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Question

In the following circuit, the resultant capacitance between A and B is 1 μF. Find the value of C.
127101.png

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Solution

The equivalent capacitance for RS is CRS=4+6×126+12=4+4=8μF
Now, CPS=1×81+8=8/9μF and CPQ=8×48+4=8/3μF
Now CPQ and CRS are in parallel and then it is series with C.
Thus, CAB=C(32/9)C+32/9=1
or C+32/9=32C/9
or 9C+32=32C or 23C=32
so, C=32/23=1.4μF


530233_127101_ans.png

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