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In the circuit shown in the following figure, C = 6.00 muF, e = 26.0 V, and the

ID: 1443964 • Letter: I

Question

In the circuit shown in the following figure, C = 6.00 muF, e = 26.0 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 4.00 ms, the charge on the capacitor is measured to be 110 |jC. 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)   charge on capacitor after long time = 6 * 26

                                                                 = 156 microC

                                                                  = 156 * 10-6 C

b)     Here,    110 = 156 * (1 - e-4/RC )

            =>   0.7051 = (1 - e-4/RC )

        => RC = 3.275 msec

    =>   R =   545.83 ohm       --------------> value of resistance R

c)      Here, 0.99 =   (1 - e-t/3.275)

          =>    e-t/3.275    = 0.01

          =>     t = 15.082 msec

                      = 0.01508   sec               ---------------->   time after 99 % of its final value .