In the circuit shown in the figure, the total charge is 750 μC and the voltage across capacitor C2 is 20 V. Then the charge on capacitor C2 is
(A)160μC
(B)650μC
(C)590μC✓ Correct
(D)450μC
Answer: (C)
Solution
The capacitor network draws a fixed total charge of 750μC from the source, and this charge divides among the branches according to how they are connected. With the potential difference across C2 fixed at the measured 20 V, the charge held on the rest of the network alongside it accounts for 160μC, so by conservation of charge the amount left on C2 is 750−160=590μC.
JEE Main · 2020 · 03 Sep 2020 · Shift 1 · Single correct option