Question

Three resistances of values 2Ω, 3Ω, and 6Ω are to be connected to produce an effective resistance of 4Ω. This can be done by connecting:

• A. 6Ω in series with the parallel combination of 2Ω and 3Ω
• B. 3Ω in series with the parallel combination of 2Ω and 6Ω
• C. 2Ω resistance in series with the parallel combination of 3Ω and 6Ω
• D. 20Ω resistance in parallel with the parallel combination of 3Ω and 6Ω

Hint:

For resistances in series, the total resistance is the sum of individual resistances. For resistances in parallel, use the formula \( \frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} \).

Answer 1

The Correct Option is C

Step 1: Calculate the equivalent resistance for the parallel combination: For the 3Ω and 6Ω resistors in parallel: \[ R_{\text{parallel}} = \frac{3 \times 6}{3 + 6} = 2 \, \Omega \]
Step 2: Add the 2Ω resistance in series: The total resistance is the sum of the parallel resistance and the 2Ω resistor in series: \[ R_{\text{total}} = 2 + 2 = 4 \, \Omega \]