Please answer part D Four capacitors are connected across a 90-V voltage source

Question

Please answer part D

Four capacitors are connected across a 90-V voltage source as shown in the figure.

(a) What is the charge on the 4.0-?F capacitor?

120

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Correct

Part B

(b) What is the charge on a 2.0-?F capacitor?

120

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Correct

Part C

(c) What is the charge on the 3.0-?F capacitor?

180

  ?C

Part D

(d) What is the potential difference across the 6.0-?F capacitor?

120

  ?C Please answer part D Four capacitors are connected across a 90-V voltage source as shown in the figure. (a) What is the charge on the 4.0-?F capacitor? 120 ?C SubmitMy AnswersGive Up Correct Part B (b) What is the charge on a 2.0-?F capacitor? 120 ?C SubmitMy AnswersGive Up Correct Part C (c) What is the charge on the 3.0-?F capacitor? 180 ?C Part D (d) What is the potential difference across the 6.0-?F capacitor? ___ V

Explanation / Answer

here, 90v is appiled across two parallel path. so, 90 v will appear across each parallel path.

Vcx=Vs*(CT/Cx), Where: Cx is the capacitance of the capacitor in question, Vs is the supply voltage across the series chain and Vcx is the voltage drop across the target capacitor

Series Capacitors Equation 1/CT=1/C1+1/C2

for upper series path, CT=4*2/(4+2)=4/3uF

a) for 4uF capacitor,

Vcx=Vs*(CT/Cx)=(90*(4/3))/4=30 V

so, Q=CV=4uF*30=120uC

b)for 2uF capacitor,

Vcx=Vs*(CT/Cx)=(90*(4/3))/2=60 V

so, Q=CV=2uF*60=120uC

now, for lower series path,

using Series Capacitors Equation 1/CT=1/C1+1/C2

CT=3*6/(3+6)=2uF

c)

for 3uF capacitor,

Vcx=Vs*(CT/Cx)=(90*2)/3=60 V

so, Q=CV=3uF*60=180uC

d)

for 6uF capacitor,

Vcx=Vs*(CT/Cx)=(90*2)/6=30 V

so,the potential difference across the 6.0uF capacitor is 30 V

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