Question

A 9.0 V battery is connected to three capacitors in series. The capacitors have the following capacitances: 4.76 $$\mu F$$ , 13.2 $$\mu F$$ , and 32.2 $$\mu F$$ . Calculate the voltage across the 32.2 $$\mu F$$ capacitor.

Answer 1

first find equivaelent capacitance, 1/C =1/4.76 + 1/13.2 + 1/32.2

C=3.16 uF

so Q = CV = 3.16*9=28.44 uC

things in series share charge so V = Q/C = 28.44/32.2=0.883 V

Answer 2

Total capacitance of the capacitor= 1/(1/C₁ +1/C₂+1/C₃)=1/(1/4.76+1/32.2+1/13.2)=3.15

Charge, Q=CV=9.0*3.15=28.4C

Ptential across 32.2C capacitor=Q/C=28.4/32.2=0.882V