A parallel plat capacitor with C= 20 F is being charged by a battery of =12 V. I

Question

A parallel plat capacitor with C= 20 F is being charged by a battery of =12 V. If total resistance of the circuit is 5.2 k, then:

a) How long does it take for the capacitor to charge to 90% of fully charged?

b) Draw the current profile and determine the current at this time

c) How much energy is stored in the capacitor by this time?

d) How much energy is provided by the battery by this time?

e) To what value should we change the resistance so that the time in part (a) is reduced by 50%?

f) What are the E and the B field values and directions between the plates of the capacitor by this time?

Explanation / Answer

a)

we know that

Q = Qmax ( 1- e^-t/RC )

Q = 0.9 Qmax

so

0.1 = e^-t/RC

substituting we get

t = 0.239 s

b)

we know that

for current

i = imax e^-t/RC

so from above problem

e^-t/Rc = 0.1

so

i = 0.1 imax

i = 0.1 (12/5200)

i = 2.3 x 10^-4 A

c)

energy stored in capacitors

energy = 0.5 Q^2/2C

Q = 0.9 ( 20 x12 uC)

Q = 216 uC

E = 1166.4 J

d)

work done by batter

W = 2E

W = 2332.8 J

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