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Question

www.webassign.net/web/Student/Assignment-Responses submit?dep 13251039 6.00 uF 9.00 (a) the equivalent capacitance of the capacitors in the figure above Your response is within 10% of the correct value. This may be due to roundoff or you could have a mis error, least four-digit accuracy to minimize roundoff error, vF (b) the charge on each capacitor on the right 28.00 VF Your response is capacitor intermediate 10% of the correct value. This may be due to roundoff error, or results to at least four-digit accuracy to minimize roundoff error. on the left 28.00 Your response is within 10% of the correct value. This may be due to roundoff error, or y capacitor intermediate results to at least four-digit accuracy to minimize roundoff error. uc capacitor HF 60 Your response is within 10% the correct may be due to roundoff error, or yo intermediate results to at least four-digit accuracy to error, Hc on the 6.00 uF Your response is within 10% of the correct value. This may be due to roundoff error, or yo intermediate results to at least four digit accuracy to minimize roundoff error. UC (c) the potential difference across each capacitor on the right 28.00 ws Your response is within 10% of the correct value. This may be due to roundoff error, or you intermediate o at least four digit accuracy to minimize roundoff error. V

Explanation / Answer

(a) given that C1=28uF

C2=22uF

V=9V

firstly we solve the circut &find the Ceq.

6uF&C2 are in prallel combination. so

Cp=(6*10^(-6))+(22*10^(-6)) F

Cp=28 uF

now Cp &both C1 are in series combination .so

1/Ceq=(1/C1)+(1/Cp)+(1/C1)

1/Ceq=(1/28 uF)+(1/28 uF)+(1/28 uF)

Ceq=28/3 uF

Ceq= 9.33 uF

we know Q=C*V

Q=(9.33*10^(-6))*9=84 uC

(c)

voltage on right 28uF capacitor

V1= (84*10^(-6))/(28*10^(-6))

V1=3 V

voltage on left 28uF capacitor

V2= (84*10^(-6))/(28*10^(-6))

V2=3 V

remaining voltage in circuit

Vr=V-(V1+V2)

Vr=9-(3+3)

Vr=3V

now this Vr distributed in 6uF & C2

voltage in 6uF capacitor

V3=3 V

voltage in C2

V4=3 V

(b)

charge on right 28uF cpacitor

Q1=(28*10^(-6))*3

Q1=84 uC

charge on left 28uF capacitor

Q2=84uC

charge on 6uF capacitor

Q3=(6*10^(-6))*3

Q3=18uC

charge on capacitor C2

Q4= (22*10^(-6))*3

Q4=66 uC

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