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) = 5.40volts 0.064 s 0.163 s (a) If the equivalent resistance of yourcircuit is 360.0 , calculate thecapacitance of the circuit.
C = 1

(b) Using this capacitance in your calculation, find the charge onthe capacitor when it is fully charged.
Q = 2 Time when V = 0 Time when V =(0.63Vmax) = 5.40volts 0.064 s 0.163 s (a) If the equivalent resistance of yourcircuit is 360.0 , calculate thecapacitance of the circuit.
C = 1

(b) Using this capacitance in your calculation, find the charge onthe capacitor when it is fully charged.
Q = 2 Time when V = 0 Time when V =(0.63Vmax) = 5.40volts 0.064 s 0.163 s

Explanation / Answer

Time constant   T = 0.163 s – 0.064 s = 0.099s

Since the time taken to reach the voltage from 0 to 0.63 V maxis called time constant

(a). Resistance R = 360 ohm

We know  T = RC

From this capacitance C = T / R

                                        = 2.75 * 10 ^ -4 F

(b). ).63 V max = 5.4 V

From this V max = 5.4 V / 0.63

                            = 8.571 V

So, Required charge   Q = V max * C

                                       = 2.357 * 10 ^ -3 Coulomb

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