A 3.75 ? F capacitor, initially charged to 30.4V , discharges when it is connect

Question

A 3.75?F capacitor, initially charged to 30.4V , discharges when it is connected in series with a resistor.

Part A

How much energy does this capacitor store when fully charged?

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Part B

What is the capacitor's voltage when it has only half of its maximum energy?

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Part C

What resistance is necessary to cause the capacitor to have only 50% of its energy left after 0.5s of discharge?

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Part D

What is the current in the resistor at this time?

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Explanation / Answer

a)

Energy stored in the capacitor is

U = ( 1/ 2) C V^2 = ( 1/2 ) ( 3.75*10^ -6 F ) ( 30.4 V) ^2 = 0.00115 J

b)

If the decreases to half

U /2 = ( 1/2 ) C V^2 ==> V = sqrt ( U / C) = sqrt ( 0.00115 J / 3.75*10^ -6 F) = 17.55 V

c)

charge

Q = CV = ( 3.75*10^ -6 F ) ( 17.55 V ) = 6.58*10^-5 C

Q_o =CV = ( 3.75*10^-6 V ) ( 30.5 V) = 0.115*10^-3 C

e ^ - t / RC = Q / Q_o

-t / RC = 0.558

R = 0.5 s / ( 0.558 ) ( 3.75*10^ -6 )

=2.3895*10^5 ohm

d)

current

I = V / R = 7.323*10^-5 A

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