Question

Four resistors each of 1.5 Q are arranged in the form of a parallelogram. The equivalent resistance between any two opposite corners is

• A. 6 Ω
• B. 3 Ω
• C. 0.66 Ω
• D. 1.5 Ω

Answer 1

The Correct Option is D

Step 1: Analyze the arrangement of resistors.

The four resistors form a parallelogram, which can also be thought of as a combination of two parallel branches, each containing two resistors in series. Let the resistors be labeled as follows:

• Resistors $ R_1 $ and $ R_2 $ are in one branch (series).
• Resistors $ R_3 $ and $ R_4 $ are in the other branch (series).

Step 2: Calculate the resistance of each branch.

In each branch, the two resistors are in series. The equivalent resistance of two resistors in series is the sum of their resistances. Thus:

\[ R_{\text{branch}} = R_1 + R_2 = 1.5 + 1.5 = 3 \, \Omega. \]

Similarly, for the second branch:

\[ R_{\text{branch}} = R_3 + R_4 = 1.5 + 1.5 = 3 \, \Omega. \]

Step 3: Combine the two branches in parallel.

The two branches are connected in parallel. The equivalent resistance of two resistors in parallel is given by:

\[ \frac{1}{R_{\text{eq}}} = \frac{1}{R_{\text{branch 1}}} + \frac{1}{R_{\text{branch 2}}}. \]

Substitute $ R_{\text{branch 1}} = 3 \, \Omega $ and $ R_{\text{branch 2}} = 3 \, \Omega $:

\[ \frac{1}{R_{\text{eq}}} = \frac{1}{3} + \frac{1}{3} = \frac{2}{3}. \]

Solve for $ R_{\text{eq}} $:

\[ R_{\text{eq}} = \frac{3}{2} = 1.5 \, \Omega. \]

Final Answer: The equivalent resistance between any two opposite corners is $ \mathbf{1.5 \, \Omega} $, which corresponds to option $ \mathbf{(4)} $.