In the circuit shown in the following figure, C = 5.80 µF, e m f = 27.0 V, and t
ID: 1409415 • Letter: I
Question
In the circuit shown in the following figure, C = 5.80 µF, e m f = 27.0 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 5.00 ms, the charge on the capacitor is measured to be 110 µC. 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
Q=Qo[1-e^(-t/(R*C))]
should read V = Vs[1-e^(-t/(R*C))]
the capcitance is known as 5.8uf , the charge is 110uC
calculate the voltage V = q/C , V = 110uC / 5.8uf
q/C =Vs[1-e^(-t/(R*C))]
plug in q , C and Vs =27V , one can calculate the required.
a)
Q = CV = 5.8*27*10^-6 = 156.6*10^-6 cOULOUMBS
B)
R = -t/C ln ( 1-q/eC)^-1
R = -5*10^-3/5.8*10^-6 * ln ( 1- 110/27*5.8)^-1
R = 711.2 oHMS
C)
1-0.99 = e^(-t/RC)
t = - 711.2 * 5.8*10^-6 ln (0.01)
t = 18.99 ms