Solve LI + RI ' + I /C = E 0 e it ,
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Solve LI∼" + RI∼' + I∼/C = E0eiωt, i = √-1, by substituting lp = Keiωt (K unknown) and its derivatives and taking the real part lp of the solution l∼p. Show agreement with (2), (4). Use (11) eiωt = cos ωt + i sin ωt; cf. Sec. 2.2, and i2 = -1.
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Substitution gives Divide this by e it on both sides and solve the resulting equation algebraically ...View the full answer
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