i am having trouble with the last question. d) what is the voltage left in the c

Question

i am having trouble with the last question. d) what is the voltage left in the capacitor in letter (c)? .
i already solved part a, b, and c. but i dont understand the last question.

a) time constant = .03 sec.

b) maximum charge = 6E-5 C

c) Time need to discharge the capacitor up to 87.0% of its maxinum (charge) value is 4.18E-3 sec.

5. pts) Discha Capacitor: A of capacitance C 5.00 uF is connected in series capacitor with a resist R 00 to a battery of emf what is the time constant? (b) What is the maximum charge stored in the capacitorn (c) What is the time needed to discharge the capacitor eo value? (d) What is the voltage left in the capacitor in up to 87.0 of its maximum (cha letter (c)? T .02 sec Equation (sysolution/Work: Answer: hre Cons 03 sec

Explanation / Answer

c) for part C

for the final charge = 0.87 * Qo

Qo is the maximum charge on the capacitor

0.87 * Qo = Qo * (e^(-t/T))

T = 0.03 s

0.87 = e^(-t/.03)

solving for t

t = 4.18 *10^-3 s

the time period is 4.18 * 10^-3 s

b) for the voltage left in capacitor

V = Vo * e^(-t/T)

V = 12 * e^(-4.18 * 10^-3/.03)

V = 1.56 V

the voltage left in the capacitor in letter C is 1.56 V

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