5. In the circuit shown in the figure below, C = 6.00 F, = 28.0 V, and the emf h
ID: 1444074 • Letter: 5
Question
5. In the circuit shown in the figure below, C = 6.00 F, = 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. (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 3.00 ms, the charge on the capacitor is measured to be 110 C. What is the value of the resistance R? Switch S in position SwitchS in position 2 (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.
Charge on the capacitor after long time
Q = C*V
Q = 6*10^-6*28
Q = 1.68*10^-4 C
2.
Q = CV*(1 - e^(-t/RC))
t = 3.00 ms = 3*10^-3 sec
Q = 110*10^-6 C
using above equation
R = t/[C*ln(CV/(CV - Q))]
R = 3*10^-3/[6*10^-6*ln[6*10^-6*28/(6*10^-6*28 - 110*10^-6)]]
R = 470.13 ohm
3.
Q = 0.99*Q0 = 0.99*CV
0.99*CV = CV*(1 - e^(-t/RC))
0.99 = 1 - e^(-t/RC)
e^(-t/RC) = 0.01
t = -RC*ln (0.01)
t = -470.13*6*10^-6*ln(0.01)
t = 0.01299 sec
t = 12.99 msec
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