(10%) Problem 4: A 1.25 Hinductor in an RL circuit (with resistance 2.15 2 is in
ID: 1569009 • Letter: #
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(10%) Problem 4: A 1.25 Hinductor in an RL circuit (with resistance 2.15 2 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 4.8 A. Randomized Variables I 4.8 A L 1.25 H R 2.15 S2 Otheexpertta.com A 33% Part (a) What is the initial energy in the inductor in J? Grade Summary Deductions Potential 100% sino cos0 tan0 7 8 9 Submissions Attempts remaining: 3 cotano asin acos0 per attempt) detailed view atan0 acotan0 sinh0 1 2 3 cosh0 tanh0 cotanh Degrees O Radians BACKSPACE Submit Hint I give up Hints 4 deduction per hint. Hints remaining Feedback 5% deduction per feedback. A 33% Part (b) How long will it take the current to decline to 25.00% of its initial value? Express this value in seconds A 33% Part (c) Calculate the average power dissipated in W.Explanation / Answer
a) energy = 0.5 Li^2
= 0.5*1.25*4.8^2
= 14.4 J
b) 1/4 = e^(-tR/L)
- ln 4 = -t *2.15/1.25
t = 1.25/2.15 * ln 4
= 0.806 s
c) average power dissipated
= iavg^2 R
average power = 2.15* (4.8^2+1.2^2)/2
= 26.3 W
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