You have an uncharged capacitor (12 x 10-6 F) and a resistor (1000 ohms) connect
ID: 2186126 • Letter: Y
Question
You have an uncharged capacitor (12 x 10-6 F) and a resistor (1000 ohms) connected to a battery of 200 V at time t=0 (the switch is closed at this time). A. What is the final current (after a very long time)? C. What is the final charge on the capacitor (after a very long time)? D. After the capacitor is fully charged, you disconnect the battery and allow the capacitor to discharge. What is the initial current in the circuit for the discharging capacitor? E. What is the current after 0.005 s of discharging? What is the charge on the capacitor at this time?Explanation / Answer
Applying Kirchoff's loop rule:
V- IR - Q/C = 0
As Q increases I decreases, but Q changes because there is a current I. As the current decreases Q changes more slowly.
I = dQ/dt, so the equation can be written:
V- R (dQ/dt) - Q/C = 0
This is a differential equation that can be solved for Q as a function of time. The solution (derived in the text) is:
Q(t) = Qo [ 1 - e-t/t ]
where Qo = CV and the time constant t = RC.
Differentiating this expression to get the current as a function of time gives:
I(t) = (Qo/RC) e-t/t = Io e-t/t
where Io = V/R is the maximum current possible in the circuit.
What happens if the capacitor is now fully charged and is then discharged through the resistor? Now the potential difference across the resistor is the capacitor voltage, but that decreases (as does the current) as time goes by.
Applying Kirchoff's loop rule:
-IR - Q/C = 0
I = dQ/dt, so the equation can be written:
R (dQ/dt) = -Q/C
This is a differential equation that can be solved for Q as a function of time. The solution is:
Q(t) = Qo e-t/t
where Qo is the initial charge on the capacitor and the time constant t = RC.
Differentiating this expression to get the current as a function of time gives:
I(t) = -(Qo/RC) e-t/t = -Io e-t/t
where Io = Qo/RC
a). final current (after a very long time):
Qo = CV =12*10-6*200 = 2.4 mC
at t = infinity
I = I0 =Q0/(RC) = 2.4*10-3/(1000*12*10-6) = 0.2 Amp
c). final charge on capacitor :
Q = Qo = CV =12*10-6*200 = 2.4 mC
d). initial current in the circuit for the discharging capacitor:
I(t=0) = - I0 = -Q0/(RC) = - 2.4*10-3/(1000*12*10-6) = - 0.2 Amp
e). current after 0.005 s of discharging:
I(t) = I0*e-t/
= RC = 1000*12*10-6 = 12*10-3 s
I(t=0.005 s) = - 0.2*e-0.005/0.012 = -0.132 Amp
-ve sign means current is decreasing.
charge at this time on the capacitor:
Q(t) = Qo e-t/
Q(t=0.005) = 2.4*e-0.005/0.012 = 1.58 mC
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.