NEET Class neet Answered
please answer this
Asked by Prashant DIGHE | 14 Mar, 2020, 10:30: PM
Expert Answer
When S1 is closed, capacitor C1 acquires charge Q = 20V × 6 μF = 120 μC
Now s1 is open and S2 is closed. Hence we have capacitors connected parallel to give equivalent capacitance = 6 μF + 3 μF = 9 μF
Now 9 μF capacitor is having charge 120 μC . Hence potential difference across plates = Q/C = 120 μC / 9 μF = 40/3 V
Charge on capacitor plates of C2 , Q = C V = 3 μF × ( 40/3) V = 40 μC
Answered by Thiyagarajan K | 14 Mar, 2020, 11:04: PM
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