I am trying to understand a problem I did once with capacitors. I am not a native English speaker so I am translating it, I hope you can understand it:

A capacitor C2 is charged with 10V, then, it is connected with C1 and C3 which are discharged. After that we connect points A and B with a cable (resistance 0). Calculate the final charge of each capacitor.


By Kirchhoff and knowing that V = Q/C I know that:

$$V_{C1} + V_{C2} + V_{C3} = 0$$ $$\frac{Q1(initial) + q}{C1} + \frac{Q2(initial) - q}{C2} + \frac{Q3(initial) - q}{C3}$$

Here Q1(initial) and Q3(initial) is 0 because they are discharged and q is the charge which is moved when we connect A and B.

I understand this:

$$\frac{Q2(initial) - q}{C2}$$

Because C2 loses charge to charge C1 and C3, but why C3:

$$\frac{Q3(initial) - q}{C3}$$

Also loses charge if it is being charged? I know the result of q is: $$31.25\mu C$$

So the equation is correct but it doesn't makes sense to me!

  • \$\begingroup\$ I believe your "x 10^-6" is redundant for the resultant q value you've provided. \$\endgroup\$
    – Samuel
    Commented Sep 26, 2013 at 2:35

3 Answers 3


Alright, it's been a while since fundamentals, but here we go. Keep in mind there are lots of different avenues to the solution for this problem.

First of all, I calculate a total charge of 100\$\mu C\$. We're going to need that.

This problem might be a little easier (like many beginner EE problems) by rearranging the way it's drawn. The clue comes from the connection of the zero resistance wire between the outside capacitors. Then, at the instant of connection you get this arrangement:


simulate this circuit – Schematic created using CircuitLab

That looks a little easier to me. The total charge is conserved and it all comes from C2. You're using voltages to solve, so we'll try that route. One approach is to calculate the new voltage across C2 and the C1/C3 combination. For that we can look at this as a single large capacitor, in which case the equivalent capacitance would be $$ C_{eq} = C_2 + ({ 1 \over C1} + {1 \over C3})^{-1} $$

I get an equivalent capacitance of just about 14.55\$\mu F\$.

As you've already pointed out the voltage across a capacitor is: \$ V = {Q \over C} \$. Now we know the total charge and the equivalent capacitance. Thus the voltage follows. I calculate 6.875V. We now know the voltage across C2 and its capacitance, so we can find the charge remaining on it. I calculate that 68.75\$\mu C\$ remains on C2. Seems reasonable.

The remaining charge, 31.25\$\mu C\$, is shared between C1 and C3.

Let's sanity check. $$ {(100\mu C - 31.25\mu C) \over 10\mu F } - { 31.25\mu C \over (5\mu F^{-1} + 50\mu F ^{-1})^{-1}} = 0 $$

Putting that all together...


simulate this circuit

Now if we look at each terminal of added capacitors (C1 and C3) which are connected to C2, we can see why your original equation would give a \$+q \$ for C1, a \$-q \$ for C3 while the Q on C2 is \$ Q2(initial)-q\$. Notice that the nets that C2 are connected to (top and bottom) sum to the original total charge while the new wire connected (between C1 and C3) has a total charge of zero. Thus charge is conserved and the universe goes on.

