I am learning to find the voltage drops across the capacitors in a DC circuits. we all know that capacitor charges till it equals the input voltage (assuming initial charge of capacitor is zero). If a DC voltage is applied

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For the above circuit Vc= Vs(1-exp(-t/rc))

Now I considered little complex circuit something like below. enter image description here

Here capacitor is not directly connected to a voltage source. After googling I found that the circuit can be solved by considering the capacitor as a load and finding the Voc and Rth by using Thevenin's theorem( Or its dual Norton's theorem). Now R value in the time constant is replaced with Rth value and Vs voltage with Vth voltage.

Finally the voltage across capacitor, Vc= Vth(1-exp(-t/RthC))

Now I considered more complex circuit. Suppose if the circuit consists of more than one capacitor in the circuit. Something like below.

enter image description here

Now I am stucked here. How do I solve for the voltages across the capacitors C1 and C2.

I am wondering what would be the capacitor voltage equations for both capacitors. If there is a single capacitor, we used Thevinin's theorem but how do I solve if I have more than one capacitor in the DC circuits.

Vc1= Vunknown1(1-exp(-t/Runknown1 C1) Vc2= Vunknown2(1-exp(-t/Runknown2 C2)

How do i solve for Vunknown1, Vunknown2, Runknown1 and Runknown2. Could anyone kindly explain me. How do I solve if we come across these type of circuits. Kindly help me through this.Thank you.

  • 2
    \$\begingroup\$ Please take into account that engineering is a science of correctness. The comment you made "(assuming initial charge of capacitor is zero)" is not correct in this context. The final voltage on the capacitor will still become equal to the input voltage even if the capacitor had some initial charge or not. The comment really only applies when using the formulas to determine the time to full charge. In that case you have to take the initial charge into account or state that it is zero to begin with. \$\endgroup\$ Feb 16, 2013 at 14:22
  • 1
    \$\begingroup\$ For DC, remove the capacitors, calculate the DC voltages, replace the capacitors. The capacitors will assume the same DC voltages in nough time as if they were never there. That makes circuit 3 trivial. If you have trouble working out the DC voltages in 3 try adding a conceptual infinite resistor to negative from any point or points as required. eg at C2 location if need to help visualisation. The answer should be intuitive and obvious from inspection once you understand the principle. \$\endgroup\$
    – Russell McMahon
    Feb 16, 2013 at 20:29

3 Answers 3


Solving ckt#3 the hard way using differential equations:

To start with, this equations always holds, for any capacitor $$i = CdV/dt$$

In the circuit you've provided, we have two unknown voltages (V1 across C1 and V2 across C2). These can be solved by applying Kirchoff's Current Laws on the two nodes.

For node V1: $$ (V_s-V_1)/R_1 = C_1 dV_1/dt + (V_1-V_2)/R_2 $$

And for node V2: $$ (V_1-V_2)/R_2 = C_2 dV_2/dt $$

Now we've got two differential equations in two unknowns. Solving the two simultaneously give us the expressions for V1 and V2. Once V1 and V2 are calculated, calculating the currents through the branches is trivial.

Solving differential equations is, of course, not trivial. What we generally do is to use Laplace Transform or Fourier Transform to convert them into algebraic equations in the frequency domain, solve the unknowns, and then do Inverse Laplace/Fourier transform to get the unknowns back into time domain.

Method 2: Use voltage divider rule:

If we recall that the impedance across a capacitor C is $$Z=1/jwC$$ and denoting the impedances of the two capacitors C1 and C2 as Z1 and Z2, we can calculate V2 using the formula for voltage division across two impedances (http://en.wikipedia.org/wiki/Voltage_divider): $$V_2 = V_1 R_2/(R_2 + Z_2)$$ V1 can also be calculated using the same rule, the only issue is that the impedance on the right side of node 1 is a bit complex: it's the parallel combination of Z1 and (R2 + Z2). V1 now becomes $$V_1 = V_s (Z_1*(R_2+Z_2)/(Z_1+R_2+Z_2))/(R_1 + (Z_1*(R_2+Z_2)/(Z_1+R_2+Z_2)))$$

What to do next is to expand Z1 and Z2 using the capacitive-impedance formula, to get V1 and V2 in terms of w. If you need the complete time response of the variables, you can do Inverse Fourier Transforms and get V1 and V2 as functions of time. If however, you just the need the final (steady-state) value, you can set $$w=0$$ and evaluate V1 and V2.

A rather simpler way:

This method can give only the final steady-state values, but it's a bit handy for quick calculations. The catch is that once a circuit has settled into a steady state, the current through every capacitor will be zero. Take the first circuit (the simple RC) for example. The fact that the current through C is zero dictates the current through R (and hence the voltage drop across it) also to be zero. Hence, the voltage across C will be equal to Vs.

For the second circuit, all the current must pass through the path R1->R2->R3 if the capacitor draws no current. This means the voltage across C (equal to the voltage across R2) is $$V_s R_2 / (R_1 + R_2 + R_3)$$

In the last circuit, current through C2 being equal to zero implies the current through R2 being zero (and hence any voltage drop across it). This means any current that flows must take the path R1->C1. However, the current through C1 is also zero, which means R1 also carries no current. So both the voltages V1 and V2 will be equal to Vs in steady state.


In my opinion, if you are familiar with analysing circuits using loop equations and Laplace transforms, it would be the best choice. Circuit analysis using Laplace transforms have the same power as that using classical differential equations, but are a lot easier.

Now in order to apply Laplace transform directly we use

1) X_L( Impedence of inductor ) as sL

2) X_C(Impedence of capacitor ) as 1/ (sC)

3) R ( Resistance ) as it is

all assuming zero initial conditions.

For your problem, assuming currents in both loops as clockwise;

V(s) = I1( R1 + 1/sC1) - I2( 1/sC2) -------loop1

0 = I1( 1/sC1) - I2( 1/(sC1) + R2 + 1/(sC2))---loop 2

Two equations for two unknowns. The answer for I1 and I2 would be in s-domain. So take the inverse Laplace transform. Once we have the currents, voltages are easy to find as well.

Alternatively, the node method may directly be applied to get voltages.

  • \$\begingroup\$ As soon as how this is an old question it'd be worth adding some more details on how to apply Laplace transforms. The other answer already mentions the technique is easier. \$\endgroup\$
    – PeterJ
    Jun 1, 2014 at 13:24
  • \$\begingroup\$ Agreed. I have modified answer accordingly. \$\endgroup\$ Jun 1, 2014 at 16:14

The simplest way to solve this problem would be to put the circuit into the laplace aka the frequency domain. In the frequency domain the dependent variable is the frequency instead of time. There are equivalent values for each of the characteristics of the circuit.

L -> LS

C -> 1/Cs

R -> R

v(t) -> V(S)

and so on...

Substitute these into your circuit design and you can use basic circuit analysis techniques; considering connection constraints. Also you can find an equivalent thevein circuit just as before.

However, its important to note that to turn the resulting functions into something that you can use you'll need to preform an inverse la'place transformation. I suggest searching for a table of identities and trying to get your function to look like the identities through algebraic manipulation.

If you have time this is a a great skill to learn and will simplify and circuit analysis you would have to do in future applications.


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