Question

In the circuit shown in the figure, the current through

• A. the \(3\Omega\) resistance is \(0.50\) A.
• B. the \(3\Omega\) resistance is \(0.25\) A.
• C. the \(4\Omega\) resistance is \(0.50\) A.
• D. the \(4\Omega\) resistance is \(0.25\) A.

Hint:

Reduce the ladder from the right-hand end by series and parallel combinations, then split the current at each junction.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
The circuit is a ladder network. Reduce it step by step from the far right to the battery.

Step 2: Right section
The \(2\ \Omega\), \(4\ \Omega\) and \(2\ \Omega\) resistors are in series, giving \(8\ \Omega\), which is in parallel with the right \(8\ \Omega\) resistor. This gives \(4\ \Omega\).

Step 3: Middle and total
This \(4\ \Omega\) is in series with the two \(2\ \Omega\) resistors of the middle, giving \(8\ \Omega\), which is in parallel with the left \(8\ \Omega\) giving \(4\ \Omega\). The total is \(3+4+2=9\ \Omega\), so the battery current is \(I=\frac{9}{9}=1\) A.

Step 4: Current division
The \(3\ \Omega\) resistor carries \(1\) A. The current splits equally at the first junction, so \(0.5\) A flows to the middle branch. At the next junction it splits equally again, so the \(4\ \Omega\) resistor carries \(0.25\) A, which is option (D).

Final Answer:
The current through the \(4\ \Omega\) resistance is 0.25 A. \[ \boxed{\text{(D) }\text{the }4\ \Omega\text{ resistance is }0.25\text{ A.}} \]