In the RC circuit pictured below, he switch has been closed for a long time, and
ID: 1295297 • Letter: I
Question
In the RC circuit pictured below, he switch has been closed for a long time, and the capacitor in the circuit is fully charged. When switch S is opened, the battery is not part of the circuit anymore and the capacitor starts discharging. The capacitance is C = 0.2 ?F, the resistance is R = 5 ?, and the voltage provided by the battery is ? = 12 V.
17.
What is the initial charge on the capacitor (before opening switch S)?
18.
What is the time constant for this circuit?
19.
What is the initial current through the circuit when switch S is opened?
20.
After how much time does the current reduce to half its initial value?
A. 1.2 ?C B. 2.4 ?C C. 3.6 ?C D. 4.8 ?CExplanation / Answer
17) Vini = E = 12
so Qinitial = Qo = CV = 12 * 0.2 uF = 2.4 uC .............(B)
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18) tau (T) = RC = 5 * 0.2uF= 1 us ........... (A)
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19) Q = Qo e^(-t/T)
=> Q = 2.4 e^(-t/1us) uC
so i(t) = dQ/dt = - 2.4 e^(-t/1us) A
so |i(0)| = 2.4 A .............. (C)
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20)
=> 0.5 = e^(-t/1us)
ln both sides :
=> -0.693 = - t /1*10^-6
=> t = 0.693 x 10^-6 = 0.7 x 10^-6 s approx ................. (A)