How Does Closing a Switch Affect Capacitors and Resistors in a Circuit?

[JosephK]

Homework Statement

A charge Q is placed on a capacitor of capacitance C. The capacitor is connected into the circuit as shown in the figure below, with an open switch, a resistor, and an initially uncharged capacitor of capacitance 3C. The switch is then closed and the circuit comes to an equilibrium.

(a) find the final potential difference between the plates of each capacitor.
(b) Find the charge on each capacitor.
(c) Find the final energy stored in each capacitor.
(d) Find the internal energy appearing in the resistor.

Homework Equations

E = 1/2 C ΔV^2
U = QΔV

The Attempt at a Solution

As we move across the charged capacitor from negative to positive, the capacitor increases the electric potential energy by exactly E= 1/2 C V^2. Energy is delivered to the resistor and the uncharged capacitor.



Conservation of energy:

$1/2 C(\Delta V_C^{2})= Q(\Delta V_R) + 1/2 (3C)(\Delta V_{3C}^{2})$

In equilibrium, $V_C = V_R+V_{3C}$
Where V_C, V_R and V_C is the voltage across the initially charged capacitor, resistor, and initially uncharged capacitor respectively.

 

[gneill]

A complete solution is offered:Part (a): Final potential difference between the plates of the capacitors
After the switch is closed and the system reaches steady state, current has ceased to flow. There is no potential drop across the resistor (no current) and the capacitors have the same potential difference. Effectively the capacitors are connected in parallel.

Since charge is conserved, charge Q will be on an effective capacitance of C + 3C, or 4C. Since the potential across a capacitor is given by V = Q/C, we have:

$V_c = \frac{Q}{4C}$

This will be the potential difference between the places of each capacitor.

Part (b): The charge on each capacitor
Using the same relationship, V = Q/C, and rearranging it as Q = CV we find the charge on each capacitor:

On the 'C' capacitor: $Q_C = C⋅\frac{Q}{4C} = \frac{1}{4}Q$

On the '3C' capacitor: $Q_{3C} = 3C⋅\frac{Q}{4C} = \frac{3}{4}Q$

Part (c): The energy stored on each capacitor
Two common expressions used for the energy stored on a capacitor are:

$E = ½ C V^2~~~~~~$ and $~~~~~~E = ½ \frac{Q^2}{C}$

We can choose either since we have previously found both the voltage and the charge on each capacitor. Both will yield:

On the 'C' capacitor: $E_C = \frac{Q^2}{32 C}$

On the '3C' capacitor: $E_{3C} = \frac{3Q^2}{32 C}$

Part (d): The "internal energy" appearing in the resistor
This will be given by the difference in energy between the initial state and the final state of the system. That is:

$ΔE = \left\{ \frac{1}{2}\frac{Q^2}{C} \right\} - \left\{ \frac{Q^2}{32 C} + \frac{3~Q^2}{32 C} \right\} = \frac{3}{8} \frac{Q^2}{C}$