Solve this:
Q). Two capacitors of 3pF and 6pF are connected in series and a potential difference of 5000 V is applied across the combination. They are then disconnected and reconnected in parallel. The potential between the plates is
(a) 2250 V
(b) 2222 V
(c) 2.25 × 10 6 V
(d) 1.1 × 10 6 V

5000v becaues in parallel connection p.d remains same current differs
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1/C = 1/3 + 1/6

so 1/C = 1/2

C= 2pF = 2 x 10-12

total charge = Q = 2 x 10-12 x 5000 = 10-8

V = 2 x 10-8 / (3 + 6 ) x 10-12
V = 2222V


 
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