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I have this example below from "Fundamentals of Electric Circuits", example 7.11.

schematic

simulate this circuit – Schematic created using CircuitLab

The question goes like:

The switch has been closed for a long time and is opened at t=0. Find i and v for all time.

What I want to ask is about the second part, which is t>0.

For t>0, the circuit becomes:

schematic

simulate this circuit

From this circuit above we find V(∞) to be 20V. We have V(0) as 10V. Inserting in the equation, we get :

$$ V(t) = 20-10e^{-0.6t} V $$

Then the solution says, to obtain i, we notice from the figure above that i is the sum of the currents through the 20 ohms and the capacitor that is,

$$ i = \frac{V}{20} + C\frac{dV}{dt} $$

Then from this we get i to be:

$$ i(t) = 1+e^{-0.6t} A$$

My question is that,

  • Is there any other way to find the i?
  • If I would consider the capacitor to be an open circuit and then calculate the i by dividing 30V with 30ohms, would I be wrong?
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  • \$\begingroup\$ V1 is marked 30u(t); is this a dependent source? \$\endgroup\$ – Hearth May 8 '18 at 15:03
  • \$\begingroup\$ @Felthry, no, that is an independent source. u(t) is indicating a unit step function, which has the definition, when t < 0 30u(t) = 0 V and when t > 0 30u(t) = 30V. So when t < 0 you can consider that source to be a short circuit and contributing nothing to overall voltage. \$\endgroup\$ – André Yuhai May 8 '18 at 15:11
  • \$\begingroup\$ Ah, I see. The connection I made was that u is often used as a symbol for voltage alongside v, but the unit step function makes more sense. \$\endgroup\$ – Hearth May 8 '18 at 15:14
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Is there any other way to find the i?

I would personally prefer to get \$i(t)\$ like so:

$$ i(t) = \frac{V_s(t) - v(t)}{R_1} = \frac{30 - 20+10e^{-.6t}}{10}u(t) = (1+e^{-.6t})u(t) $$

Current \$i(t)\$ is the current through \$R_1\$, which is given by Ohm's law from the voltage accross it \$V_s(t) - v(t)\$.

If I would consider the capacitor to be an open circuit and then calculate the i by dividing 30V with 30ohms, would I be wrong?

Yes, you would be wrong. The current varies with time until the capacitor voltage settles. This method would result in a current that is constant.

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  • \$\begingroup\$ Exactly! That was the answer I was looking for. I just couldn't get my mind around it. I was trying to sum the two voltage sources, but then I figured the two voltage sources are not in series so that I thought it was not possible to sum them up. But now, if I am not mistaken, I figured from your solution that you just used the voltage across the 20 ohm resistor which is equal to the capacitor voltage anyway. About the second question, so dividing 30 V by 30 ohms would give me the current when the capacitor is fully charged right? \$\endgroup\$ – André Yuhai May 8 '18 at 15:56
  • \$\begingroup\$ @AndréYuhai Yes, exactly as you put it. \$\endgroup\$ – Vicente Cunha May 8 '18 at 16:00

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