For the following circuit, at time 0, the switch switches from A to B, figure out the voltage wave form at B.

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The question I am concerned with is whether the voltage across the capacitors will change instantaneously or not? Will the output waveform at B be a step function? (0 for t<0 and C1/(C1+C2) after t>0).


4 Answers 4


Ideally, yes. In reality, no. The impedance of the wires will slow down the transfer of current, making the output waveforms resemble the normal R-C curves (albeit very shortened).

  • \$\begingroup\$ If an interviewer asked this question, would there be anything important missing in the answer? \$\endgroup\$ Jan 28, 2016 at 10:32
  • \$\begingroup\$ Yes. Loop inductance for one thing. \$\endgroup\$
    – user16324
    Jan 28, 2016 at 12:11

Everything is fine with the answers. I agree:

  • There is no instantaneous change in nature (most cases)
  • There is always resistance in wires
  • There is the Equivalent Series Resistance of the capacitor.
  • etc...

But what if there was no resistance at all? Or to place it different, what if we eliminate as possible as we can the resistance? To what behavior the system tends to?

To answer that lets assume the next circuit. Lets call

  • t=0- and t=0+ the time just before and after the switch closes.
  • C1 and C2 the capacitance @t=0-
  • Vc1 and Vc2 the voltage of the capacitors @t=0-


simulate this circuit – Schematic created using CircuitLab


@t=0- :
The total electric charge Q(0-) = C1*Vc1(0-) + C2*Vc2(0-) (1) and 
The Total energy                   1                 1
                          W(0-) = -- C1*Vc1(0-)^2 + -- C2*Vc2(0-)^2 (2)
                                   2                 2
@t=0+ :
The voltages are "instantaneously" equal so Vc(0+) = Vc1(0+) = Vc2(0+)
The charge, Q(0+) = (C1+C2)*Vc(0+) (3)

The charge @t=0- and t=0+ are the same so from (1) and (3) we have:

C1*Vc1(0-) + C2*Vc2(0-) = (C1+C2)*Vc(0+) <=> ... <=>
Vc(0+) = ------- * [C1*Vc1(0-) + C2*Vc2(0-)] (4)

So the energy @t=0+ is
W(0+) = -------- * [C1*Vc1(0-) + C2*Vc2(0-)]^2 (5)

If we examine the (2) and (5) We will see a difference in Energy

ΔW = W(0+) - W(0-) = (-) --------*[Vc1(0-) - Vc2(0-)]^2 => ΔW < 0 (6)

Where has this energy gone? Simply. Is the amount of energy that the circuit radiates to the environment. Remember in order for the Kirchhoff laws to apply the circuit has to be lumped [1]. In our case, at t=0, the circuit is not lumped, so the energy is not flowing only inside the circuit and thus the circuit radiates to the environment.

The (6) also means that the final total energy is always less than the initial.

Edit - Lumped circuits:

[1]: We have 2 major sets of laws to study the electrical circuits. The laws of Maxwell and the laws of Kirchhoff. The Maxwell laws are functions of space and time and thus are complicated. The Kirchhoff laws are functions of time only, not space. In order to apply the Kirchhoff laws we accept that in an electrical element of 2 terminals the current flow to the element equals the current flow out from the element at every time. This assumption is valid only if the dimension of the electrical elements are much smaller than the wave length of the current. These circuits are called lumped. You can also see here. In addition, if the wave length is smaller or comparable to the dimension of the circuit, the current in and out is not guaranteed equal and the circuit is called distributed.

  • \$\begingroup\$ could you please tell me what do you mean by "circuit has to be centralized"? \$\endgroup\$ Jan 31, 2016 at 3:01
  • \$\begingroup\$ @GyananshuUpadhyay Sorry for the term "centralized" It was wrong. I'm from Greece and we call them "Συγκεντρωμένα κυκλώματα". I've edited the answer to include the correct term which is Lumped circuits. And I also add a small explanation of what a lumped circuit is:-) \$\endgroup\$
    – hoo2
    Jan 31, 2016 at 13:35

The question I am concerned with is whether the voltage across the capacitors will change instantaneously or not?

Nothing changes instantly.

In addition to this, if you assumed that the net energy contained in the pre-charged capacitor would still be the same net energy but distributed between the two capacitors, you'd be wrong because infinities are involved i.e. with zero impedance connections (impossible of course) an infinite current would flow.

You can rightly assume that net charge is unchanged (Q = CV) so, with two identical capacitors the "final" voltage will be half the "original" voltage. This is at odds with the assumption that energy will be maintained. If you assumed that energy is conserved and the capacitances were equal then you might expect the final voltage to be 0.7071 V. As said above this would be incorrect.


Not only wire impedances but also capacitor ESR (Equivalent Series Resistance) Equivalent Circuit, will affect the charge time DC characteristics.


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