A 2.70 uF capacitor is charged to 80 uC and then placed in series with a 6.4 M o

Question

A 2.70 uF capacitor is charged to 80 uC and then placed in series with a 6.4 M ohms resistor and an open switch. At time t = 0, the switch is closed and the capacitor starts to discharge.


a) What is the initial voltage across the capacitor?



b) What is the initial current in the circuit?



c) What is the time constant of this circuit?




d) What is the current in the circuit after 9.0 s have passed?




e) What is the charge on the capacitor after 9.0 s have passed?




f) What is the voltage across the resistor at this time?




g) How long until the capacitor has discharged to half of its maximum value?

Explanation / Answer

(a) q= CV ; V= q/C =29.63 V

(b) initial current = dq/dt = q/RC*e^(-t/RC) = q/RC = 4.63 uA

(c) time conctant = 17.28 s

(d)I after 9 sec = 2.75uA

(e)Q= 47,522 UC

(f) V(resistor) = V(capacitor) = q/C = 17.6 V

(g) t= 11.97 s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.