# How do I calculate the current in this circuit using Laplace transforms, with a time shift?

In this exercise I have to calculate i(t) and the switch occurs when t=0.01 s, I tried doing it using Laplace but I always get the wrong answer. I did an exercise that is exactly like this one except the switch occurs when t=0 s and I got the correct answer, but I can't seem to be able to solve with this time shift. I know that the initial condition for the voltage in the capacitor is V1(0.01). I solved the exercise exactly the same way but with a different initial condition and got the wrong answer.

– Chu
Jun 23, 2018 at 7:46
• @Chu I have updated the question with my attempt Jun 23, 2018 at 14:34
• If it makes it easier, say switch closes at t=0, and offset the voltages. Jun 23, 2018 at 15:41
• @Andrés the thing is I can't find my error Jun 23, 2018 at 15:49
• What's the answer given by the book? Jun 23, 2018 at 17:45

Not an answer! (This should be a comment but I don't have enough points). And the first part of the comment is a question:

Are you to assume that v1 has been active for a long time, so that the voltages on the inductor and capacitor have reached their steady state sinusoidal values? Or are you to assume that all voltages and currents were zero for all values of negative time?

If the second case, you must include both transient and steady state components to determine the cap voltage at t=.01. Too much work for me. If you are to assume the steady state sinusoidal situation, I give some rough calculations (using phasors, not Laplace, sorry)

zC = 1/jωC = 1/(j*2*π*100*.001) = -j*1.59 Ω

zL = jωL = j*2*π*100*.1 = j*62.8 Ω

zLC = 1/(jωC + 1/jωL) = -j*1.57 Ω

zRLC = 10.12 Ω /_-8.9° (-0.156 radians)

iR = (14.14V/10.12Ω) cos(100t + .156)

vC = v1 - vR

for t = .01, iR = +0.563A, so vR = 5.63V

vC = 7.64V - 5.63V = 2.01V

So I'm saying that (if we assume v1 has been steady state for a long time) the voltage on the cap just before the switch changes would be 2.01V. I didn't calculate it for the other case but I would not expect the full 7.64V across the cap. You might take another look at the value you calculated to be infinite.

 Good catch and apologies Mr. Pina. Revision below. So the LC is resonant at 100 rad/sec. What a strange analysis problem.

zC = 1/jωC = 1/(j*100*.001) = -j*10 Ω
zL = jωL = j*100*.1 = j*10 Ω
zLC = 1/(jωC + 1/jωL) = 1/0 Ω
zRLC = infinite
iR = zero
vC = v1 - vR

for t = .01, iR = 0A, so vR = 0V
vC = 7.64V - 0 = 7.64V


I cannot review your Laplace analysis but I believe that (for first case above) the textbook solution is correct. The instantaneous current just after the switch is thrown should be

iC = (15.28V - 7.64V)/15Ω = 0.509A


If I understand, t' = 0 just after the switch is thrown? For t' = 0 that reduces to

0.509A = 1.56A*cos(91°) + transient(t'=0)
0.509A = -0.027A + transient(t'=0)

• in the cosine function the 100 value is already the angular frequency so there is no need to multiply it by 2pi Jun 24, 2018 at 16:31

I can’t see a transient analysis for $\small 0\le t \le 0.01$ to determine the capacitor voltage at $\small t=0.01$. Note, the system is not at (sinusoidal) steady state at $\small t=0.01$, so $s\rightarrow j\omega$ doesn’t apply.

• Why isn't the system at steady state? Jun 24, 2018 at 16:32
• $\small 10 \sqrt {2} cos (100t)$ is applied at t=0, hence there’s a transient. What would you have done if a step were applied at t=0 instead of a cosine?
– Chu
Jun 24, 2018 at 18:59
• But nothing happens at t=0, and since for t<0.01s it is an open circuit the voltage across the capacitor is v1(0.01). I don't think I understan where you are getting at Jun 24, 2018 at 20:41
• Can you translate the full question - an initial condition does not make any sense.
– Chu
Jun 24, 2018 at 21:50
• The question is, assuming that at t<0.01 s the switch is connected to node 1 for a long time, and so we can assume it has reached a steady state. Calculate i(t) for t>0.01s. Jun 24, 2018 at 22:15