In the circuit shown, two capacitors C1 and C2 are charged to a potential difference of V1 and V2 separately. After charging switches S1, S2 and S3 are closed. Calculate the charge (in µC) passed through switch S1. Take capacitance C1 = 1µF, C2 = 2µF and their respectively potential difference V1 = 100 V and V2 = 20 V.
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q1 = C1V1 = 1µF × 100V = 100µC
q2 = C2V2 = 2µF × 20V = 40µC
Initially
–q1 + q2 = –60µC
After closing the switch
q'1 + q'2 = –60µC
Also + = 0 (from Kirchoff's loop law)
⇒ q'2 = –2q'1
Thus, q'1 = 20µC and q'2 = –40µC
Charged passed through switch S1 = (100 – 20)µC = 80µC