Problem 1. If a current of 80.0 mA exists in a metal wire, how many electrons flow past a
given cross section of the wire in 10.0 min? Sketch the directions of the current and the
electrons’ motion.
Solution: The charge that moves past the cross section is ΔQ = I( Δt) , and the
number of electrons is
n=
ΔQ
e
=
I( Δt)
e
( 80.0 × 10
−3
=
)
C s ⎡⎣(10.0 m in) ( 60.0 s m in) ⎤⎦
= 3.00 × 1020 electrons
1.60 × 10-19 C
The negatively charged electrons move in the direction
opposite to the conventional current flow.
v
Problem 2. A small sphere that carries a charge q is whirled in a circle at the end of an
insulating string. The angular frequency of rotation is ω. What average current does this
rotating charge represent?
Solution: The period of revolution for the sphere is T = 2π ω , and the average
current represented by this revolving charge is I=
q qω
=
2π
T
Problem 3. In a particular television picture tube, the measured beam current is 60.0 μA.
How many electrons strike the screen every second?
Solution: ΔQ = I( Δt)
and the number of electrons is
I( Δt) ( 60.0 × 10 C s) (1.00 s)
n=
=
=
= 3.75 × 1014 electrons
-19
1.60 × 10 C
e
e
ΔQ
−6
Problem 4. A total charge of 6.0 mC passes through a cross-sectional area of a wire in 2.0
s. What is the current in the wire?
Solution: The current is I=
ΔQ 6.0 × 10−3 C
=
= 3.0 × 10−3 A = 3.0 m A
2.0 s
Δt
I
–
Problem 5. In the Bohr model of the hydrogen atom, an electron in the lowest energy state
moves at a speed of 2.19 × 106 m/s in a circular path having a radius of 5.29 × 10–11 m. What
is the effective current associated with this orbiting electron?
Solution: The period of the electron in its orbit is T = 2π r v , and the current
represented by the orbiting electron is
I=
e ve
ΔQ
= =
Δt T 2π r
( 2.19 × 10 m s)(1.60 × 10
2π ( 5.29 × 10 m )
−19
6
=
−11
C
) = 1.05 × 10
−3
C s = 1.05 m A
Problem 6. If 3.25 × 10–3 kg of gold is deposited on the negative electrode of an
electrolytic cell in a period of 2.78 h, what is the current through the cell in this period?
Assume that the gold ions carry one elementary unit of positive charge.
Solution: The mass of a single gold atom is
m atom =
197 g m ol
M
=
= 3.27 × 10-22 g = 3.27 × 10-25 kg
N A 6.02 × 1023 atom s m ol
The number of atoms deposited, and hence the number of ions moving to the
negative electrode, is
n=
m
m atom
=
3.25 × 10−3 kg
= 9.93 × 1021
-25
3.27 × 10 kg
Thus, the current in the cell is
(
)(
)
21
−19
ΔQ ne 9.93 × 10 1.60 × 10 C
I=
=
=
= 0.159 A = 159 m A
Δt Δt
( 2.78 h) ( 3600 s 1 h)
Problem 7. A 200-km-long high-voltage transmission line 2.0 cm in diameter carries a
steady current of 1 000 A. If the conductor is copper with a free charge density of 8.5 × 1028
electrons per cubic meter, how long (in years) does it take one electron to travel the full
length of the cable?
Solution: The drift speed of electrons in the line is vd =
vd =
4(1000 A )
( 8.5 × 10
28
m
3
)(1.60 × 10
)
I
I
, or
=
nqA n e ( π d2 4)
C π ( 0.020 m
−19
)2
= 2.3 × 10-4 m s
The time to travel the length of the 200-km line is then
Δt=
1 yr
L
200 × 103 m ⎛
⎞
=
= 27 yr
⎜⎝
-4
7 ⎟
vd 2.34 × 10 m s 3.156 × 10 s⎠
Problem 8.An aluminum wire with a cross-sectional area of 4.0 x 10–6 m2 carries a current
of 5.0 A. Find the drift speed of the electrons in the wire. The density of aluminum is 2.7
g/cm3. (Assume that one electron is supplied by each atom.)
Solution:Assuming that, on average, each aluminum atom contributes one
electron, the density of charge carriers is the same as the number of atoms per
cubic meter. This is
n=
density
N ρ
ρ
=
= A ,
m assper atom M N A
M
( 6.02 × 10
n=
23
or
)(
m ol ⎡⎣ 2.7 g cm
3
)(10
6
cm
3
26.98 g m ol
1m
3
) ⎤⎦ = 6.0 × 10
28
m
3
The drift speed of the electrons in the wire is then
vd =
I
=
n e A ( 6.0 × 1028 m
5.0 C s
3
)(1.60 × 10
−19
)(
C 4.0 × 10−6 m
2
)
=
1.3 × 10− 4 m s
Problem 9. If the current carried by a conductor is doubled, what happens to the (a)
charge carrier density and (b) electron drift velocity?
Solution: (a)
The carrier density is determined by the physical
characteristics of the wire, not the current in the wire. Hence, n is
unaffected .
