Question

Two capacitors of capacitance 6 uF and 4 uF are connected in parallel with a battery. The charge on 6 uF
capacitor is 24 uC, then find the charge on another capacitor and voltage of battery.


please solve i will mark u ​

Answer 1

Explanation:

Cp =2+4=6μF

1/C = 1/6 + 1/6 = 2/6 = 1/3

orC=3μF

Total charge

Q=CV=3×12=36μC

voltage across 6μF capacitor = 36μC / 6μF =6V

∴ Voltage across each of 2μF and 4μF capacitors

=12V−6V

V=6V