Example 18-9 Power in a Single-Loop Circuit
A battery with emf 12.0 V and internal resistance 1.00  is connected to a resistor with resistance 19.0 . Find (a) the
rate at which energy is supplied by the emf, (b) the rate at which energy flows into the internal resistance, and (c) the rate
at which energy flows into the resistor.
Set Up
We can find the current in the circuit using
Kirchhoff’s loop rule (the sum of the changes
in potential around a closed loop in a circuit
= 12.0 V
Relationship among potential
difference, current, and resistance: a
must equal zero). Equation 18-9 then tells us
the voltage across either resistance. We’ll use
Equation 18-23 to determine the power for
each element of the circuit, and check our
results for the two resistances using Equations 18-24.
Power for a circuit element:
Solve
(a) First apply Kirchhoff’s loop rule to determine the current i. Start at the point a in the
circuit and go around in the direction of the
current.
r = 1.00 Ω
(18-9)
V = iR
R = 19.0 Ω
i
(18-23)
P = iV
Power for a resistor:
P = i 2R =
V2
R
(18-24)
There is a voltage rise of e = 12.0 V going through the emf,
a voltage drop of ir going through the internal resistance r =
1.00 , and a voltage drop of iR going the resistor with resistance
R = 19.0 . The sum of the potential changes around the circuit
is zero:
+e + (2ir) + (2iR) = 0
Rearrange this to solve for the current i:
e 2 i(r + R) = 0, so i(r + R) = e and
i =
e
12.0 V
12.0 V
=
=
= 0.600 A
r + R
1.00  + 19.0 
20.0 
(Recall that 1 A = 1 V> .)
Use Equation 18-23 to find the power extracted
from the emf.
The voltage across the emf itself is e = 12.0 V. The rate at which
energy is extracted from the emf is
Pemf = ie = (0.600 A)(12.0 V) = 7.20 A # V
Since 1 A = 1 C>s, 1 V = 1 J>C, and 1 W = 1 J>s,
Pemf = 7.20
(b) First use Equation 18-9 to find the potential
difference across the internal resistance r =
1.00 . Then find the power in the internal
resistance using Equation 18-23.
J
C# J
= 7.20 = 7.20 W
s C
s
From Equation 18-9, the voltage across the internal resistance is
Vr = ir = (0.600 A)(1.00 ) = 0.600 V
From Equation 18-23, the rate at which energy flows into the
internal resistance is
Pr = iVr = (0.600 A)(0.600 V) = 0.360 W
This power into the internal resistance goes into heating the
battery.
(c) Repeat part (b) for the resistor of
resistance R = 19.0 .
The voltage across the resistor is
VR = iR = (0.600 A)(19.0 ) = 11.4 V
The rate at which energy flows into the resistor is
PR = iVR = (0.600 A)(11.4 V) = 6.84 W
Reflect
Note that the rate at which energy is extracted
from the source of emf is equal to the net rate
at which energy flows into the internal resistance and the resistor. This is equivalent to saying that energy is conserved in the circuit.
The net rate of energy flow into the two resistances is
We can check our result PR = 6.84 W for the
power into the resistor by showing that we get
the same results using Equations 18-24. Can
you use the same approach to check the result
Pr = 0.360 W for the internal resistance?
From the first of Equations 18-24, the power into the 19.0-
resistor is
Pr + PR = 0.360 W + 6.84 W = 7.20 W
This is the same as the rate at which energy flows out of the source
of emf, Pemf = 7.20 W.
PR = i2R = (0.600 A)2(19.0 ) = 6.84 W
(Note that 1 A2 #  = 1 A # V = 1 W.)
To use the second of Equations 18-24, use the voltage VR = 11.4 V
across the resistor:
PR =
111.4 V2 2
V 2R
=
= 6.84 W
R
19.0 
(Note that 1 V 2 >  = 1 V # 1V> 2 = 1 V # A = 1 W.)
This agrees with the result for PR found above.