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 V1,V2,V3,V4V_1, V_2, V_3, V_4 be the capacitor voltages. To get a solution, declare the value of V1 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):

V2V1V4+V3=0V2+V3=VV4V3=VV_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 V1 V_1 , move it over to the other side:

V2V4+V3=V1V2+V3=VV4V3=VV_2 - V_4 + V_3 = V_1 \\ V_2 + V_3 = V \\ V_4 - V_3 = V

Recall that VV is the battery voltage. For V1=0V_1 = 0 , (V2,V3,V4)=(V,0,V) (V_2, V_3, V_4) = (V,0,V) . For V1=2VV_1 = 2V , (V2,V3,V4)=(3V,2V,V) (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+V1V+V3=0V1+V3=0 V + V_1 - V + V_3 = 0 \\ V_1 + V_3 = 0

So if V1V_1 is chosen as a positive value, V3V_3 will be negative. If one wishes to impose an additional constraint that none of the capacitor voltages can be negative, it follows that V1V_1 and V3V_3 are both zero.

Note by Steven Chase
2 months, 2 weeks ago

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@Steven Chase now I am little bit more confused in this probelm.
Why we are doing V1=0V_{1}=0

Talulah Riley - 2 months, 2 weeks ago

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If all capacitor voltages must be positive (or at least not negative), V1V_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.

Steven Chase - 2 months, 2 weeks ago

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

Talulah Riley - 2 months, 2 weeks ago

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

Steven Chase - 2 months, 2 weeks ago

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@Lil Doug

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.

Steven Chase - 2 months, 2 weeks ago

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@Lil Doug

Notice how the V1=0 V_1 = 0 case corresponds to the expected solution.

Steven Chase - 2 months, 2 weeks ago

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