Problem + Hard 3.3   Source:   Problem 25-28 - pag 135 -    HALLIDAY, RESNICK, WALKER, Jearl - Book: Fundamentals of Physics - Vol 3 - Ed. LTC - 8th edition - 2009.

The capacitor C₃ of the circuit in the figure below, on the left, is a capacitor whose capacitance can be varied. O graph in the figure below, on the right, represents the electric potential V₁ between the plates of the capacitor C₁ as a function of C₃. For different values of C₃ we have different values for V₁. We know that the electric potential V₁ tends asymptotically to 10 volts when C₃ → ∞.

a) The electric potential V of the source;

b) The values of C₁ and C₂.

Solution of the Problem + Hard 3.3

Item a

For the solution of this problem it is very important that we keep in mind the equation that relates charge, capacitance and voltage in a capacitor. Below, we present this equation.

From the statement of the problem, it is known that when C₃ → ∞ , the voltage across the capacitor C₁ approaches the value of 10 volts. If C₃ → ∞, it can be said that the association parallel of C₂ and C₃ is equal to C₃ . Then the equivalent capacitance of the entire circuit will be equal to the value of C₁, as the capacitors are in series. And since the equation shown above should be satisfied, obviously it can be said that the voltage V₂₃ drops asymptotically to zero. Then all the tension of the voltage source will be applied to the capacitor C₁. Therefore, it is concluded that:

Item b

To answer this item, two particularities must be taken into account:

1) Capacitors in series have equal charges, that is, q₂₃ = q₁.

2) From the graph provided in the problem, it can be seen that when C₃ = 6 µF the voltage across the capacitor C₁ is V₁ = 5 volts. So it's obvious that V₂₃ = 5 volts. Now, notice that if the charges are equal on the capacitors and the voltage across them is also equal, it follows that:

In other words: the capacitance of the parallel association of C₂ and C₃ must be equal to capacitance of C₁.

Another information that can be taken from the graph: when C₃ = 0 we have V₁ = 2 volts. Well, if V₁ = 2 volts then V₂₃ = 8 volts, because the sum of these voltages must be equal to voltage from the source, (V = 10 volts). And, once again, as the charges must be equal, that is, q₁ = q₂, it follows that:

Doing the simplification:

Therefore, a system of two equations with two unknowns with an extremely easy solution was obtained. Therefore, the values ​​of C₁ and C₂ are: