PHY294H - Lecture 17

Capacitor/Resistor networks

Kirchhoff's laws still apply and are the basis for analysis of RC circuits. In analysing these circuits we need to relate the charge on the capacitors to the current flow into the capacitor. The relationship we need is simply i = dQ/dt. This is combination with Kirchhoff's laws enable us to solve RC circuits. We shall be interested in the way that capacitors in a circuit charge and discharge. We shall also be interested in the long time behavior (steady state) in which each capacitor has reach a final charge state. To study charging, we will close a switch in the circuit and will analyse what happens as charge either flows onto or off of the capacitors.

A resistor and capacitor in series

Consider Kirchhoff's loop law for a capacitor, C, resistor R and an emf $\cal{E}$ in series. At long times, this circuit carries no current and the emf is dropped across the capacitor. The circuit is only interesting if we have a switch in the circuit which we can open and close as we wish. Analysis of the time dependence of the current in the circuit and the charge across the capacitor is a study of transients.

Charging a capacitor

Consider placing a switch in the circuit and the situation in which the capacitor initially has charge Q(t=0) = 0. We close the switch at t=0 and the capacitor begins to charge.

$${\cal{E}} - iR - V_c = {\cal{E}} - iR - {Q\over C} = 0.$$ (1)

In order to proceed we use the relation between current and charge, ie. i = dQ/dt. From this we find,

$${\cal{E}\over R} - {dQ \over dt} - {Q\over R C} = 0$$ (2)

In the steady state $t\rightarrow \infty$, i = dQ/dt=0 and ${\cal{E}} = Q/C$. The time dependence of the charging of the capacitor is given by the solution to Eq. (2), which is,

$$Q(t) = {\cal{E} C} (1 - e^{-t/\tau})$$ (3)

where the time constant $\tau = RC$. Note that Q(0) = 0 and that $Q(t\rightarrow\infty) = \cal{E} C$. The current as a function of time is given by,

$$i = {dQ\over dt} = {{\cal{E}} C\over \tau } ( e^{-t/\tau}) = {{\cal{E}}\over R } e^{-t/\tau}$$ (4)

Discharging a capacitor

Now consider a capacitor which has charge Q₀ in a circuit with a resistor and an open switch. The switch is closed at t=0. Find the current in the circuit and the charge on the capacitor as a function of time. Kirchhoff's loop law gives,

$$V_c - I R = 0 = {Q\over C} - R {dQ\over dt}$$ (5)

The solution to this equation is,

$$Q(t) = Q(0) e^{-t/\tau}$$ (6)

The current in the circuit is,

$$i(t) = -{Q(0)\over \tau } e^{-t/\tau} = -{Q(0)\over RC} e^{-t/\tau}$$ (7)

Energy storage and dissipation in an RC circuit

In the charging process above, the energy supplied by the emf (battery) is

$$W_{battery} = Q(\infty) {\cal{E}} = {C \cal{E}}^2.$$ (8)

half of this energy is stored in the capacitor and half of this energy is dissipated in the resistor.

$$E_c = {1\over 2} C {\cal{E}}^2$$ (9)

The energy dissipated in the resistor is

$$E_R = \int_0^{\infty} I^2 R dt = \int_0^{\infty} ({{\cal{E}}\over R })^2 e^{-2t/\tau} R dt ={C {\cal{E}}^2 \over 2}$$ (10)

FGT 28.59

Consider an emf $\cal{E}$ in parallel with a resistor R₂ and in parallel with a series combination of a resistor R₁ and a capacitor C. A switch is placed near the emf. At times t<0 the capacitor is uncharged and the switch is open. At t=0 the switch is closed. Find an expression for the current through the battery, I(t), as a function of time and the charge on the capacitor Q₁(t), also as a function of time.

Kirchhoff's junction law gives,

I = I₁ + I₂ (11)

where I₁ and I₂ are the currents through resistors R₁ and R₂ respectively. Kirchhoff's loop law gives,

$${\cal{E}} - V_c - I_1R_1 = {\cal{E}} - {Q_1\over C} - I_1R_1 = 0$$ (12)

and,

$${\cal{E}} - I_2R_2 = 0$$ (13)

For times t<0, there is no current in the circuit as the switch is open. At time t=0 the switch is closed. At that time, there is no charge on the capacitor, so there is no voltage across the capacitor. Therefore the circuit acts like a parallel combination of resistors R₁ and R₂. We thus have,

$$I_1(0) = {{\cal{E}} \over R_1} \ ; \ \ I_2(0) = {{\cal{E}}\ov... ...= {{\cal{E}}\over R_1} + {{\cal{E}}\over R_2}\ ;\ \ Q(0) = 0.$$ (14)

At long times, the capacitor is fully charged and we have,

$$I_1(\infty) = 0\ ;\ \ I_2(\infty) = I(\infty) = {{\cal{E}}\over R_2} \ ;\ \ Q_1(\infty) = C{\cal{E}}$$ (15)

To find the time dependence of the current I₁ we use I₁ = dQ₁/dtin Eq. (12). This yields,

$${\cal{E}} - {Q_1\over C} - R_1{dQ_1\over dt} = 0$$ (16)

This has the solution,

$$Q_1(t) = {\cal{E}} C(1-e^{-t/R_1C})$$ (17)

The current I₁(t) is then,

$$I_1(t) = {dQ_1\over dt} = {{\cal{E}}\over R_1}e^{-t/R_1C}$$ (18)

Using Kirchhoff's junction rule, we then find,

$$I(t) = I_1 + {{\cal{E}}\over R_2} = {{\cal{E}}\over R_2} + {{\cal{E}}\over R_1}e^{-t/R_1C}$$ (19)