I do not know if this is an appropriate place to ask this question here but this is the only website that I am a member of so I hope that it is okay. If not, I can delete it.

This circuit question was brought to me while tutoring yesterday that I was not able to answer physically. Initially the capacitor had been charged to 2V so $v_C(0^−) = 2V$. Before the switch is thrown there is no current through either the resistor or inductor, so $i_L(0^−) = 0A$.

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Since the inductor and capacitor cannot have an instantaneous change in their values just before and after the switch is thrown, these values must remain the same:

$i_L(0^−) = i_L(0^+−) = 0A$, $v_C(0^−) = v_C(0^+) = 2V$

This implies that at the moment the switch is closed, the inductor shorts out the capacitor and resistor, and no current occurs along any branches of the circuit. That is, the voltage drop across all three elements is zero.

What is happening to the capacitor’s and inductor’s current and voltage in the circuit? How can the capacitor have a voltage drop of 2V and yet the inductor appears to short it out simultaneously?


3 Answers 3


Inductors resist changes in current. So in a circuit like the one you describe, for short times after the switch is closed, the inductor acts like a broken wire. This is consistent with the statement you made that there is 2 V across the inductor. For these short times the circuit is essentially reduced (i.e. literally just cut the inductor out of the circuit) to a discharging RC circuit. In the effective RC circuit there will be 2 V across the resistor and capacitor, and a (decreasing) current will flow through the resistor. So far, this is just for short times after the switch is closed.

In general, all 3 elements are in parallel, which is another way of saying that all 3 should have the same potential at all times, so if the capacitor has some voltage (2V or otherwise) across it, so should the resistor and inductor. Now to find the instantaneous current through the resistor and inductor as a function of time one will have to solve a set of differential equations, which is not very difficult, but is likely outside the scope of your original question so I will hold off on that.

  • $\begingroup$ Your answer makes me very happy! My question was in trying to understand the solution from a problem in Dorf-Svoboda on Circuits, which I now realize is incorrect. I absolutely agree and understood everything you wrote, which was beautifully described. Many thanks! $\endgroup$
    – Carlos
    May 8, 2014 at 17:25
  • $\begingroup$ @Carlos - your welcome, glad I could be of assistance! $\endgroup$
    – DJBunk
    May 8, 2014 at 17:33

This implies that at the moment the switch is closed, the inductor shorts out the capacitor and resistor,

Your reasoning is incorrect. Remember, the voltage across the inductor is proportional to the rate of change of current,

$$v_L(t) = L \frac{di_L}{dt} $$

not the instantaneous current. So you can't say that the voltage across the inductor is zero when the inductor current is zero.

Also, the voltage across the capacitor cannot change instantly but the voltage across the resistor and inductor can.

Finally, while the current through an inductor must be continuous, the rate of change of the current does not.

The instant after the switch closes, we have:

$$v_C(0+) = v_R(0+) = v_L(0+) = 2V$$

From this, we can deduce

$$i_R(0+) = \frac{v_R(0+)}{R} = \frac{2V}{0.75 \Omega} $$


$$\frac{d}{dt}i_L(0+) = \frac{v_L(0+)}{L} = \frac{2V}{1H}$$

Thus, the initial conditions for the inductor current are:

$$i_L(0+) = 0A$$

$$\frac{d}{dt}i_L(0+) = 2\frac{A}{s}$$

I assume you already know the general solution to the homogeneous differential equation for $i_L$ and now you can find the particular solution given the above initial conditions.

  • $\begingroup$ +1 for great answer. Please see my comments above with DJBunk. I wish I could also give you credit for answering this question too. $\endgroup$
    – Carlos
    May 8, 2014 at 17:28

No, the inductor and resistor don't "short out" the capacitor.

The current thru the resistor is simply the voltage accross it divided by its resistance. This is instantaneous. The initial current thru the resistor will therefore be (2 V)/(750 mΩ) = 2.67 A.

The current thru the inductor is the integral of the voltage applied to it over time divided by its inductance. The inductor current is therfore initially 0, but will start to build up by (2 V)/(1 H) = 2 A/s.

This circuit is easier to analyze by noting it is just a damped tank circuit, for which is well explained out there, along with lots of ways of deriving the solution. Just the capacitor and inductor alone, assuming ideal components, would be a undamped tank circuit and would oscillate forever. The energy would continually slosh back and forth between being voltage on the capacitor and current thru the inductor. The result is a sine for both signals at the resonant frequency, which is 876 Hz in this case.

The resistor dissipates some of the energy every cycle, so with the resistor there the sine waves will be damped over time. The resulting voltage is a cosine with exponentially decaying envolope, and the current is a sine with the same exponentially decaying envelope.

Again, this is a classic circuit with lot's of ways of analyzing it and thinking about it out there, so there is little point going into all the details here. Every undergrad EE will have seen this as a test question at some point.


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