Final Answer:
Current drawn is 0.2 A. \[ \boxed{0.2\ \text{A}} \]