Question

Three capacitors 2 μF, 6 μF, and 24 μF are connected in parallel to battery of 240 V. Calculate the capacitance of the combination and the charge on each capacitor.

Answer 1

a) For parallel combination,
 Ceq​=C1​+C2​+C3
​=(2+6+24)= 32μF
(b) q=V×C
for parallel array, V remains the same across each capacitor.
for C=2 μF
q1​=VC=240×2×10^-6
=480*10^(-6)C
for C=6*10^-6
q2​=VC=240*6*10^(-6)
=1440*10^(-6)C
for C=24*10^(-6)
q3​=VC=240*24*10^(-6)
=5760*10^(-6)C