12:04 PM ? portal-mail.rcgc.edu .\'Il Sprint? 4. Consider a series RC circuit (s

Question

12:04 PM ? portal-mail.rcgc.edu .'Il Sprint? 4. Consider a series RC circuit (see Figure) for which R- 2.00 ??, C-5.00pE and E-100 V. Assume that the capacitor is initially free of charge. (a) Find the charge on the capacitor two time constants after the switch is closed and express it in coulombs as well as a percentage of the max charge on the capacitor. (b) Determine the power delivered to the resistor at that time (two time constants after the switch is closed). Multiple choice questions 1. The electrons in the beam of a TV set initially move horizontally from west to east. The vertical component of the earth's magnetic field points down, toward the surface of the earth. In what direction are the electrons deflected by this field component? a. due south b. due west c. due east d, due north 2. The rectangular loop is placed in a uniform magnetic field directed along the ty axis (see figure). The loop consists of N turns and carries a current I in the direction indicated. The loop is free to rotate about the z axis. Initially, the loop makes an angle 0 with y axis as the drawing shows. State what will happen to the angle 0: a. it will increase b. it will decrease c, it will stay the same d. it will decrease until it reaches 0 and then it will start to increase

Explanation / Answer

Given,

R = 2 x 10^6 Ohm ; C = 5 x 10^-6 F ; e = 100 V

(a)We know that,

Q = Qm (1 - e^-t/tau)

t = 2 tau

Q = 5 x 10^-6 x 100 (1 - e-2tau/tau) = 432 x 10^-6 C

Hence, Q = 432 x 10^-6 C

Q/Qm = 432 uC/500 uC x 100 = 86.4 %

Hence, percentage = 86.4 %

b)We know that

I = Vb/R e^-t/tau

I = 100/2 x 10^6 e^-2 = 6.77 x 10^-6 A

P = I^2 R = (6.77 x 10^-6)^2 x 2 x 10^6 = 9.17 x 10^-5 W

Hence, P = 9.17 x 10^-5 W

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