RC Circuits
Suppose we connect a C in series with an R and a battery or power supply and include a
switch which can be closed to complete the circuit. We wish to study the properties of this
circuit when the switch is closed. In particular we will see that an electric current will only flow
briefly in this circuit when the switch is closed at time = 0.
Charging a Capacitor:
C
R
+
_
∆V=ε
To analyze this circuit we can start at the switch and imagine, when it is closed, going around the
circuit in a clockwise direction and adding up all the voltage increases and decreases. Upon
returning to the starting point, the net voltage difference around the entire loop must be zero. We
call this Kirchhoff’s loop rule and for this circuit it is written as
Q
+ε − − IR = 0 ,
C
where Q/C is the voltage drop across the capacitor and IR is the voltage drop across the resistor.
Are you clear why both of these are drops?
From this equation we can see the main features of the circuit. At time zero, the
capacitor is initially uncharged and so the initial voltage across it is zero. Then at t = 0 the above
equation yields an initial current Io = ε/R. As time goes on the charge on the capacitor increases
as positive charges pile onto the left plate, making the right plate negative since positive charges
will leave the right plate. The final situation will have I = 0 and Qf = Cε. So what must happen
is that the electric current decreases over time and the charge on the capacitor increases with
time. But exactly how? To find this out we need to solve the above equation for Q and I at any
time.
Substituting the definition of I = dQ/dt into the above equation,
Q
dQ
dQ ε
Q
Q − Cε
ε − −R
= 0 , and re-arranging this we have
= −
=−
. We can
C
dt
dt R RC
RC
solve this equation for Q(t) by re-writing it and integrating once:
dQ
1
=−
dt , so that on integrating and noting that Q = 0 at t = 0,
(Q − Cε )
RC
Q (t )
∫
0
t
dQ
1
t
 Q(t ) − Cε 
=−
dt or ln 
=−
∫
(Q − Cε )
RC 0
RC
 −Cε 
Solving for Q(t), we find Q(t ) = Cε (1 − e− t / RC ) = Q f (1 − e− t / RC ) , where Qf = Cε.
To find I(t) we can simply differentiate: I (t ) =
current in the circuit (since Q(0) = 0 on C.
ε − t / RC
e
= I o e − t / RC , where Io = ε/R, the initial
R
Note that the time constant RC is the time for the current to drop to 1/e of its starting value Io, or
0.368 Io. Similarly 2RC is the time for the current to drop another factor of 0.368 x I(t = RC) =
(0.368)2Io = Ioe-2. Etc,. Plots of I(t) and Q(t) vs t (see p. 649) We will study this in detail in the
lab this week.