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I was trying to do some first order circuits and there was this one that kind of confused me. Correct me if I'm wrong but if $$V_s=u(t)V$$ this means that for t<0 Vs=0 and for t>0 Vs=1 right?

So therefore my question. For t<0 since technically there is no voltage generator does that mean that the current ix=0 and therefore the dependent source 2ix=0? So in total there is no current whatsoever in the circuit? which means the capacitor is not charged initially?

I am not searching for someone to solve this. I'm just asking for some direction.

Picture of the circuit

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  • \$\begingroup\$ I think it is safe to assume that for t < 0 all voltages and current are equal to 0. \$\endgroup\$
    – mooshoomatt
    Commented May 6, 2022 at 0:22
  • \$\begingroup\$ Ani Lici avata, since the dependent current source is always just twice what's flowing through the \$2\:\Omega\$ resistor, couldn't you just remove the dependent current source (cut it out of the circuit) and just place an added \$1\:\Omega\$ in parallel with the \$2\:\Omega\$ resistor and have exactly the same effect? \$\endgroup\$
    – jonk
    Commented May 6, 2022 at 0:37
  • \$\begingroup\$ You can't assume that a voltage source of 0V has no current flowing through it. For an ideal voltage source there is no fixed relationship between the voltage across the source and the current through it...you must solve the entire circuit to determine the current. \$\endgroup\$
    – Elliot Alderson
    Commented May 6, 2022 at 9:43
  • \$\begingroup\$ You can find the thevenin equivalent of the rest of the circuit and then solve a simple RC circuit. \$\endgroup\$
    – Miss Mulan
    Commented May 6, 2022 at 11:12
  • \$\begingroup\$ Ani Lici, Almost all of your question is well-spoken and arrives at the right conclusion. The only remaining thing seems to be that you are looking for some help about how to proceed starting at \$t_{0+}\$ and for \$t\gt 0\:\text{s}\$. For that, you offer nothing to us about what tools you already know that might be of some use here (except you don't know how.) What tools do you know that you feel might be needed? Or are you okay, now? Or do you feel you need something more? (I'm tentatively assuming you are supposed to solve this using diff-eq and KCL.) \$\endgroup\$
    – jonk
    Commented May 8, 2022 at 17:26

2 Answers 2

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The current through \$\small R\$ and \$\small C\$ is \$\small3i_x\$, so the differential equation relating \$v_s\$, \$v_u\$ and \$i_x\$ can be derived, e.g. via nodal analysis. Also, assuming zero initial conditions, the capacitor is a short circuit at \$\small t=0^+\$.

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  • \$\begingroup\$ basically what i was thinking. Just the fact that everything was 0 in the beginning confused me a bit \$\endgroup\$
    – Ani Lici
    Commented May 6, 2022 at 14:10
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The step function is 1 for times greater than 0 and 0 for times less than 0, so it is correct. Therefore for t<0, there is no voltage source, so no current exists. Therefore, the capacitor is assumed to be not charged before t=0. This is due to the fact that the circuit is assumed to be at steady state.

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