In the circuit shown in the following figure, C = 6.25 F, = 29.0 V, and the emf
ID: 1635546 • Letter: I
Question
In the circuit shown in the following figure, C = 6.25 F, = 29.0 V, and the emf has negligible resistance. Initially the capacitor is uncharged and the switch S is in position 1. The switch is then moved to position 2, so that the capacitor begins to charge. Switch S in position 1 Switch S in position 2 (a) What will be the charge on the capacitor a long time after the switch is moved to position 2? (b) After the switch has been in position 2 for 4.00 ms, the charge on the capacitor is measured to be 110 C, what is the value of the resistance R? (c) 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] After a long time, the charge on the capacitor will be:
Q = CV = 6.25 x 10-6 x 29 = 181.25 x 10-6 C
b] The charge on the capacitor while discharging will be:
Q = Qoe-t/RC
110 x 10-6 = 181.25 x 10-6 e-0.004/RC
=> 0.6068 = e-0.004/RC
taking ln on both sides gives
ln(0.068) = -0.004/RC
substitute for C to get:
R = 1281.546 ohms.
c]
0.99 = e-t/RC
take ln on both sides
ln(0.99) = - t/RC
=> t = 8.05 x 10-5s .