PLEASE ANSWER ALL 3 PARTS AND SHOW ALL WORK NEATLY! THANK YOU :) Consider the ci

Question

PLEASE ANSWER ALL 3 PARTS AND SHOW ALL WORK NEATLY! THANK YOU :)

Consider the circuit shown on the attached figure where E = 100 V, R1 =3.0k,R2 =4.0k,C1 =5.5F,andC2 =8.5F.Notethatthe charge on both capacitors is initially zero before the switch S is closed.

a) Determine the maximum charge on each capacitor long after the switch has been closed.

b) How long will it take the charge on C1 to be at 0.49 mC?
c) What current will be coming out of the EMF at the time determined

in part b? If you cannot solve part b use t = 200 ms.

Explanation / Answer

a)

We know that

The Charge on the capacitor C1 is (q) =C1V =5.5F*100V =5.5*10-4C =0.55mC

The charge on the capacitor C2 is (q) =C2V =8.5F*100V =8.5*10-4C =0.85mC

b)

Now the time taken for the chage up to 100V by C1 is given by

=5RC =5(R1+R2)C1 =5(3k+4k)*5.5uF=0.1925s

Therefore for q1 0.55mC it takes 0.1925s

ND FOR 0.49mC the time taken is t= 0.49mC*0.1925/0.55 =0.1715 =171.5ms

c)

For q2 qt 100V =5RC =5(3k+4k)*8.5uF =0.2975s

Now Q =0.49mC+171.5mC+0.25mC/0.2975 =0.98mC

Then current I =q/t =0.98mC/171.5m =5.7143mA

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