Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

For the circuit diagram as Figure 4 of Assignment 7, sigma = 15 kV, R = 4.5 k Oh

ID: 1456016 • Letter: F

Question

For the circuit diagram as Figure 4 of Assignment 7, sigma = 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^(-5)C. 2.61 Times 10^(2)C. 7.51 Times 10^(-9)C. 1.17 Times 10^(2)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. 35.8s. 24.3 s. 7.51 Times 10^(-2)s. 15.4s. A point mass consisting of 34 charged panicles travels at a velocity of magnitude 4.80 Times 10^(-1) m/s through a uniform magnetic field of magnitude 4.8 T. Each charged particle has a charge of -12e and a mass of 3 60 Times 10^29 kg. The angle between the

Explanation / Answer

5) a) time constant = RC

                                   = 4.5 * 103 * 7800 * 10-6   = 35.1 sec

      => option a) is correct .

   b)   maximum charge = 7800 * 10-6 * 15 * 103 =   117 C

=> option e) is correct .

c)    Here, 1/2 = (1 - e-t/35.1)

          => t = 24.3 sec     

=>    option c) is correct .