# Why is capacitor equilibrium based on balanced voltages and not balanced charges?

The problem below is from the electrical circuits chapter of Essential University Physics Volume 4. I am having trouble understanding the essential idea behind the solution.

I initially tried solving this with the assumption that charges would balance between the capacitors since they are in series. The solutions manual tells me that equilibrium means the voltages balancing. This doesn't make sense to me, since this would imply the capacitors have different charges. I was under the impression that in parallel, voltages are equal, and in series, charges are equal. Why would voltages be inclined to find balance?

• Only one addition to the good anwers you got. Capacitance is the ability to accumulate electric charges. Then if the capacitance is different, the charge stored is different. Commented Feb 26 at 9:59

The final state of "equilibrium" of this circuit will be when there's no longer any current flowing. That state can only be achieved when the voltage across the resistor is zero.

If it's non-zero, then current is flowing, according to Ohm's law, and one of the capacitors will still be charging while the other is still discharging.

It seems you are stuck on the idea of series and parallel, which are useful concepts in some circumstances, but not here. The appropriate concepts at work here are Ohm's law, and Kirchhoff's Voltage Law (KVL). I'll apply them here:

simulate this circuit – Schematic created using CircuitLab

KVL states that the sum of voltage changes encountered as one traverses each element around a closed loop must be zero. For your circuit that's written:

$$+ V_{C1} - V_{R1} - V_{C2} = 0$$

Ohm's law tells us what voltage $$\V_{R1}\$$ exists across the resistor. In the state of equilibrium, no current is flowing, so:

\begin{aligned} V_{R1} &= I \times R_1 \\ \\ &= 0A \times 2.2k\Omega \\ \\ &= 0V \end{aligned}

Plugging that result into the KVL equation:

\begin{aligned} V_{C1} - 0V - V_{C2} &= 0 \\ \\ V_{C1} = V_{C2} \\ \\ \end{aligned}

To summarise, since when this equilibrium state is achieved, there's no voltage across the resistor, KVL implies that the two capacitors must have exactly the same voltage across them. Therefore it is capacitor voltage that equalises in this scenario.

Charge is not balanced. This is evident when one considers the charge equation:

$$Q = CV$$

Given that the voltages $$\V\$$ are eventually equal, but the capacitances $$\C\$$ are not, in that state of equilibrium these capacitors do not have the same charge.

This doesn't make sense to me, since this would imply the capacitors have different charges.

They do have different charges. Because they have different capacitances.

• Q = CV.
• V ends up the same for both after some time (when they equalize, and the current drops to 0)
• C is different: twice the capacitance on the left than on the right
• Therefore, Q ends up different: twice the charge on the left than on the right

I was under the impression that in parallel, voltages are equal

Correct.

and in series, charges are equal.

Undetermined. The charge and the voltage of one capacitor are unrelated to the charge and the voltage of another capacitor.

since they are in series

Kind of. Once the capacitor are balanced, there is no current flow through the resistor, which means the voltage across it is zero, which means that you could replace it and the switch with a direct connection with no effect. At that point, the capacitor can be viewed as being in parallel: same voltage.

• So current in series is equal and charges are based off capacitance & current? Also could you expand on why we can treat it like parallel. Is it because the loop rule would show that the voltages must cancel eachother out? Commented Feb 26 at 1:21
• "current in series is equal" Yes. "and charges are based off capacitance & current?" Yes; the end charge is the initial charge plus the integral of the current in or out. " could you expand on why we can treat it like parallel." Because they end up at the same voltage. "the voltages must cancel each other out?" No. They don't. They are equal, not zero. Commented Feb 26 at 14:14

It looks like you have trouble with intuition, thinking the charges are going to be equal. Instead of doing the problem with commensurate capacitor sizes, which complicates things, use a much smaller capacitor for the 1 μF one, perhaps 1 nF or 1 pF. I'll bet your intuition now has no trouble seeing that the final voltages will be equal, but that the final charges will be very different.

The solutions manual tells me that equilibrium means the voltages balancing. This doesn't make sense to me

"Equilibrium" means "steady state": all circuit parameters are constant, which means voltage and current are constant. This does not necessarily mean current or voltage are zero, but they all have to be constant.

So what is "equilibrium" in the case of your circuit?

All three components are the same, so current through them is the same.

If V1 is not equal to V2, current i will flow in resistor R, and thus in the capacitors:

$$\ i = \frac{V1-V2}{R} = C_1 \frac{-dv_1}{dt} = C_2 \frac{dv_2}{dt} \$$

So the capacitor with higher voltage will discharge into the capacitor with lower voltage.

If current flows in a capacitor, its voltage will change. Thus, if current is not zero, equilibrium has not been reached yet. Thus, in this circuit's case, equilibrium means current is zero, and therefore V1=V2.

Another way to say it is: at equilibrium, everything is constant so derivatives are zero, thus $$\ \frac{dv}{dt} = 0 \$$ thus i=0.

For any circuit with capacitors, "equilibrium" means current through the capacitors is zero. Otherwise voltage across the caps is changing, thus it's not equilibrium.

Capacitors are insulating, so while current can flow through them, charges (in this case electrons) cannot pass. Thus your circuit is split in two halves (red and blue) and the charge in each is constant. Thus $$\ Q_1+Q_2 = Constant \$$.

Charges are equal in series because the same current (which is charge/time) passes through each device. Note that if you start from a start where both capacitors are not at 0 V, then it's only the additional charge (from an additional current flowing) that remains equal.

Voltages (in parallel) are equal because, even if you start with different voltages, any voltage difference will lead to current flowing. Current will reach zero when the voltage difference becomes zero.

In your example, the 2 uF has an initial charge (getting to 150 V), but also consider that the current (i.e. charge) that flows is not 'in the same direction' -- it is out of one capacitor and into the other.

• I understand the first two paragraphs but not the last one. Would that mean its parallel or series? Commented Feb 26 at 1:22
• In some sense it's both; it's a circuit, and 'parallel' or 'series' is just a way of simplifying some connections for analysis. Note that if you consider 'series', these are more in 'anti-series', since the 'tops' of each capacitor are connected; not the top of one to the bottom of the other. Commented Feb 26 at 4:45

I was under the impression that in parallel, voltages are equal, ...

No, the two capacitors are in series ... through R1.

I am having trouble understanding the essential idea behind the solution.

Find the total energy dissipated in the resistor ... until equilibrium

The integral of energy flowing through the resistor is "constant" ...
Whatever the value of the resistor.

NB: Time is logarithmic.

Here is a Maple sheet for calculating the "loss" of energy across R.

Note that Energy_R is "independent" of R ... Energy is dissipated, even if R -> Zero Ohm!
Note that variable time is calculated with "tau", so independent also of R.

I was under the impression that in parallel, voltages are equal, ...

You actually took the first step to the solution already.

When the circuit is in equilibrium the current flowing is zero which basically means no drop at all across the resistor, as others have pointed out.

So, at equilibrium, you can think of the capacitors as connected in parallel. Therefore the voltage across them will be equal.

simulate this circuit – Schematic created using CircuitLab

Note that this is not what physically happens, but I replaced the resistor with a short for simplicity. At equilibrium, the resistor has no "function" so any resistance can be used there.

Alternatively, you can think of the entire circuit as an RC network excited by a voltage source but the main difference is that the source here is the 2-uF capacitor. Since in an RC circuit the voltage across the capacitor becomes equal to the supply voltage, same will happen in the circuit in your question. The main difference is that the source capacitor's voltage will decrease over time.