Question:
Five equal resistances are connected in a network as shown in figure. The net resistance between the points A and B is :
Five equal resistances are connected in a network as shown in figure. The net resistance between the points A and B is :
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A and B are joined by wires to the centre O and to the corner C, so find the resistance between O and C; the network is a balanced bridge.
Updated On: Nov 4, 2026
- \(\frac{3R}{2}\)
- \(\frac{R}{2}\)
- \(2R\)
- \(R\)
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The Correct Option is D
Solution and Explanation
From the figure, A is the centre junction O (wire connection) and B is the corner C (wire connection). The five resistors R are: O-D, O-E, D-E, D-C and E-C.
Between O and C we have a bridge: O-D and O-E on the source side, D-C and E-C on the other side, with D-E as the bridge arm.
The two branches O-D-C and O-E-C are identical (R+R each), so D and E are at the same potential and no current flows in the D-E resistor (balanced Wheatstone bridge).
The two paths of resistance 2R are in parallel: \(R_{AB} = \frac{2R\cdot 2R}{2R+2R} = R\).
Answer: (4) \(R\)
Between O and C we have a bridge: O-D and O-E on the source side, D-C and E-C on the other side, with D-E as the bridge arm.
The two branches O-D-C and O-E-C are identical (R+R each), so D and E are at the same potential and no current flows in the D-E resistor (balanced Wheatstone bridge).
The two paths of resistance 2R are in parallel: \(R_{AB} = \frac{2R\cdot 2R}{2R+2R} = R\).
Answer: (4) \(R\)
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