An RC circuit, hooked up to a battery as shown in the figure, starts with an unc
ID: 1321307 • Letter: A
Question
An RC circuit, hooked up to a battery as shown in the figure, starts with an uncharged capacitor. The resistance in the circuit is R = 633.0 ? the capacitor has capacitance of C = 47.0 ?F and the battery maintains the emf of ? = 31.0 V. The switch is closed at time t = 0.0 s and the capacitor begins to charge.
What is the time constant for this circuit?
What is the charge on the capacitor after the switch has been closed for t = 4.0810-2 s?
What is the current through the circuit after the switch has been closed for t = 4.0810-2 s?
What is the voltage across the capacitor after the switch has been closed for t = 4.0810-2 s?
An RC circuit, hooked up to a battery as shown in the figure, starts with an uncharged capacitor. The resistance in the circuit is R = 633.0 ? the capacitor has capacitance of C = 47.0 ?F and the battery maintains the emf of ? = 31.0 V. The switch is closed at time t = 0.0 s and the capacitor begins to charge. What is the time constant for this circuit? What is the charge on the capacitor after the switch has been closed for t = 4.08½10-2 s? What is the current through the circuit after the switch has been closed for t = 4.08½10-2 s? What is the voltage across the capacitor after the switch has been closed for t = 4.08½10-2 s?Explanation / Answer
Maximum Current
Imax=E/R =31/633 =0.04897 A
Maximum Charge
Qmax=CE =(47*10-6)*31=1.457*10-3 C
a)
Time Constant
T=RC=633*(47*10-6)
T=0.02975 s or 29.75 ms
b)
Charge on the capacitor as a function of time is given by
Q=Qmax[1-e-t/T] =(1.475*10-3)[1-e-0.0408/0.02975]
Q=1.1*10-3 C or 1.1 mC
c)
Current in a RC circuit as a function of time is given by
I=Imaxe-t/T =(0.04897)*e-0.0408/0.02975
I=0.01243 A or 12.43 mA
d)
Voltage across capacitor is
V=Vmax[1-e-t/T]=31[1-e-0.0408/0.02975]
V=23.13 Volts
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