Question

In the following network, the current flowing through \(15 \Omega\) resistance is

• A. \(2.1\) A
• B. \(0.9\) A
• C. \(1.2\) A
• D. \(1.5\) A

Hint:

Check if the bridge is balanced (15/3 = 20/4). If yes, split 2.1 A between the 18 ohm and 24 ohm paths.

Answer 1

The Correct Option is C

Step 1: Use the potential difference:
The bridge is balanced because $15\times4 = 3\times20 = 60$. So no current passes through BD, and the same voltage acts across both parallel paths.

Step 2: Equivalent resistance:
The paths are $18\ \Omega$ and $24\ \Omega$ in parallel. $R_{eq} = \frac{18\times24}{18+24} = \frac{432}{42} = \frac{72}{7}\ \Omega$.

Step 3: Voltage:
$V = IR_{eq} = 2.1 \times \frac{72}{7} = 21.6$ V.

Step 4: Current in the upper path:
The upper path has $18\ \Omega$, so $I = \frac{21.6}{18} = 1.2$ A. The same current passes through the 15 ohm resistor.

Final Answer:
The current in the 15 ohm resistor is 1.2 A. \[ \boxed{1.2\ \text{A}} \]