You monitor the voltage difference across a capacitor in an RCcircuit as time pa

Question

You monitor the voltage difference across a capacitor in an RCcircuit as time passes and find the following results:
Time when V = 0 Time when V =(0.63Vmax) = 5.22volts 0.082 s 0.11 s (a) If the equivalent resistance of yourcircuit is 450.0 , calculate thecapacitance of the circuit.
C =

(b) Using this capacitance in your calculation, find the charge onthe capacitor when it is fully charged.
Q =
Time when V = 0 Time when V =(0.63Vmax) = 5.22volts 0.082 s 0.11 s (a) If the equivalent resistance of yourcircuit is 450.0 , calculate thecapacitance of the circuit.
C =

(b) Using this capacitance in your calculation, find the charge onthe capacitor when it is fully charged.
Q = Time when V = 0 Time when V =(0.63Vmax) = 5.22volts 0.082 s 0.11 s

Explanation / Answer

Time taken to develop 0.63 V max voltage is t = 0.11s– 0.082 s= 0.028 s

Time constant T = t = 0.028 s

We know T = RC

From this capacitance C = T / R

Where R = 450 ohm

Plug the values weget   C = 6.222* 10 ^ -5 F

(b). Maximum charge Q = CVmax

Where   V max = V / 0.63 = 5.22 /0.63

                        = 8.285 V

So. Q = 5.153 *10 ^ -4 C

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