When some potential difference is maintained between A and B, current / enters the network at A and leaves at B:
A
the equivalent resistance between A and B is 8Ω
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B
C and D are at the same potential
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C
no current flows between C and D
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D
current (3/5) flows from D to C
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Solution
The correct options are A C and D are at the same potential B the equivalent resistance between A and B is 8Ω D current (3/5) flows from D to C As C and D are connected by a wire they are at same potential. So the circuit can be replaced by that given in figure. Between AC [Req=(20×5)/20+5] ohm Between CB [Req=(20×5)/20+5] ohm So between AB [Req=4+4] ohm [Req=8] ohm By symmetry the amount of current passing through 20 ohm resistance between AC will be equal to the current passing through the 20 ohm resistance between CB [20×I1=5×(I−I1)] [I1=I/5] Therefore, [I−2×I1=(3/5)I]