(1096) Problem 4: A 1.7 H inductor in an RL circuit (with resistance 2.1 ) is in

Question

(1096) Problem 4: A 1.7 H inductor in an RL circuit (with resistance 2.1 ) is initially fully charged. A switch is then thrown (between positions 1 and 2, see figure), which causes the inductor to discharge. The initial current through this discharging circuit is 4.95 A. Randomized Variables 2 1 = 4.95 A L=1.7H R= 2.1 ©theexpertta.com 33% Part (a) What is the initial energy in the inductor in J? Grade Summa Deductions Potential ind 0% 100% cos0 7 8 9 tan() | acosO Submissions Attempts r (490 per attempt) Sin cotanasin0 emaining: 3 acotanOsinh atan detailed view cotanhO END Degrees Radians Submit Hint I give up! Hints: 4% deduction per hint. Hints remaining: 3 Feedback: 5% deduction per feedback. 33% Part (b) How long will it take the current to decline to 25.00% of its initial value? Express this value in seconds - 33% Part (c) Calculate the average power dissipated in W

Explanation / Answer

Part (a):

Initial energy in the inductor -

U = (1/2)*L*I^2 = 0.5*1.7*4.95^2 = 20.83 J

Part (b):

The expression for the current in the case of R-L Circuit -

I = I0*[1 - e-Rt/L]

As per the condition -

0.25*I0 = I0*[1 - e-2.1t/1.7]

=> 0.25 = 1 - e-1.24t

==> 0.75 = e-1.24t

=> -1.24t = -0.288

=> t = 0.288/1.24 = 0.23 s

Part (c):

RMS Current, I = 4.95 / root(2) = 3.50 A

So, Average power, P = I^2*R = 3.50^2*2.1 = 25.73 W

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