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: Understanding the Figure:
The figure is a diamond-shaped bridge with corners A, B, C and D. A current of 2.1 A enters at A and leaves at C. Arm AB has 15 ohm, arm BC has 3 ohm, arm AD has 20 ohm and arm DC has 4 ohm. The middle branch BD carries a resistor and a galvanometer G.

Step 2: Key Formula or Approach:
A Wheatstone bridge is balanced when \(\frac{R_{AB}}{R_{BC}} = \frac{R_{AD}}{R_{DC}}\). Then points B and D are at the same potential and no current flows through branch BD. The circuit reduces to two parallel paths.

Step 3: Check balance:
\[ \frac{15}{3} = 5, \qquad \frac{20}{4} = 5 \] The ratios are equal, so the bridge is balanced and BD can be ignored.

Step 4: Branch resistances:
Upper path A-B-C: \(R_1 = 15 + 3 = 18\ \Omega\). Lower path A-D-C: \(R_2 = 20 + 4 = 24\ \Omega\).

Step 5: Current divider:
Current splits in the inverse ratio of resistance. The 15 ohm resistor lies in the upper path: \[ I_{15} = 2.1 \times \frac{R_2}{R_1 + R_2} = 2.1 \times \frac{24}{42} = 1.2\ \text{A} \]

Step 6: Why the other options are wrong:
The value 0.9 A is the current in the lower path (\(2.1 \times 18/42\)). The value 2.1 A is the total current. The value 1.5 A does not follow from any split of the circuit.

Final Answer:
The current through the 15 ohm resistor is 1.2 A, option (C). \[ \boxed{1.2\ \text{A}} \]