Question #190238
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.
Expert's answer
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
