The resistance between points A and B in the circuit shown in figure is:
A
4Ω
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B
6Ω
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C
10Ω
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D
8Ω
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Solution
The correct option is A10Ω In this case, the 8 ohms and 4 ohms resistances at the extreme right are connected in series. So their combined resistance is 8+4 = 12 ohms. The 12 ohms and adjacent 6 ohms resistance are connected in parallel. So the combined resistance is 116+112=10.25=4ohm The two 8 ohms resistances at the bottom are connected in parallel. so, the combined resistance is 118+18=10.25=4ohm This combined 4 ohms resistance is again connected in parallel with another 4 ohms resistance at the bottom end. So the total resistance at this point is 114+14=10.5=2ohm Now, all the combined resistances are connected in series. Therefore, the total resistance in the circuit is 4+4+2=10 ohms.