Example 18-11 A Charging Series RC Circuit
A 10.0-M resistor is connected in series with a 5.00-mF capacitor. When a switch is thrown, these circuit elements are
connected to a 24.0-V battery of negligible internal resistance. The capacitor is initially uncharged. (a) What is the current
in the circuit immediately after the switch is moved so that charging begins? (b) What is the charge on the capacitor once it
is fully charged? (c) Find the capacitor charge, current, power provided by the battery, power taken in by the resistor, and
power taken in by the capacitor at t = 50.0 s. (d) When the capacitor is fully charged, find the total energy that has been
delivered by the battery and the total energy that has been delivered to the capacitor.
Set Up
We are given R = 10.0 M = 10.0 * 106 ,
C = 5.00 mF = 5.00 * 1026 F, and e = 24.0 V.
Equations 18-27 tell us the capacitor charge
and current at any time, including at t = 0
(when the switch is first closed) and t S 
(long after the switch is closed, so the capacitor is fully charged). We’ll use Equations 18-23
and 18-24 to find the power out of the battery
and into the resistor and capacitor. (Equation
17-14 will help us in this.) In order to charge
the capacitor to its maximum charge qmax, the
total charge that must pass through the battery
is qmax; we’ll use this and Equation 17-6 to
find the total energy delivered by the battery.
Equation 17-17 tells us the total energy that is
stored in the charged capacitor.
Capacitor charge and current
in a charging series RC circuit:
q1t2 = Ce11 - e -t>RC 2
e
i1t2 = e -t>RC
R
= 24.0 V
R = 10.0 MΩ
switch closed
at t = 0
C = 5.00 µF
(18-27)
Power for a circuit element:
(18-23)
P = iV
Power for a resistor:
V2
(18-24)
R
Charge, voltage, and capacitance for a capacitor:
P = i 2R =
(17-14)
q = CV
Electric potential difference related to electric
potential energy difference:
V =
Uelectric
q0
(17-6)
Electric potential energy stored in a capacitor:
Uelectric =
Solve
(a) Find the current at t = 0.
q2
1
1
qV = CV 2 =
2
2
2C
(17-17)
From the second of Equations 18-27, the current when the switch is
first closed at t = 0 is
i102 =
e -102>RC
e
e
= e0
R
R
Since any number raised to the power 0 equals 1, we have e0 = 1 and
e
24.0 V
=
R
10.0 * 106 
= 2.40 * 1026 A = 2.40 mA
i102 = i max =
(b) Find the capacitor charge long after the
switch is closed (t S  ).
The first of Equations 18-27 tells us the capacitor charge q(t).
As t S  , the exponent -t>RC S -  . Any number raised to
the power -  is zero, so
e -t>RC S 0
q1t2 = Ce11 - e -t>RC 2 S q max = Ce11 - 02 = Ce
= (5.00 * 1026 F)(24.0 V)
= 1.20 * 1024 C = 0.120 mC
(c) The time constant for this circuit is RC =
50.0 s, so we are actually being asked about the
behavior of the circuit one time constant after
the switch is closed. Use this to find charge q,
current i, and the power out of or into each
circuit element.
The time constant for this circuit is
RC = (10.0 * 106 )(5.00 * 1026 F) = 50.0 s
so at t = 50.0 s, t>RC = 150.0 s2 > 150.0 s2 = 1.00
From Equations 18-27,
q = Ce(1 2 e21.00) = (5.00 * 1026 F)(24.0 V)(1 2 0.368)
= 7.58 * 1025 C = 0.632qmax
e
24.0 V
i = e -1.00 =
10.3682
R
10.0 * 106 
= 8.83 * 1027 A = 0.368imax
The voltage across the battery is e = 24.0 V, so from Equation 18-23
the power out of the battery is
Pbattery = ie = (8.83 * 1027 A)(24.0 V)
= 2.12 * 1025 W = 21.2 mW
From the first of Equations 18-24, the power into the resistor is
PR = i2R = (8.83 * 1027 A)2(10.0 * 106 )
= 7.80 * 1026 W = 7.80 mW
Equation 17-14, q = CV, tells us that the voltage across the capacitor
is V = q>C. Combining this with Equation 18-23 gives the power
into the capacitor:
q
7.58 * 10-5 C
b = 18.83 * 10-7A2 a
b
C
5.00 * 10-6 F
= 1.34 * 1025 W = 13.4 mW
PC = i a
Note that the net power into the resistor and capacitor combined
equals the power out of the battery:
PR + PC = 7.80 mW + 13.4 mW = 21.2 mW = Pbattery
(d) Use the maximum charge stored by the
­capacitor, which is the total charge moved
across the battery, and Equation 7-6 to calculate the change in electric potential energy
imparted by the battery. Use Equations 17-7 to
calculate the electric potential energy stored in
the capacitor.
Long after the switch is closed, the total amount of charge that
has passed through the battery and to the positive capacitor plate
is qmax = 1.20 * 1024 C. From Equation 17-6, the potential energy
change that was imparted by moving this charge across the 24.0-V
emf of the battery is
Ubattery = qmax e = (1.20 * 1024 C)(24.0 V)
= 2.88 * 1023 J = 2.88 mJ
The last of Equations 17-17 tells us the amount of energy that went
into the capacitor to store charge qmax there:
11.20 * 10-4 C2 2
q 2max
=
2C
215.00 * 10-6 F2
= 1.44 * 1023 J = 1.44 mJ
UC =
Reflect
So exactly one-half of the energy taken from the battery goes into the
capacitor: UC = 11>22Ubattery.
Our results for charge q and current i in part (c) agree with Figure 18-18: After one time constant the capacitor charge has
reached 3 1 - 11>e2] = 0.632 = 63.2% of its fully charged value qmax, and the current has decreased to 1>e = 0.368 =
36.8% of its initial value imax. Can you show that after 5 time constants (t = 5RC = 250 s) the charge will have reached
99.3% of qmax and the current will have decreased to just 0.674% of imax?
The power calculations in part (c) show that all of the power extracted from the battery is accounted for: Part of
the energy extracted from the battery goes into the resistor, and the rest goes into adding to the electric potential energy
stored in the capacitor. We’ve shown this for a specific instant, but it’s true at all times during the charging process.
These results from part (c) also help us understand our calculations in part (d): Only one-half of the energy extracted from the battery goes into the capacitor, so the other half must have gone into the resistor. This is a general result for
charging any series RC circuit.