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2 resistors
1. Two resistors
R
1
= 12
Ω
and
R
2
= 18
Ω
can be connected to an ideal battery with
emf
ε
= 100 V in two different ways: in series or in parallel.
Find the total power dissipated in the system when the two resistors are connected
in series.
When the resistors are connected in series , the equivalent reistance of the circuit is:
R = R1 + R2
R = 12 + 18
R = 30
ΩΩ
Ω
The voltage across this equivalent resistor is
ε
= 100 V. Thus, the power dissipated is:
22
(100 V)
333 W
30
V
P
R
==
=
Ω
2. Find the total power dissipated in the system when the two resistors are connected
in parallel.
The equivalent resistane of the circuit when the resistors are in parallel is:
1
2
12
1
2
111
(12 )(18 )
7.2
30
RR
R
R
R
R
R
=+
=
+
Ω
Ω
The voltage across this equivalent resistor (and across
R
1
and
R
2
too, since they are in
parallel) is
ε
= 100 V. Thus, the power dissipated is:
(100 V)
1390 W
7.2
V
P
R
=
Ω
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View Full Document3. Now imagine that the battery is not ideal. Its internal resistance is
r
= 5.0
Ω
.
Use
this nonideal battery for the next two questions.
What is in this case the power dissipated in the circuit (outside the battery) when the
resistors are in series?
For a nonideal battery there will be an internal resistance which will be in series with
battery, and also in series with the external circuit (which had an equivalent resistance of
30 Ohms, see question 1).
This time the voltage across the 30 Ohm resistor is less than 100 V, because there is a
potential drop inside the battery, due to its internal resistance.
The current in the system is
ext
100 V
2.86 A
35
V
I
Rr
==
=
+Ω
Thus, the power dissipated in the circuit outside the battery (so excluding the internal
resistance of the battery) is
( )( )
2
2.86 A
30
245 W
PIR
Ω
=
4. And what is in this case the power dissipated in the circuit (outside the battery)
when the resistors are in parallel?
In the case when the resistors in parallel, we have
This time the voltage across the 7.2 Ohm resistor is less than 100 V, because there is a
potential drop inside the battery, due to its internal resistance.
The current in the system is
ext
100 V
8.2 A
12.2
V
I
Rr
===
+Ω
Thus the power dissipated in the equivalent resistance outside the battery (so excluding
the internal resistance of the battery) is,
2
(8.196 )(7.2 )
484
0.48
pi
pi
pi
pi
PI
R
A
P
W
P
Pk
W
=
Ω
=
=
=
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View Full Documentcircuit
1. Four resistors and an ideal battery have been assembled in the circuit below. The
value of
R
3
is unknown (but greater than zero and finite).
R
1
= 10
Ω
,
R
2
= 5
Ω
,
R
4
=
5
Ω
.
Compare the magnitude of the voltage across
R
1
with the magnitude of the voltage
across
R
2
.
a. 
V
1
 < 
V
2

b. 
V
1
 = 
V
2

*c. 
V
1
 > 
V
2

R
1
>
R
2
.
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This homework help was uploaded on 03/26/2008 for the course PHYS 221 taught by Professor Herrerasiklody during the Spring '08 term at Iowa State.
 Spring '08
 HerreraSiklody
 Power

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