D
IRECT
C
URRENT
C
IRCUITS
26.7.I
DENTIFY
:
First do as much seriesparallel reduction as possible.
S
ET
U
P
:
The 45.0
Ω and 15.0

Ω resistors are in parallel, so fi
rst reduce them to a single equivalent resistance.
Then find the equivalent series resistance of the circuit.
E
XECUTE
:
1/
R
p
= 1/(45.0 Ω) + 1/(15.0 Ω) and
R
p
= 11.25 Ω. The total equivalent resistance is
18.0 Ω + 11.25 Ω + 3.26 Ω = 32.5 Ω. Ohm’s law gives
I
= (25.0 V)/(32.5 Ω) = 0.769 A.
E
VALUATE
:
The circuit appears complicated until we realize that the 45.0
Ω and 15.0

Ω resistors are in
parallel.
26.8.
I
DENTIFY
:
Eq.(26.2) gives the equivalent resistance of the three resistors in parallel. For resistors in parallel,
the voltages are the same and the currents add.
(a) S
ET
U
P
:
The circuit is sketched in Figure 26.8a.
E
XECUTE
:
parallel
eq
1
2
3
1
1
1
1
R
R
R
R
eq
1
1
1
1
1.60
2.40
4.80
R
eq
0.800
R
Figure 26.8a
(b)
For resistors in parallel the voltage is the same across each and equal to the applied voltage;
1
2
3
28.0 V
V
V
V
E
1
1
1
28.0 V
so
17.5 A
1.60
V
V
IR
I
R
2
3
2
3
2
3
28.0 V
28.0 V
11.7 A and
5.8 A
2.40
4.8
V
V
I
I
R
R
(c)
The currents through the resistors add to give the current through the battery:
1
2
3
17.5 A
11.7 A
5.8 A
35.0 A
I
I
I
I
E
VALUATE
:
Alternatively, we can use the equivalent resistance
eq
R
as shown in Figure 26.8b.
eq
0
IR
E
eq
28.0 V
35.0 A,
0.800
I
R
E
which checks
Figure 26.8b
(d)
As shown in part (b), the voltage across each resistor is 28.0 V.
(e) I
DENTIFY
and
S
ET
U
P
:
We can use any of the three expressions for
2
2
:
/
.
P
P
VI
I R
V
R
They will all
give the same results, if we keep enough significant figures in intermediate calculations.
26
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E
XECUTE
:
Using
2
/
,
P
V
R
2
2
2
2
1
1
1
2
2
2
28.0 V
28.0 V
/
490 W,
/
327 W, and
1.60
2.40
P
V
R
P
V
R
2
2
3
3
3
28.0 V
/
163 W
4.80
P
V
R
E
VALUATE
:
The total power dissipated is
out
1
2
3
980 W.
P
P
P
P
This is the same as the power
in
2.80 V
35.0 A
980 W
P
I
E
delivered by the battery.
(f)
2
/
.
P
V
R
The resistors in parallel each have the same voltage, so the power
P
is largest for the one with the
least resistance.
26.10.
(a)
I
DENTIFY
:
The current, and hence the power, depends on the potential difference across the resistor.
S
ET
U
P
:
/
P
V
R
E
XECUTE
:
(a)
(5.0 W)(15,000
)
274 V
V
PR
(b)
/
(120 V)
/(9,000
)
1.6 W
P
V
R
S
ET
U
P
:
(c)
If the larger resistor generates 2.00 W, the smaller one will generate less and hence will be safe.
Therefore the maximum power in the larger resistor must be 2.00 W. Use
2
P
I R
to find the maximum current
through the series combination and use Ohm’s law to find the potential difference across the combination.
E
XECUTE
:
P
I R
gives
I
=
P
/
R
= (2.00 W)/(150 Ω) = 0.0133 A. The same cu
rrent flows through both
resistors, and their equivalent resistance is 250 Ω. Ohm’s law gives
V
=
IR
= (0.0133 A)(250 Ω) = 3.33 V.
Therefore
P
150
= 2.00 W and
100
P
I R
= (0.0133 A)
2
(100 Ω) = 0.0177 W.
E
VALUATE
:
If the resistors in a series combination all have the same power rating, it is the
largest
resistance
that limits the amount of current.
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 Spring '08
 HKMIET
 Current, Resistance, Resistor, Electrical resistance, Series and parallel circuits

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