The initial energy stored in the inductor in the following circuit is 36 mA. At t=0, a dc current source of 24 mA is applied to the circuit. The value of resistor is 1000 N. Find i for t > 0.ict = 0,25 nF25 mH RO iL (t) = (24 - 24e-32,000f cos 24, 000t – 32e-32.000t sin 24, 000t) mA, t 20O iL(t) = (24 + 12e 20.000t cos 34, 641t + 6.92e-20,000t sin 34, 641t) mA, t >0O iL (t) = (24 + 12e-20,000t + 11lte-20,000t) mA, t 2 0O iL (t) = (24 –- 2.2 x 10 te-40,000f + 5e-40,000t) mA, t 20

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The initial energy stored in the inductor in the following circuit is 36 mA. At t-0, a dc current source of 24 mA is applied to the circuit. The value of resistor is 1000 2. Find i for t > 0. iR ic 25 nF 325 mH R t= 0, I O iL (t) = (24 - 24e-32.000t cos 24, 000t – 32e-32.0004 sin 24, 000t) mA, t 20 O iL (t) = (24 + 12e-20,000t cos 34, 641t + 6.92e-20,0001 sin 34, 641t) mA, t 20 O iL (t) = (24 + 12e-20,000 + 11te 20,0004) mA, t 20 O iL (t) = (24 – 2.2 x 10 te-40,000t + 5e 40.000t ) mA, t 20

The initial energy stored in the inductor in the following circuit is 36 mA. At t-0, a dc current source of 24 mA is applied to the circuit. The value of resistor is 1000 2. Find i for t > 0. iR ic 25 nF 325 mH R t= 0, I O iL (t) = (24 - 24e-32.000t cos 24, 000t – 32e-32.0004 sin 24, 000t) mA, t 20 O iL (t) = (24 + 12e-20,000t cos 34, 641t + 6.92e-20,0001 sin 34, 641t) mA, t 20 O iL (t) = (24 + 12e-20,000 + 11te 20,0004) mA, t 20 O iL (t) = (24 – 2.2 x 10 te-40,000t + 5e 40.000t ) mA, t 20

Delmar's Standard Textbook Of Electricity

7th Edition

ISBN:9781337900348

Author:Stephen L. Herman

Publisher:Stephen L. Herman

Chapter24: Resistive-inductive-capacitive Parallel Circuits

Section: Chapter Questions

Problem 1RQ: An AC circuit contains a 24 resistor, a 15.9-mH inductor, and a 13.3F capacitor connected in...

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