In the circuit shown in the figure (Figure 1), C = 5.90 muF, epsilon = 28.0 V, a
ID: 1416016 • Letter: I
Question
In the circuit shown in the figure (Figure 1), C = 5.90 muF, epsilon = 28.0 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. What will be the charge on 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? Part 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 ?Explanation / Answer
part A
the charge on the capacitor after a long time is maximum
Q = C * V
Q = 5.9 * 10^-6 * 28
Q = 165.2 * 10 ^-6 C
part B
t = 0.003 s
Q' = 110 * 10^-6 C
we have relation
Q' = Q ( 1 - e^(-t/Rc) )
110 = 165.2 ( 1 - e^(-0.003/RC) )
R = 463.85 ohm
part C
Q' = 0.99 * Q
Q' = Q ( 1 - e^(-t/Rc) )
t = 12.6 ms