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: Understand the figure
A Wheatstone bridge ABCD with a galvanometer between B and D. Arms: AB = \(10\ \Omega\), BC = \(10\ \Omega\), AD = \(30\ \Omega\), DC = \(30\ \Omega\). The battery is \(3\) V.

Step 2: Check the balance
\(\frac{R_{AB}}{R_{BC}}=\frac{10}{10}=1\) and \(\frac{R_{AD}}{R_{DC}}=\frac{30}{30}=1\). The ratios are equal, so the bridge is balanced and no current flows in the galvanometer branch BD.

Step 3: Equivalent resistance
Upper path ABC: \(10+10=20\ \Omega\). Lower path ADC: \(30+30=60\ \Omega\). They are in parallel: \[ \frac{1}{R_{eq}}=\frac{1}{20}+\frac{1}{60}=\frac{4}{60}\ \Rightarrow\ R_{eq}=15\ \Omega \]

Step 4: Current
\[ I=\frac{V}{R_{eq}}=\frac{3}{15}=0.2\ \text{A} \]

Step 5: Other options
0.1 A would need 30 ohm, 0.15 A would need 20 ohm (only the upper branch, 3/20), and 0.25 A would need 12 ohm. None matches 15 ohm.

Final Answer:
The battery current is 0.2 A, option (C). \[ \boxed{0.2\ \text{A}} \]