Problem 8: Six parts A resistor with resistance 12 is connected in series with a
ID: 1458806 • Letter: P
Question
Problem 8: Six parts
A resistor with resistance 12 is connected in series with a capacitor with capacitance 1.2µF and a battery with emf 13.0V. Before the switch is closed at t = 0, the capacitor is uncharged. (a) What is the time constant? (b) What fraction of the final charge is on the plates at time t = 44 s? (c) What fraction of the initial current remains at t = 44 s? Suppose the capacitor is originally given a charge of 4.0µC and the discharged by closing the switch at t = 0. (d) At what time will the charge be equal to 0.5µC? (e) What is the current at this time?
Explanation / Answer
1 . time constant T = Rc
T = 12 e6* 1.2 e -6
T = 14.4 secs
-------------------------------------
use the equation Q = Qo *(1- e^-t/RC)
Q/Qo = (1-e ^-44/14.4)
Q/Qo = 95.2 %
------------------------------------------------------
use the current eqn as I = Io e^-t/RC
I/Io = e^-(44/14.4)
I/Io = 4.7%
-----------------------------------------------
Q/Qo = (e^-t//RC)
1-e^-t/Rc = 0.5/4 = 0.125
e^-t/RC = 0.125 = 0.875
-t/RC = ln(0.875)
-t/RC = -1.1333
t = 1.134 * RC
t = 1.134 * 14.4
t = 16.32 secs
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.