The circuits (a) and (b) below caontain capacitors that can be discharged. Answe

Question

The circuits (a) and (b) below caontain capacitors that can be discharged. Answer the following questions for both capacitors.

(a) When the capacitor is fully charged, the battery is removed from the circuit. What is the charge on the capacitor as a function of time as the capacitor is discharging? Assume that the battery is removed at time t = 0.

(b) What is the current through the resistor as a function of time as the capacitor is discharging?

(c) At what time when the capacitor is discharging will the current through the resistor be 1 A?

                         (a)   

                       (b)

Explanation / Answer

After a long time, the capacitor will be charged to the battery voltage. When you remove the battery, the capacitor will act like a voltage source.

Discharging equation: Vc= V0 * e-t/RC and IR=I0*e-t/RC with the initial current I0=V0/R

Now, plug in all the numbers.

For the first circuit:

a) Vc = 9 e-t/5 * 2F = 9e-100000t

b) IR = 9/5*e-100000t = 1.8e-100000t

c) IR = 10-6A = 1.8e-100000t

>> t = ln(10-6/1.8) / (-100000)= 14.4 milisec

Similarly, for the second circuit:

a) Vc = 1.5 e-t/4 * 2F = 1.5e-125000t

b) IR = 1.5/4*e-125000t = 0.375e-125000t

c) IR = 10-6A = 0.378e-125000t

>> t = ln(10-6/0.378) / (-125000)= 10.3 milisec

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