RC Circuits
I. Kirchhoff’s Rules
Circuits are combinations of closed closed loops through which charges
can continuously move. In Electromagnetism, we use Kirchhoff’s Rules
to reduce a circuit to a single loop:
The Junction rule is best explained as follows:
The loop rule is used along with Ohm’s Law ∆V = IR to reduce the
circuits as follows:
In a single loop circuit only the loop rule is needed to find the current
in the circuit, for instance:
2
In a more complicated circuit, we use for of Kirchoff’s rules to reduce
the system.
In order to find currents I1 , I2 , I3 , we set up one equation for each unknown.
3
II. Charging a Capacitor
Consider a simple RC circuit with a capacitor that is initially uncharged.
4
Applying Kirchhoff’s loop rule to a counterclockwise loop:
q
Σ∆V = 0 =⇒ ε − IR − = 0.
C
dq
Using the definition of current, I =
, we obtain the following differdt
ential equation for the charge q on the capacitor:
dq
q
ε
q − Cε
=
− =−
dt
RC R
RC
To solve (1), we separate the variables to obtain:
(1)
dq
dt
=q − Cε
RC
−t
−t
Exercise 1: Show that the solution (1) is q(t) = Cε(1−e τ ) = Q(1−e τ ),
where τ = RC is the time constant of the circuit. It represents the time
interval during which the current decreases to 1/e of its initial value. Q
= Cε represents the maximum charge on the capacitor.
Exercise 2: Derive the expression from Exercise 1 to obtain:
ε
−t
I(t) = Ii e τ , where Ii =
is the current in the circuit at time t = 0.
R
III. Discharging a Capacitor
Imagine that the capacitor is now completely charged. A potential difference Q/C exists across the capacitor, and there is zero potential difference across the resistor because. If the switch is now thrown to position
5
b at t = 0, the capacitor begins to discharge through the resistor. At
some time t during the discharge, the current in the circuit is I and the
charge on the capacitor is q. The circuit is the same as the circuit when
charging a capacitor except for the absence of the battery. Therefore,
we eliminate the emf ε from Equation (1) to obtain the appropriate loop
equation for the circuit:
q
− − IR = 0
(2)
C
dq
q
dq
=⇒ −R =
Again using, I =
dt
dt
C
We separate the variables once again:
dq
dt
=q
RC
t
Exercise 3: Show that the solution (2) is q(t) = Qe− τ .
Exercise 4: Derive the expression from Exercise 3 to obtain:
Q
t
I(t) = −Ii e τ , where Ii =
is the initial current. The negative sign inRC
dicates that as the capacitor discharges, the current direction is opposite
its direction when the capacitor was being charged.
IV. More exercises
Exercise 5:
In the circuit illustrated above, switch S is initially open and the battery
has been connected for a long time. (Hint: Capacitors are fully charged)
6
(a) What is the steady-state current through the ammeter?
(b) Calculate the charge on the 10 mF capacitor.
(c) Calculate the energy stored in the 5.0 mF capacitor.
(d) The switch is now closed, and the circuit comes to a new steady
state. Calculate the steady-state current through the battery.
(e) Calculate the final charge on the 5.0 mF capacitor.
(f) Calculate the energy dissipated as heat in the 40 W resistor in one
minute once the circuit has reached steady state. (Hints: Power =
Energy/time and If a potential difference ∆V is maintained across a
circuit element, the power, or rate at which energy is supplied to the
V2
element, is P =IV =
= I 2 R. Because the potential difference
R
across a resistor is given by ∆V = IR, we can express the power
V2
= I 2 R. The energy delivered
delivered to a resistor as P = IV =
R
to a resistor by electrical transmission appears in the form of internal
energy in the resistor.)
Exercise 6: Problem 32 on page 822 of your book.
Exercise 7: Problem 37 on page 823 of your book.
Exercise 8: Problem 49 on page 824 of your book.
7