# Voltages in Capacitor Network

Here is an interesting problem recently shared by Neeraj.

As far as I can tell, there are infinitely many solutions, because when you write a system of four linear equations, the system matrix always ends up being singular. Let $V_1, V_2, V_3, V_4$ be the capacitor voltages. To get a solution, declare the value of $V_1$ and solve a system of three linear equations to get the other three voltages. The system of three linear equations is non-singular.

The following equations describe the system (being mindful of the capacitor polarities):

$V_2 - V_1 - V_4 + V_3 = 0 \\ V_2 + V_3 = V \\ V_4 - V_3 = V$

Since we are declaring the value of $V_1$, move it over to the other side:

$V_2 - V_4 + V_3 = V_1 \\ V_2 + V_3 = V \\ V_4 - V_3 = V$

Recall that $V$ is the battery voltage. For $V_1 = 0$, $(V_2, V_3, V_4) = (V,0,V)$. For $V_1 = 2V$, $(V_2, V_3, V_4) = (3 V,-2 V,-V)$. These are just two solutions out of infinitely many possibilities.

One other thing to note. The following relation also holds:

$V + V_1 - V + V_3 = 0 \\ V_1 + V_3 = 0$

So if $V_1$ is chosen as a positive value, $V_3$ will be negative. If one wishes to impose an additional constraint that none of the capacitor voltages can be negative, it follows that $V_1$ and $V_3$ are both zero.

Note by Steven Chase
10 months, 1 week ago

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@Steven Chase now I am little bit more confused in this probelm.
Why we are doing $V_{1}=0$

- 10 months, 1 week ago

If all capacitor voltages must be positive (or at least not negative), $V_1$ has to be zero. See the last part of the note. Otherwise, it is just an arbitrary choice, given that the general four by four system is singular.

- 10 months, 1 week ago

@Steven Chase how do you know the polarities of capacitor?

- 10 months, 1 week ago

The curved side is negative, and the flat side is positive

- 10 months, 1 week ago

The final comment that I just added seems relevant also. Perhaps the question assumes that all capacitor voltages must be positive, in which case their solution is indeed the only allowable one.

- 10 months, 1 week ago

Notice how the $V_1 = 0$ case corresponds to the expected solution.

- 10 months, 1 week ago