6.2.1  RC Circuits

When we combine resistors and capacitors in a circuit, we have to take account of time in a serious way for the first instance of our study of electrodynamics. The reason is that capacitors store charge by having work done on them and then release that energy by doing work themselves. The time rate of change for the charge on the capacitor's plates depends on the current, the time rate of change of charge, that is impinging on or leaving the plates of the capacitor. This current, in turn, has a well-defined value only in the presence of some specified resistance. Hence, even though we could describe the process of charging a capacitor before knowing about resistance, we could not quantitatively describe it before knowing about resistance so that we can specify the current and how it changes with time. Let's examine the specific case of a capacitor discharging through a resistor.

Figure 6.10: A charged capacitor which can be connected to a resistor through a switch S.

At time t = 0, the switch S is closed. Before that time, the capacitor has an initial charge q₀  on its plates and no current flows because the circuit is not complete. When S is closed, the capacitor's potential difference across its plates sends positive current through the resistor in the clockwise direction. The positive charge, after going through the resistor, lands on the negative plate of the capacitor and begins to neutralize some of the negative charge on that plate. Since a potential difference exists as long as any charge remains on the capacitor and since a current flows through the resistor as long as any nonzero potential difference exists across it, this process of discharging continues until the capacitor has no charge left on its plates. Since the resistor dissipates energy thermally due to the current being pushed through it, and energy is conserved, it is reasonable to expect that the energy appearing as heat in the resistor comes from the initial store of potential energy in the capacitor. We can use this fact to determine how the charge on the capacitor changes as a function of time. First, consider that the energy stored in the capacitor at any time is

UC = 1 2 q2 C

(6.2.1.18)

The rate at which energy is dissipated in the resistor is

PR = I2R

(6.2.1.19)

Note that all the current going through the resistor represents charge per unit time lost from the capacitor, so, we have two assumptions

dq dt = I    and   PR = - dUC dt

(6.2.1.20)

where the minus sign for P_R represents that energy is being lost from the capacitor. Let's make use of these to produce a differential equation, then solve that equation in the usual fashion.

PR = - dUC dt I2R = d dt æ ç è - 1 2 q2 C ö ÷ ø = - q C dq dt I2R = - q C I Þ I = dq dt = - q RC Þ - RC q dq = dt -RC ó õ q(t¢) q0  dq q = ó õ t¢ 0  dt -RClnq|q0q(t¢) = t¢ -RCln q(t¢) q0 = t¢Þ q(t¢) = q0e-t¢/RC (6.2.1.21)

where t¢ represents the time after the switch S is closed. The formula does indeed predict that the charge on the capacitor goes to zero after a long time.

6.2.2  RC Charging

We can now ask what happens when a capacitor is charged. Consider the picture below.

Figure 6.11: A capacitor, resistor, and EMF source which can be all connected by the switch S.

The capacitor is initially uncharged and the switch S is open. When the switch is closed, the EMF source begins pushing charge through the resistor and onto the capacitor top plate (assuming positive charges are being pushed). If we follow the same line of reasoning as in the discharging case, then we must have the rate of energy expended by the battery equal the energy dissipated as heat in the resistor plus the rate at which energy is stored in the capacitor. The current through the resistor (and through the battery) must equal the rate at which charge collects on the capacitor plates, thus

I = dq dt     and   Pb = PR + dUC dt

(6.2.2.22)

We now produce and solve the corresponding differential equation

Pb = PR + dUC dt IVB = I2R + d dt æ ç è q2 2C ö ÷ ø IVb = I2R + q C dq dt Vb = dq dt R + q C CVb = dq dt RC + q Þ CVb - q RC = dq dt Þ dt = RC dq CVb - q ó õ t¢ 0  dt = RC ó õ q(t¢) 0  dq CVb - q t¢ = -RC ó õ CVb - q(t¢) CVb  du u -t¢ = RClnu|CVbCVb - q(t¢) Þ - t¢ RC = ln CVb - q(t¢) CVb Þ CVbe-t¢/RC = CVb - q(t¢) Þ q(t¢) = CVb(1 - e-t¢/RC) (6.2.2.23)

where again t¢ represents the time after the switch S is closed. Now note some interesting properties to this solution. First, it predicts that the charge on the capacitor at t¢ = 0 is zero. That's good. For a long time later (t¢ approaches infinity), we find that the charge approaches CV_b. That's good also as we know that current will flow through the circuit until the potential across the capacitor equals the potential difference across the poles of the EMF source. When this happens, we expect the charge on the capacitor to be CV_b  as predicted. The quantity RC is referred to as the time constant for the circuit.

6.2.3  RC circuits applied

The equations for charging and discharging above stand as they are for calculating the charge as a function of time for charging or discharging RC circuits. Note that, if we can reduce a circuit down to equivalent resistors and capacitors, then the formulas above are modified just to reflect this. Note the examples below

Figure 6.12: Examples of various R and C combinations and their corresponding time constant values.

More often in exam problems you are not asked to do calculations with the specific formulas as given, but to understand the physics of the situation both just after and a long time after some change has been made to the circuit, e.g. opening or closing a switch. For such problems, what is important to remember is that the specific formula to apply may not be easily derived, but that the behavior of a charging or discharging capacitor remains the same no matter what. If a capacitor is initially uncharged, then it allows current to``pass" (actually the current deposits charge on one plate and the resulting electric field induces a current in the opposite plate) with no opposition in the instant that the current is first applied. After a long time, a capacitor charges up to produce whatever potential difference is needed across its plates so as to prevent any more current from flowing onto it. Consider the following circuit

Figure 6.13: A circuit with an initially uncharged capacitor and open switch.

The capacitor is initially uncharged. You are asked to find the current through resistor R₂ both immediately after the switch S is closed and a long time after it is closed. The solution is easy to derive if you reason as follows.

1. Before switch S is closed, no current flows through the circuit at all. There is no closed path which allows any EMF source to deliver charges from negative to positive pole (remember that initially the capacitor is uncharged and therefore has no potential difference across its place and hence is not an EMF source).
2. the capacitor initially gives no opposition to the current through the leg of the circuit containing R₃  immediately after the switch S is closed.
3. the capacitor is fully charged a long time after switch S is closed and therefore allows no current to pass through the leg containing R₃.

With these in mind, it is easy to see that the equivalent circuits for the time immediately after and a long time after switch S  is closed look as follows.

Figure 6.14: The effective appearance of the circuit in figure 6.13 just after and a long time after the switch S is closed.

Now it is clear how to find the current through each of the resistors as in previous cases of resistors and EMF source circuits.

Send comments to larryg@upenn5.hep.upenn.edu.
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