Three capacitors connected in series have an effective capacitance of 4μF. If one of the capacitance is removed, the net capacitance of the capacitor increases to 6μF. The removed capacitor has a capacitance of
2μF
4μF
10μF
12μF
24μF
A
12μF
B
24μF
C
4μF
D
10μF
E
2μF
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Solution
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Let there are three capacitors with capacitances C1,C2,C3 respectively and C1 is removed.
In first case, 1Ceq1=1C1+1C2+1C3 ...(1)
In second case, 1Ceq2=1C2+1C3 ...(2)
From (1) and (2), 1Ceq1=1C1+1Ceq2
or 1/4=1/C1+1/6
or C1=12μF
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