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Question

Physics Question on electrostatic potential and capacitance

The charge deposited on 4 μF4\, \mu F capacitor in the circuit is

A

6×10−6C6\times 10^{-6}C

B

12×10−6C12\times 10^{-6}C

C

24×10−6C24\times 10^{-6}C

D

36×10−6C36\times 10^{-6}C

Answer

24×10−6C24\times 10^{-6}C

Explanation

Solution