Example 18-10 Power in Series and Parallel Circuits
Two resistors, one with R1 = 2.00  and one with R2 = 3.00 , are both connected to a battery with emf e = 12.0 V and
negligible internal resistance. Find the power delivered by the battery and the power absorbed by each resistor if the resistors are connected to the battery (a) in series and (b) in parallel.
Set Up
We’ll use the same tools as in the previous
example, plus Equations 18-11 and 18-20
for the equivalent resistance of resistors in
series and in parallel. In each case we’ll use
the equivalent resistance to find the current
through the battery, then use Equation 18-23 to
determine the power provided by the battery. If
the resistors are in series, the current through
each is the same as the current through the
battery, so we’ll use the first of Equations 18-24
(P = i 2 >R) to determine the power into each
resistor. For resistors in parallel the voltage is
the same across each resistor, so in that case
we’ll find the power into each resistor using the
second of Equations 18-24 (P = V 2 >R).
series:
Relationship among potential
difference, current, and resistance: = 12.0 V
(a) The two resistors in series are equivalent to
a single resistor Requiv. This is connected directly
to the terminals of the battery, across which the
voltage is e (we’re told to ignore the internal
resistance). So the voltage across Requiv is also
equal to e, which tells us the current through
the circuit. Equation 18-23 then tells us the
power delivered by the battery.
R2 = 3.00 Ω
(18-9)
V = iR
Equivalent resistance of two
resistors in series:
Requiv = R1 + R2
(18-11)
Equivalent resistance of two
resistors in parallel:
1
1
1
=
+
R equiv
R1
R2
parallel:
R1 = 2.00 Ω
= 12.0 V
R2 = 3.00 Ω
(18-20)
Power for a circuit element:
(18-23)
P = iV
Power for a resistor:
P = i 2R =
Solve
R1 = 2.00 Ω
V2
R
(18-24)
From Equation 18-11, the equivalent resistance of the two resistors in
series is
Requiv = R1 + R2 = 2.00  + 3.00  = 5.00 
The voltage drop across this equivalent resistance is V = iRequiv from
Equation 18-9, which is also equal to the emf e = 12.0 V of the battery. So the current through the equivalent resistance is given by
e = iR equiv or i =
e
12.0 V
=
= 2.40 A
R equiv
5.00 
This is also the current through the battery. Since the voltage across
the battery is equal to e = 12.0 V, the power delivered by the battery is
Pbattery = ie = (2.40 A)(12.0 V) = 28.8 W
This must be equal to the power that goes into the two resistors.
The current i = 2.40 A is the same through
both resistors in this series circuit. Use this to
calculate the power that goes into each resistor.
Using the first of Equations 18-24, the power into resistor
R1 = 2.00  is
P1 = i2R1 = (2.40 A)2(2.00 ) = 11.5 W
The power into resistor R2 = 3.00  is
P2 = i2R2 = (2.40 A)2(3.00 ) = 17.3 W
The net power into the two resistors is equal to the power supplied
by the battery, as it should be:
P1 + P2 = 11.5 W + 17.3 W = 28.8 W = Pbattery
(b) Follow the same steps as in part (a) to find
the equivalent resistance of the two resistors in
parallel, the current through the battery connected to that parallel arrangement, and the
power delivered by the battery in this situation.
From Equation 18-20, the equivalent resistance of the two resistors in
parallel is given by
1
1
1
1
1
+
= 0.833  -1
=
+
=
R equiv
R1
R2
2.00 
3.00 
1
R equiv =
= 1.20 
0.833  -1
Equation 18-9 tells us that the voltage drop across this equivalent
­resistance is V = iRequiv; since this equivalent resistance is connected
to the terminals of the battery, the voltage drop is also equal to the
battery emf e = 12.0 V. So the current through the equivalent resistance and through the battery is given by
e = iR equiv or i =
e
12.0 V
=
= 10.0 A
R equiv
1.20 
From Equation 18-23, the power delivered by the battery is
Pbattery = ie = (10.0 A)(12.0 V) = 1.20 * 102 W
Note that this is more than four times as much power as the same
battery delivers to the resistors in series.
The voltage is the same across resistors in
parallel. Each resistor is effectively connected
directly to the terminals of the battery, so the
voltage across each resistor is V = e = 12.0 V.
Use this to calculate the power that goes into
each resistor.
Using the second of Equations 18-24, we find that the power into
resistor R1 = 2.00  is
P1 =
112.0 V2 2
V2
e2
=
=
= 72.0 W
R1
R1
2.00 
The power into resistor R2 = 3.00  is
P2 =
112.0 V2 2
V2
e2
=
=
= 48.0 W
R2
R2
3.00 
As for the series case, the net power into the two resistors is equal to
the power supplied by the battery:
P1 + P2 = 72.0 W + 48.0 W = 1.20 * 102 W = Pbattery
Reflect
Although the same battery and same resistors are used in both circuits, the power provided by the battery is much different in the two circuits. That’s because the current through the battery is different for the two circuits: i = 2.40 A for the
series circuit, i = 10.0 A for the parallel circuit.
What’s more, the power into each resistor is
very different in the two circuits, because the
voltage and current for each resistor are greater
for the parallel circuit than for the series circuit.
If the resistors are light bulbs that take in
power and use it to produce light, the lights
V1 = iR1 = (2.40 A)(2.00 ) = 4.80 V
V2 = iR2 = (2.40 A)(3.00 ) = 7.20 V
(compared to V = e = 12.0 V for each resistor in the parallel circuit)
will glow brighter in the parallel circuit than in
the series circuit.
Use Equation 18-9 to find the current through each resistor in the
parallel circuit:
Use Equation 18-9 to find the voltage across each resistor in the
series circuit:
e
12.0 V
=
= 6.00 A
R1
2.00 
e
12.0 V
i1 =
=
= 4.00 A
R2
3.00 
(compared to i = 2.40 A for each resistor in the series circuit)
i1 =
Note that the current is the same for both resistors in the series circuit, so more power goes into the resistor with the
greater resistance (P2 = 17.3 W into R2 = 3.00  versus P1 = 11.5 W into R1 = 2.00 ). That follows from the relationship P = i2R for resistors. By contrast, the voltage is the same for the two resistors in the parallel circuit, so more power
goes into the resistor with the smaller resistance (P1 = 72.0 W into R1 = 2.00  versus P2 = 48.0 W into R2 = 3.00 ).
That agrees with the relationship P = V 2 >R for resistors. So if the resistors are light bulbs, the light bulb with R2 =
3.00  is the brighter one in the series circuit, but the bulb with R1 = 2.00  is the brighter one in the parallel circuit!
When comparing the power in resistors, choose wisely among the relationships P = iV, P = i2R, and P = V 2 >R.