You monitor the voltage difference across a capacitor in an RC circuit as time p

Question

You monitor the voltage difference across a capacitor in an RC circuit as time passes and find the following results.

Use the exact values you enter to make later calculations You monitor the voltage difference across a capacitor in an RC circuit as time passes and find the following results. Time when V = 0 Time when V= (0.63 Vmax)-5.50 volts 0.070 s 0.160 s (a) If the equivalent resistance of your circuit is 200.0 , calculate the capacitance of the circuit. (b) Using this capacitance in your calculation, find the charge on the capacitor when it is fully charged.

Explanation / Answer

a) let T is the time constant of the circuit.

we know, at time t = 0.16 - 0.07 = 0.09 s

V = Vmax*(1 - e^(-t/T))

0.63*Vmax = Vmax*(1 - e^(-0.09/T))

0.63 = 1 - e^(-0.09/T)

e^(-0.09/T) = 1 - 0.63

-0.09/T = ln(1 - 0.63)

T = -0.09/ln(1 - 0.63)

= 0.0905 s

we know,

T = R*C

C = T/R

= 0.0905/200

= 4.53*10^-4 F

b) Vmax = 5.5/0.63 = 8.73 V

Q_max = C*Vmax

= 4.53*10^-4*8.73

= 3.95*10^-3 C

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