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25.1
ELECTRIC CIRCUITS
25
EXERCISES
Section 25.1 Circuits, Symbols, and Electromotive Force
14.
A literal reading of the circuit specifications results in connections like those in sketch
(a)
. Because the connecting
wires are assumed to have no resistance (a real wire is represented by a separate resistor), a topologically
equivalent circuit diagram is shown in sketch
(b)
.
15.
INTERPRET
This problem is about how various circuit elements can be connected to form a closed series circuit.
DEVELOP
In a series circuit, the same current must flow through all elements.
EVALUATE
One possibility is shown below. The order of elements and the polarity of the battery connections are
not specified.
ASSESS
An important feature about a series circuit is that the current through all the components must be the
same. With two batteries, the direction of the current flow is determined by the polarity of the larger of the two
voltage ratings.
16.
The circuit has three parallel branches: one with
and
in series; one with just
and one with the battery
(an ideal emf in series with the internal resistance).
1
R
2
R
3
;
R
17.
INTERPRET
This problem explores the connection between the emf of a battery and the energy it delivers.
DEVELOP
Electromotive force, or emf, is defined as work per unit charge,
/.
Wq
ε
=
EVALUATE
Substituting the values given in the problem statement, we find the emf to be
27 J
9 V
3 C
W
q
==
=
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View Full Document 25.2
Chapter 25
ASSESS
For an ideal battery with zero internal resistance, the emf is equal to the terminal voltage (potential
difference across the battery terminals).
18.
The average power, supplied by the battery to the bulb, multiplied by the time equals the energy capacity of the
battery. For an ideal battery,
,
PI
ε
=
therefore
4.5 kJ,
It
=
or
3
4.5 kJ/(1.5 V)(0.60 A)
5 10 s
1.39 h.
t
==
×
=
19.
INTERPRET
This problem is about the chemical energy used up in the battery for the work done.
DEVELOP
The power delivered by an emf is
.
=
Therefore, if the voltage and current remain constant, then
the energy converted would be
.
WP
tI
t
EVALUATE
Substituting the values given, the energy used in
(5 A)(12 V)(3600 s)
216 kJ
WI
t
=
ASSESS
The result makes sense; the energy used up is proportional to the current drawn, the emf, and the
duration the headlights were left on.
Section 25.2 Series and Parallel Circuits
20.
From Equations 25.1 and 25.3b,
22 k
47(39 k )/(47 39)
43.3 k .
R
=Ω
+
Ω+
=
Ω
21.
INTERPRET
This problem is about connecting two resistors in parallel.
DEVELOP
The equivalent resistance of two resistors connected in parallel can be found by Equation 25.3a:
parallel
1
2
11
RR
=+
1
R
The equation allows us to determine
when
2
R
parallel
R
and
are known.
1
R
EVALUATE
The solution for
R
2
in Equation 25.3a is
1
parallel
2
1
parallel
(56 k )(45 k )
229 k
56 k
45 k
R
ΩΩ
=
−Ω
−
Ω
Ω
ASSESS
Our result shows that
This is consistent with the fact that the equivalent resistance
2
parallel
.
>
parallel
R
is
smaller than
and
1
R
2
.
R
22.
The starter circuit contains all the resistances in series, as in Figure 25.10. (We assume
L
R
includes the resistance
of the cables, connections, etc., as well as that of the motor.) With the defective
starter,
int
6 V
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This note was uploaded on 10/21/2010 for the course PHYSICS 2131441 taught by Professor Pheong during the Fall '10 term at University of California, Berkeley.
 Fall '10
 Pheong

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