Question

The circuit shown in Figure is equivalent to a load of

• A. \(\frac{4}{3}\) Ohms
• B. 4 Ohms
• C. \(\frac{8}{3}\) Ohms
• D. 2 Ohms

Hint:

When combining resistances, always use the series and parallel formulas step by step to simplify the network.

Answer 1

The Correct Option is C

The circuit consists of resistors in series and parallel combinations. We can solve this step by step:
- First, combine the 2\(\Omega\) resistor and the 4\(\Omega\) resistor in parallel.
Using the parallel combination formula: \[ R_{\text{eq}} = \frac{R_1 R_2}{R_1 + R_2} = \frac{2 \times 4}{2 + 4} = \frac{8}{6} = \frac{4}{3} \, \Omega \] - Next, combine the equivalent resistance of \( \frac{4}{3} \, \Omega \) in series with the 2\(\Omega\) resistor.
The total equivalent resistance is: \[ R_{\text{total}} = \frac{4}{3} + 2 = \frac{4}{3} + \frac{6}{3} = \frac{10}{3} \, \Omega \] Thus, the correct answer is option (3).