  • \$\begingroup\$ Very interesting way of solving it, I have a few questions. Can you calculate equivalent capacitor even if one is charged and the other discharged? What about if I have for example C1 and C2 charged with different Q? Also the equivalent capcitance of C1,C2 and C3 is 14,53uF not 50uF and V=Q/C. Thank you very much for your answer! :) \$\endgroup\$
    – Andres
    Commented Sep 26, 2013 at 3:32
  • \$\begingroup\$ The capacitance does not depend on charge, only on the physical properties. So the equivalent capacitance can always be calculated regardless of charge. The charge remains constant for the entire system, you just need to determine how it gets distributed into the system. My mistake on the two things you mentioned, I'll correct them. I used the correct numbers in the calculations, I just wrote down the wrong thing. \$\endgroup\$
    – Samuel
    Commented Sep 26, 2013 at 4:25
  • \$\begingroup\$ Nice! Thank you! :) One more question, why the voltage you calculate for the equivalent capacitor (6.875V) is the voltage on C2? If I have C1 and C2 charged, I only have to add the charges to get the total? \$\endgroup\$
    – Andres
    Commented Sep 26, 2013 at 15:02
  • \$\begingroup\$ The voltage on the equivalent capacitor is the same as on that C2 because we're talking about the same two points that are being calculated. In the picture the calculated voltage is across the top and bottom nets. So the same voltage that is across the equivalent capacitor (C2 in parallel with C1 and C3 in series) is across C2 and the C1/C3 combination. For DC circuits, the voltage between one net (wire) and another is the same anywhere along that net. \$\endgroup\$
    – Samuel
    Commented Sep 26, 2013 at 15:47

I think you may be having a tough time understanding the concept of "losing charge", and I feel for you because I messed up a sophomore college test question due to not understanding capacitors and Kirchhoff's Current Law (KCL).

KCL states that the sum of current flowing into/out of a node is equal to 0, and this includes charging a capacitor: the current flowing into one side of the cap will be matched by an equal amount of current flowing out of the other side of the capacitor.

The charge moves from C2 to C1 and C3 in the form of electrical current. There is only one loop in this circuit, so when you connect nodes A and B, current flows through all the capacitors, and the current is the same through the whole loop at any point in time (this is always true for loops with current flowing in them, implied by KCL).

So when you say "losing charge", think of it in terms of "the capacitor is becoming more and more negatively charged, so it has a negative voltage" and NOT "the capacitor had 0 charge to begin with and so how could it have any charge to give up?"

  • \$\begingroup\$ "The current flowing into one side of the cap will be matched by an equal amount of current flowing out of the other side of the capacitor." This might be true for DC analysis but not AC. \$\endgroup\$ Commented Sep 26, 2013 at 1:29
  • 1
    \$\begingroup\$ First, this is a DC problem. Second, yes, actually, this holds for all circuit problems: you can't have current going in one side of a wire and not out the other in a closed loop. A phase shift doesn't mean you break KCL or KVL rules. \$\endgroup\$
    – Bob
    Commented Sep 26, 2013 at 1:49

So the equation is correct but it doesn't makes sense to me!

It doesn't make sense to me either and I believe your equation is incorrect. You have:

$$\frac{Q1(initial) + q}{C1} + \frac{Q2(initial) - q}{C2} + \frac{Q3(initial) - q}{C3} = 0$$

Now, consider the simple case where \$C_1 = C_3 = 2C_2\$. We know in advance that the final charge on C2 will be one-half the initial charge.

But, according to your equation, we get:

$$\dfrac{q}{2C_2} + \dfrac{Q2(initial) - q}{C2} - \dfrac{q}{2C_2} = 0$$

which yields

$$q = Q2(initial)$$

which is clearly not correct.

Now, another way to look at this is to consider that C1 and C3 are series connected and thus form an equivalent capacitor with capacitance \$C_{EQ} = C_1||C_3 = \dfrac{1}{\frac{1}{C_1}+\frac{1}{C_3}}\$.

Then we have:

$$\dfrac{Q2(initial) - q}{C_2} = \dfrac{q}{C_{EQ}} \rightarrow q = \dfrac{Q2(initial)}{1 + \frac{C_2}{C_{EQ}}} $$

When \$C_1 = C_3 = 2C_2\$, \$C_{EQ} = C_2 \$ and we get the correct result

$$q = \dfrac{Q2(initial)}{2} $$.

But look, we can rewrite the equation as:

$$\dfrac{Q2(initial) - q}{C_2} - \dfrac{q}{C_{EQ}} = 0 = \dfrac{Q2(initial) - q}{C_2} - q(\dfrac{1}{C_1} + \dfrac{1}{C_3})$$

so there is a sign error in your equation.


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