In the circuit shown in the figure (Figure 1).C = 5.90 mu F, epsilon = 28.0 V, a
ID: 1621075 • Letter: I
Question
In the circuit shown in the figure (Figure 1).C = 5.90 mu F, epsilon = 28.0 V, and the emf has the negligible resistance. Initially the capacitor is uncharged and the switch S is in position 1. The switch is moved to position 2, so that the capacitor begins to charge. What will be the charge the capacitor a long time after the switch is moved to position 2? After the switch has been in position 2 for 3.00 ms, the charge on the capacitor is measured to be 110 mu C. What is the value of the resistance R? How long after the switch is moved to position 2 will the charge on the capacitor be equal to 99.0% of the final value found in part A?Explanation / Answer
(a)
Q = CV = 5.9 * 10^-6 ( 28) = 1.65 * 10^-4 C
(B)
q= Q ( 1- e^-t/RC)
e^-t/RC = 1- q/Q
R = - t/ C ln ( 1- q/Q)
R = - 3* 10^-3 s/ 5.9 * 10^-6 ln ( 1- 110/165)
=463 ohm
(c)
t= -RC ln ( 1- q/Q)
= - 463 ( 5.9 * 10^-6) ln ( 0.01)
=0.0126 s