I'm having confusion with applying superposition to simple RC circuit with non-zero initial capacitor voltage.


simulate this circuit – Schematic created using CircuitLab

Consider a simple RC circuit series connected with voltage source of step voltage \$V_s=8u(t)\$, and let's say initial voltage of the capacitor is \$V_c = 2V\$.

It's obvious that the value of \$V_c\$ over time is

\$ V_c = 2+(8-2)(1-e^{t/RC}) \$.

But if I split the voltage source \$V_s=8u(t)\$ with two voltage sources \$V_{s1}=4u(t), V_{s2}=4u(t)\$, then the corresponding \$V_{c1}\$ and \$V_{c2}\$ will be

\$ V_{c1} = 2+(4-2)(1-e^{t/RC}) \$.

\$ V_{c2} = 2+(4-2)(1-e^{t/RC}) \$, respectively.

Then, applying superposition yields \$V_{c1}+V_{c2}=4+4(1-e^{t/RC})\$, which is different from

\$ V_c = 2+(8-2)(1-e^{t/RC}) \$.

Where am I doing wrong?

Is it wrong to apply superposition theorem to non-zero state of capacitor?

  • \$\begingroup\$ Draw the circuit? \$\endgroup\$
    – MathieuL
    Commented Aug 12, 2015 at 15:10
  • \$\begingroup\$ @MathieuL Edited the post \$\endgroup\$
    – asqdf
    Commented Aug 12, 2015 at 15:14
  • \$\begingroup\$ The problem here that you are using a solution to the differential equation describing the circuit, which is including the boundary conditions and is not exhibiting the linear properties by itself. To benefit the linear nature of the circuit and use the superposition rule, the differential equations should be written using it, and then solved. \$\endgroup\$
    – Eugene Sh.
    Commented Aug 12, 2015 at 15:20
  • \$\begingroup\$ A somewhat related concept: you actually can use superposition with capacitors if you are working with signals that are all at a single frequency. To read more, look up "phasor notation" and "impedance", en.wikipedia.org/wiki/Electrical_impedance. It does not apply to the unit step function though, as that is not a single frequency. \$\endgroup\$
    – Justin
    Commented Aug 12, 2015 at 18:08
  • \$\begingroup\$ @Justin You can use the superposition theorem with capacitors only when you are dealing with a sinusoidal "steady state" response of the circuit. When you are dealing with the problem of transient response of RLC circuit, you cannot apply the impedance approach and cannot apply the superposition theorem naively because the RLC circuit's differential equation can be nonlinear when capacitors or inductors in the circuit have non-zero state. \$\endgroup\$
    – asqdf
    Commented Aug 13, 2015 at 7:43

2 Answers 2


Yes, you can use superposition -- once for the V source, and once for the cap decay.

The cap decay is (set the other V to 0) Vc=2+(0−2)(1−e(-t/RC)), or 2.e(-t/RC) [you missed a '-' in the exponential]

The 8 V source adds Vc += 8(1-e(-t/RC))

Two 4 V sources would each add Vc += 4(1-(e-t/RC)).

They all add together as expected.


While applying superposition you have to treat the initial voltage across capacitor as an independent source.Then find the response due to 4V source alone by making the other 4V Source and the initial voltage across capacitor as zero.Then find the response due to another 4V source following the same procedure.Then it time to find response due to the initial condition treated as source but we can't find the response due yo initial condition treating it as source. So we consider it as initial condition only and find the response due to it.Then finally add up the three responses


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.