Question

helpppp

5. A 250 Q resistor is in series with 200 uF capacitor. These are connected across a 20 V battery. The switch is turned on, and the capacitor begins to fill. After the capacitor is completely filled the battery is taken out of the circuit and the leads to the battery are shorted together. (a) (6 pts) What is the time constant, the maximum current, and the maximum charge on the capacitor? (b) (6 pts) When t = 0.04 seconds, find the current through the resistor, the voltage on the capacitor, the voltage on the resistor, and the charge on the capacitor. (c) (8 pts) When the current through the resistor is 0.02A, find t. (d) (5 pts) When the current is 0.02 A find the voltage and charge on the capacitor.

Answer 1

a) Time constant, T = R*C

= 250*200*10^-6

= 0.05 s

maximum current, I_max = V/R

= 20/250

= 0.080 A

maximum charge on the capacitor, Q_max = C*V

= 200*10^-6*20

= 4.0*10^-3 C

b) at t = 0.04 s

I = I_max*e^(-t/T)

= 0.08*e^(-0.04/0.05)

= 0.0359 A

Voltage across the capacitor,

Vc = V_max*e^(-t/T)

= 20*e^(-0.04/0.05)

= 8.99 V

charge on the capacitor, Qc = Qmax*e^(-t/T)

= 4*10^-3*e^(-0.04/0.05)

= 1.80*10^-3 C

c) use, I = I_max*e^(-t/T)

0.02 = 0.08*e^(-t/0.05)

==> t = 0.069 s

d) at t = 0.069 s

Voltage across the capacitor,

Vc = V_max*e^(-t/T)

= 20*e^(-0.069/0.05)

= 5.03 V

charge on the capacitor, Qc = Qmax*e^(-t/T)

= 4*10^-3*e^(-0.069/0.05)

= 1.01*10^-3 C