For the circuit diagram as Figure 4 of Assignment 7 epsilon

For the circuit diagram as Figure 4 of Assignment 7, epsilon = 15 kV, R = 4.5 k Ohm, and C = 7800.0 mu F. Determine the time constant for the circuit. 3.51 Times 10^(1)s. 2.58 Times 10^(-1)s. 1.61 Times 10^(-3)s. 7.51 Times 10^(-6)s. 1.54 Times 10^(2)s. Determine the maximum charge that will appear on the capacitor. 1.51 Times 10^(-2)C. 2.58 Times 10^(2)C. 2.61 Times 10^(-9)C. Assuming that the capacitor is initially uncharged, how long will it take for the capacitor to charge to half the maximum charge? 1.51 Times 10^(-1)s. 35.8 s. 24.3 s. 7.51 Times 10^(-2)s.

Solution

Emf E = 15 kV = 15000 volt

Resistance R = 4.5 k ohm = 4500 ohm

Capacitance C = 7800 x10 ^-6 F

(a).Time constant T = RC

                            = 4500(7800 x10 ^-6 )

                            = 35.1 s

i.e., option(a) is correct

(b).Maximum charge appear on the capacitor Q = CE

                                                                      = 7800 x10 ^-6 x 15000

                                                                      = 117 coulomb

i.e., option(e) is correct

(c). charge q = Q [1- e ^-t/T]

             Q/2 = Q [1- e ^-t/T]

              1/2 = [1- e ^-t/T]

          [e ^-t/T] = 1-(1/2) =0.5

            -t/T = ln(0.5)

                  = -0.6931

                t = 0.6931 T

                  = 0.6931(35.1)

                  = 24.3 s