Series RLC with 10 sin t source, classical

Given: R = 10 Ω, L = 1 H, C = 1/9 F, v = 10 sin t applied at t = 0; i(0) = 0, v_C(0) = 0.

Differential equation

L di/dt + R i + (1/C)∫i dt = 10 sin t
Differentiate:  d²i/dt² + 10 di/dt + 9 i = 10 cos t

Complementary solution

s² + 10 s + 9 = 0  →  s = −1, −9
i_c = A e^(−t) + B e^(−9t)

(overdamped)

Particular solution (phasor method, ω = 1 rad/s)

Z = R + j(ωL − 1/ωC) = 10 + j(1 − 9) = 10 − j8
|Z| = 12.806 Ω,  angle = −38.66°
I_m = 10/12.806 = 0.7809 A
i_p = 0.7809 sin(t + 38.66°) = 0.7809 sin(t + 0.6747)

Initial conditions

• i(0) = 0.
• At t = 0⁺: L di/dt = v(0) − R i(0) − v_C(0) = 0, so di/dt(0⁺) = 0.

i_p(0) = 0.7809 sin(38.66°) = 0.4878
i_p′(0) = 0.7809 cos(38.66°) = 0.6098

A + B = −0.4878
−A − 9B = −0.6098

Adding: −8B = −1.0976, so B = 0.1372; A = −0.625.

Result

i(t) = 0.7809 sin(t + 38.66°) − 0.625 e^(−t) + 0.1372 e^(−9t) A,   t > 0

Check by substitution: i(0) = 0.4878 − 0.625 + 0.1372 = 0 ✓; i′(0) = 0.6098 + 0.625 − 1.2348 = 0 ✓.

Answer: i(t) = 0.781 sin(t + 38.66°) − 0.625 e^(−t) + 0.137 e^(−9t) A.