Overtime, a capacitor that is hooked up to a voltage cource will charge until it
ID: 1833863 • Letter: O
Question
Overtime, a capacitor that is hooked up to a voltage cource will charge until its voltage matches the source voltage. The time this takes is dependent on the capacitance and the resistance of the circuit. Answer the following questions for circuits (a) and (b) in the figures below.
(a) If the battery is connected to the circuit at time t =0, what is the charge on the capacitor as a function of time as the capacitor is charging?
(b) What is the current through the resistor as a function of time as the capacitor is charging?
(c) At what time when the capacitor is charging will the current through the resistor be 1 A?
(a)
(b)
Explanation / Answer
(a) charge as a function of time
Q(t) = Qmax(1-e-t/RC), Qmax = CV
circuit a:
Q(t) = 0.000002F*9V(1-e-t/(4*0.000002F)) = 0.000018(1-e-t/0.000008) = 0.000018(1-e-125,000t)
circuit b:
Q(t) = 0.000003F*9V(1-e-t/(4*0.000003F)) = 0.000027(1-e-t/0.000012) = 0.000027(1-e-83,333.33t)
(b) current as a function of time - for this you differentiate Q(t)
I(t) = (Qmax/RC)e-t/RC = Imaxe-t/RC, Imax = V/R
circuit a:
I(t) = (9V/4)e-t/(4*0.000002F) = 2.25e-t/0.000008 =2.25e-125,000t
circuit b:
I(t) = (9V/4)e-t/(4*0.000003F) = 2.25e-t/0.000012 = 2.25e-83,333.33t
(c) At what time when the capacitor is charging will the current through the resistor be 1 A?
circuit a:
0.000001 = 2.25e-125,000t
0.00000044 = e-125,000t
2,250,000 = e125,000t
ln (2,250,000) = 125,000t = 14.626
t = 14.626/125,000 = 0.000117 = 117uS
circuit b:
0.000001 = 2.25e-83,333.33t
0.00000044 = e-83,333.33t
2,250,000 = e83,333.33t
ln (2,250,000) = 83,333.33t = 14.626
t = 14.626/83,333.33 = 0.0001755 = 175.5uS
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