An RC circuit, hooked up to a battery as shown in the figure, starts with an unc
ID: 1318650 • 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 = 936.0 ? the capacitor has capacitance of C = 57.0 ?F and the battery maintains the emf of ? = 14.0 V. The switch is closed at time t = 0.0 s and the capacitor begins to charge.
A) What is the time constant for this circuit?
B) What is the charge on the capacitor after the switch has been closed for t = 5.76?10-2 s?
C) What is the current through the circuit after the switch has been closed for t = 5.76?10-2 s?
D) What is the voltage across the capacitor after the switch has been closed for t = 5.76?10-2 s?
Explanation / Answer
A)
time constant=RC
=936*57*10^-6
=0.053352 sec
=53.352 msec ........................
B)
q0=V0*C
=14*57*10^-6 C
=798*10^-6 V
=798 uC
now
q=qo*[1-e^-t/Rc]
=798*10^-6*[1-e^-(5.76*10^-2/53.352*10^-3)]
=798*10^-6*[1-e^-(1.07)]
=798*10^-6*0.66
=526.9 uC .....is answer
c)
i=i0*(e^-t/RC)
=q0/RC*(e^-(5.76*10^-2/53.352*10^-3))
=(798*10^-6/53.352*10^-3)*(0.33)
=0.00493 A
=4.93 mA ..................................
D)
V=q/C
=526.9*10^-6/57*10^-6
=9.24 Volt ...................................
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