Question:
The equivalent resistance of the circuit shown in the figure between points A and B is:
The equivalent resistance of the circuit shown in the figure between points A and B is:
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In symmetric resistor networks, always check for zero current in central branches due to equal potential points.
Updated On: Jun 19, 2026
- \(1\,\Omega\)
- \(0.5\,\Omega\)
- \(2\,\Omega\)
- \(4\,\Omega\)
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The Correct Option is A
Solution and Explanation
Step 1: Identify symmetry in the circuit.
The given circuit is symmetric about the vertical axis passing through the top node. Hence, potential at symmetric points is equal, allowing simplification by combining equal potential nodes.
Step 2: Reduce middle bridge effect.
Due to symmetry, no current flows through the central vertical \(2\Omega\) resistor (balanced Wheatstone-type condition). Hence it can be ignored.
Step 3: Simplify upper branches.
Each side branch from A to B becomes series combinations of \(2\Omega + 2\Omega = 4\Omega\). These two branches are in parallel.
Step 4: Include bottom resistor.
The bottom \(2\Omega\) resistor is directly connected between A and B, forming a parallel path with the upper network.
Step 5: Compute equivalent resistance.
Upper path equivalent: \[ 4\Omega \parallel 4\Omega = 2\Omega \] Now: \[ 2\Omega \parallel 2\Omega = 1\Omega \]
Step 6: Final conclusion.
Thus, equivalent resistance between A and B is: \[ \boxed{1\,\Omega} \]
The given circuit is symmetric about the vertical axis passing through the top node. Hence, potential at symmetric points is equal, allowing simplification by combining equal potential nodes.
Step 2: Reduce middle bridge effect.
Due to symmetry, no current flows through the central vertical \(2\Omega\) resistor (balanced Wheatstone-type condition). Hence it can be ignored.
Step 3: Simplify upper branches.
Each side branch from A to B becomes series combinations of \(2\Omega + 2\Omega = 4\Omega\). These two branches are in parallel.
Step 4: Include bottom resistor.
The bottom \(2\Omega\) resistor is directly connected between A and B, forming a parallel path with the upper network.
Step 5: Compute equivalent resistance.
Upper path equivalent: \[ 4\Omega \parallel 4\Omega = 2\Omega \] Now: \[ 2\Omega \parallel 2\Omega = 1\Omega \]
Step 6: Final conclusion.
Thus, equivalent resistance between A and B is: \[ \boxed{1\,\Omega} \]
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