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Series R-L with sinusoidal source: particular solution

Short answer

The particular solution is the steady-state sinusoidal response. Using phasors, impedance magnitude is 100.02 Ohms and angle 88.85 degrees. Current magnitude is 0.09998 A, lagging voltage by 88.85 degrees. i_p(t) = 0.09998 sin(10^4 t - 58.85 degrees) A.

In a series R-L circuit the applied voltage is v(t) = 10 sin(10⁴t + π/6) with R = 2 Ω, L = 0.01 H. v(t) is applied at t = 0. Obtain the particular solution for current i(t) through the circuit. Assume zero initial current through the inductor. [Use classical method].

Given: v(t) = 10 sin(10⁴t + π/6), R = 2 Ω, L = 0.01 H, i(0) = 0.

Differential equation

L di/dt + R i = v(t)
0.01 di/dt + 2 i = 10 sin(10⁴ t + 30°)

Particular solution (steady-state phasor method)

X_L = ωL = 10⁴ × 0.01 = 100 Ω
Z = R + jX_L = 2 + j100,   |Z| = 100.02 Ω,   θ = tan⁻¹(100/2) = 88.85°
I_m = V_m/|Z| = 10/100.02 = 0.09998 A

The current lags the voltage by θ:

i_p(t) = 0.09998 sin(10⁴ t + 30° − 88.85°)
       = 0.09998 sin(10⁴ t − 58.85°) A

(0.09998 ≈ 0.1 A.)

Check by substitution: take i_p = I sin(ωt + β) with I = 0.09998 and β = −1.0272 rad. Then L ω I cos(…) + R I sin(…) reproduces 10 sin(ωt + 30°) ✓.

Complete solution (for completeness)

Complementary solution: s = −R/L = −200, so i_c = K e^(−200t).

i(0) = 0:  0.09998 sin(−58.85°) + K = 0  →  K = 0.0856
i(t) = 0.09998 sin(10⁴ t − 58.85°) + 0.0856 e^(−200t) A

This was verified numerically (for example i(1 ms) = 0.1137 A).

Answer: particular solution i_p(t) = 0.09998 sin(10⁴t − 58.85°) A ≈ 0.1 sin(10⁴t − 58.85°) A.

✓ Answer checked by Claude Opus 5.5 (high effort)

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