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Suppose a series RLC circuit in series with a 12 V battery and a switch initially opened. R = 100 Ω, L = 100mH and C = 10uF. L and C are initially discharged.

A t=0 the switch is closed.

I want to know the current equation.

I calculate \$ \alpha = \frac{R}{2L} = \frac{100}{2 \times 100x10^{-3}} = 500 \$

then \$ \omega_0 = \sqrt{\frac{1}{LC}} = = \sqrt{\frac{1}{100 \times 10^{-3} \times 10 \times 10^{-6}}} = 1000 \$

\$ \alpha < \omega_0 \$ , so we are dealing with an underdamped circuit and the current equation has the form

\$ i(t) = e^{- \alpha t} [K_1 \thinspace Cos(\omega_d t) + K_2\thinspace Sen(\omega_d t)] + I_{\infty}\$

I need to determine \$ K_1 \$ and \$ K_2 \$

I apply the first initial conditions. Current at time 0 is 0

\$ i(t=0) = 0 = e^{0} [K_1 \thinspace Cos(0) + K_2\thinspace Sen(0)] + 0\$

0 = [K_1 + 0] + 0 \$

\$ K_1 = 0\$


Second condition, \$ i'(0) = 0 \$

The derivate of the i equation gives me

\$ \frac{V_L}{L} = -\alpha e^{- \alpha t} [K_1 \thinspace Cos(\omega_d t) + K_2\thinspace Sen(\omega_d t)] - \omega_d e^{- \alpha t} \thinspace K_1 \thinspace Sen(\omega_d t) + \omega_d e^{- \alpha t} \thinspace K_2\thinspace Cos(\omega_d t) \$

when I apply t=0 that gives me \$ K_2 = 0\$

With K1 and K2 equal to zero there is no current equation.

How do I get these unknowns?

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If \$K_1 = 0\$ and steady-state current is \$i_{ss}=0\$: $$ i(t) = K_2e^{-\alpha t}sin(\omega t)$$

Then, $$ \frac{di(t)}{dt} = K_2 \left[ -\alpha e^{-\alpha t}sin(\omega t) + \omega cos(\omega t)e^{-\alpha t} \right ]$$

$$ K_2 = \frac{1}{\omega}\frac{di(0_+)}{dt} $$

$$ \frac{di(0_+)}{dt} = \frac{v_{L_{(0+)}}}{L}$$

But \$ v_{L_{(0+)}} = 12 V \$ and \$L=0.1 H\$ and $$K_2 = 0.12 A$$

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  • \$\begingroup\$ I guess you have a typo there on the first uknown. You mean K1 = 0. Anyway, you say that \$ i(t) = K_2e^{-\alpha t}sin(wt) \$, shouldn't it be \$ i(t) = K_1e^{-\alpha t}cos(wt) \$ ? sin(wt) would be zero, because sin(0) = 0 \$\endgroup\$ – MLL Aug 31 '18 at 0:35
  • \$\begingroup\$ I forgot to delete the third line of original text. \$\endgroup\$ – Dirceu Rodrigues Jr Aug 31 '18 at 1:39
  • \$\begingroup\$ @MLL: The statement in your question: "K2 = 0" is not correct. Have you noticed that VL (0+) = 12 V? \$\endgroup\$ – Dirceu Rodrigues Jr Sep 1 '18 at 3:24

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