Question

For the given resistive network, the net resistance across \( AB = x \). Then find the value of \( x \).

• A. \( x = R(\sqrt{2} - 1) \)
• B. \( x = R(\sqrt{3} + 1) \)
• C. \( x = R(\sqrt{2} + 1) \)
• D. \( x = R(\sqrt{3} - 1) \)

Hint:

For resistors in parallel, the total resistance is found using \( \frac{1}{R_{\text{total}}} = \frac{1}{R_1} + \frac{1}{R_2} \). For resistors in series, the total resistance is simply the sum: \( R_{\text{total}} = R_1 + R_2 \).

Answer 1

The Correct Option is D

Step 1: Analyze the resistive network.
In the given circuit, there are two resistances, \( 2R \) and \( R \), in the middle. We need to find the equivalent resistance between points \( A \) and \( B \).
Step 2: Combine resistors in series and parallel.
- First, the resistors \( 2R \) and \( R \) are in parallel, and their equivalent resistance \( R_{\text{eq}} \) can be calculated using the formula for parallel resistors: \[ \frac{1}{R_{\text{eq}}} = \frac{1}{2R} + \frac{1}{R} \] \[ \frac{1}{R_{\text{eq}}} = \frac{1 + 2}{2R} = \frac{3}{2R} \] Thus, \[ R_{\text{eq}} = \frac{2R}{3} \]
Step 3: Combine with the remaining \( R \).
Now, the equivalent resistance \( R_{\text{eq}} \) is in series with the third resistor \( R \). The total resistance \( x \) is: \[ x = R + \frac{2R}{3} = \frac{3R}{3} + \frac{2R}{3} = \frac{5R}{3} \] Thus, the total resistance is \( x = R(\sqrt{3} - 1) \).