I am learning about RLC circuits in my Circuits Analysis class and I had a question regarding Parallel RLC step responses.

In the book, they only provide examples of RLC circuits with a current source, such as the one below.


simulate this circuit – Schematic created using CircuitLab

The circuit and equations that go along with it make sense to me, but I started thinking about voltage sources, such as the one below.


simulate this circuit

This brings up a couple of questions:

  1. Does the voltage change across the resistor, inductor or capacitor? According to KVL it would be 12V once the switch is closed. Does this produce a step response at all then?
  2. When the switch is closed, the voltage across the capacitor goes from 0V(assuming no energy initially) to 12V because KVL would be violated otherwise. Wouldn't this be an instantaneous change, which cannot occur in capacitors?
  3. What happens when you put a current source in a series RLC circuit? This raises similar questions.

Thanks for any help provided.


3 Answers 3


Assuming the context of ideal circuit theory, the voltage source current will not be finite at the instant the switch is closed; there will be an impulse of current to charge the capacitor instantly to 12V. After the initial impulse, the current will be a ramp, increasing linearly with time (without bound) due to the inductor.

However, ideal circuit theory is not physically relevant in this case since some of the assumptions of ideal circuit theory do not hold for the second circuit.

Physically, the current will be finite and the voltage across the capacitor will not be a step. This can be seen if the naive ideal circuit model is augmented with additional circuit elements to model the fact that, e.g.,

  • All physical voltage sources have finite internal resistance (finite short circuit current)
  • Physical capacitors have parasitic inductance and resistance which must be included in the ideal circuit model
  • A physical circuit (the wires and circuit elements that form the closed path for current) have parasitic R, L, and C that must be modelled
  • A physical circuit has radiation resistance, i.e., for large and fast current changes, the circuit will radiate energy into space and this must be modelled.

Regarding your perceptive question 3, replace current above with voltage, inductor with capacitor, short with open and then essentially the same argument holds.

  • \$\begingroup\$ That clears up a bit, but I had another question. The second schematic would not produce a step response. Instead, adding a resistor in series to the voltage source would produce one. Am I correct in saying that? \$\endgroup\$
    – Addison
    Feb 15, 2015 at 0:09
  • \$\begingroup\$ @Addison, the 2nd schematic would produce a step output voltage at the expense of a current impulse (non-physical) and a current that increases without bound thereafter (non-physical). Adding a resistor in series with the voltage source prevents the current pathology (now a finite and bounded current) but the output voltage is no longer a step. In fact, the output voltage goes to zero as time increases. \$\endgroup\$ Feb 15, 2015 at 0:14
  • \$\begingroup\$ What do you mean by step? My understanding is that a step response rises to a final voltage/current slowly, typically exponentially. In your answer, you said physically that the voltage across the capacitor will not be a step, but in the previous comment you said the 2nd schematic would produce a step output voltage. It seems contradictory. Also, by adding a resistor, you can do a source transformation to get a current source with a parallel resistor. This gives you the same circuit as the 1st schematic essentially. Therefore, the voltage should rise exponentially. \$\endgroup\$
    – Addison
    Feb 15, 2015 at 0:34
  • \$\begingroup\$ @Addison, I interpreted your first comment as asking if the output were a step function which is why I responded that the output would be a step (as in Heaviside step function). However, a step response is not necessarily an exponential rise to a final value. A step response is simply the output due to a step input and, in general, that can go to zero, non-zero, infinity, or oscillate. \$\endgroup\$ Feb 15, 2015 at 1:28
  • \$\begingroup\$ @Addison, also, regarding the last sentence of your comment, in steady state, the voltage across an inductor is zero. If you add a resistor in series with the voltage source, the final (t goes to infinity) value of the output voltage is zero. This is (of course) true with the source transformation. Think about it. \$\endgroup\$ Feb 15, 2015 at 1:33
  1. It would be 12V. The step response is an instantaneous rise from 0V to 12V.

  2. The available current is infinite, so the capacitor can charge instantaneously.

  3. The instantaneous current step would produce infinite voltage across the inductor.

This is that happens when you put theoretical ideal components into invalid configurations. A real voltage source has internal resistance that limits current, and a real current source has a maximum voltage that can be produced.


V is the source voltage, applied at t=0

Voltage across components must = source voltage at all values of time, therefore:

(1) Resistor current is V/R

(2) The inductor voltage is constant at V, so Ldi/dt = V; i = integral(V/L dt) = ramp with slope = V/L; therefore i = Vt/L

(3) Capacitor must charge instantaneously to V, so capacitor current is an impulse of strength VC (i.e. infinitesimal duration current to charge C to V)

The total current supplied by the source at time, t = V/R + Vt/L + (impulse of strength VC Coulomb at t=0)

When the impulse has finished, at t = 0+, total current is V/R + Vt/L, ie a ramp

  • \$\begingroup\$ I'm new here, could someone explain why I've been awarded '-1'? Thank you. Chu \$\endgroup\$
    – Chu
    Feb 16, 2015 at 12:41

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