(b) The drift velocity of the electrons is vd = I nqA . Thus, the drift velocity is
doubled when the current is doubled.
Problem 10. A lightbulb has a resistance of 240 Ω when operating at a voltage of 120 V.
What is the current through the lightbulb?
ΔV 120 V
Solution: I=
=
= 0.500 A = 500 m A
240 Ω
R
Problem 11. A person notices a mild shock if the current along a path through the
thumb and index finger exceeds 80 μA. Compare the maximum allowable
voltage without shock across the thumb and index finger with a dry-skin
resistance of 4.0 × 105 Ω and a wet-skin resistance of 2 000 Ω.
Solution: ( ΔV ) m ax = Im axR = ( 80 × 10−6 A ) R
Thus, if R = 4.0 × 105 Ω , ( ΔV ) m ax = 32 V ,
and if R = 2000 Ω , ( ΔV ) m ax = 0.16 V
Problem 12. Suppose that you wish to fabricate a uniform wire out of 1.00 g of copper. If
the wire is to have a resistance of R = 0.500 Ω, and if all of the copper is to be used, what will
be (a) the length and (b) the diameter of this wire?
Solution: The volume of the copper is
V=
1.00 × 10−3 kg
m
=
= 1.12 × 10−7 m
3
3
density 8.92 × 10 kg m
Since, V = A ⋅ L , this gives A ⋅ L = 1.12 × 10−7 m 3 .
(a) From R =
3
(1)
ρL
, we find that
A
⎛ 1.70 × 10−8 Ω ⋅ m
ρ
A = ⎛⎜ ⎞⎟ L = ⎜
⎝ R⎠
0.500 Ω
⎝
⎞
L = 3.40 × 10−8 m L .
⎠⎟
(
)
Inserting this expression for A into Equation 1 gives
( 3.40 × 10
−8
)
m L2 = 1.12 × 10−7 m 3 , which yields
(b) From equation (1), A =
d=
(
4 1.12 × 10−7 m
πL
π d2
4
3
=
1.12 × 10−7 m
, or
L
) = 4(1.12 × 10
= 2.80 × 10−4 m = 0.280 m m
3
−7
m
π (1.82 m )
3
)
L = 1.82 m
Problem 13. Calculate the diameter of a 2.0-cm length of tungsten filament in a small
lightbulb if its resistance is 0.050 Ω.
Solution: From R =
d=
4ρ L
πR
=
ρL
π d2 ρL
, we obtain A =
=
, or
A
4
R
(
4 5.6 × 10−8 Ω ⋅ m
)( 2.0 × 10
−2
m
π ( 0.050 Ω )
) = 1.7 × 10
−4
m = 0.17 m m
Problem 14. Eighteen-gauge wire has a diameter of 1.024 mm. Calculate the resistance of
15 m of 18-gauge copper wire at 20°C.
4 1.7 × 10−8 Ω ⋅ m (15 m )
ρL
ρL
Solution: R = = 2 =
= 0.31 Ω
2
A πd 4
π 1.024 × 10−3 m
(
(
)
)
Problem 15.A potential difference of 12 V is found to produce a current of 0.40 A in a
3.2-m length of wire with a uniform radius of 0.40 cm. What is (a) the resistance of the wire
and (b) the resistivity of the wire?
ΔV
12 V
Solution: (a)
R=
=
= 30 Ω
I 0.40 A
(b) From, R =
ρL
,
A
2
−2
⎡
⎤
R ⋅ A ( 30 Ω ) ⎣π ( 0.40 × 10 m ) ⎦
−4
ρ=
=
= 4.7 × 10 Ω ⋅ m
L
3.2 m
Problem 16. A length L0 of copper wire has a resistance R0. The wire is cut into three
pieces of equal length. The pieces are then connected as parallel lengths between points A and
B. What resistance will this new “wire” of length L0/3 have between points A and B?
Solution: The new “wire” has length L = L0 3 and cross-section A = 3A 0 . Thus, its
resistance is
R=
ρL ρ ( L0 3) 1 ⎛ ρL0 ⎞ R 0
=
= ⎜
=
9
3A 0
9 ⎝ A 0 ⎟⎠
A
Problem 17. A wire 50.0 m long and 2.00 mm in diameter is connected to a source with a
potential difference of 9.11 V, and the current is found to be 36.0 A. identify the metal of the
wire.
ΔV 9.11 V
=
= 0.253 Ω , so the resistivity of the
Solution: The resistance is R =
I 36.0 A
metal is
−3
2
R ⋅ A R ⋅ ( π d 4) ( 0.253 Ω ) π ( 2.00 × 10 m )
ρ=
=
=
= 1.59 × 10−8 Ω ⋅ m
L
L
4( 50.0 m )
2
Thus, the metal is seen to be silver .
Problem 18. A 4.0-Ω resistor, an 8.0-Ω resistor, and a 12-Ω resistor are connected in
series with a 24-V battery. What are (a) the equivalent resistance and (b) the current in each
resistor? (c) Repeat for the case in which all three resistors are connected in parallel across the
battery.
Solution: (a)
R eq = R1 + R 2 + R 3 = +4.0 Ω + 8.0 Ω + 12 Ω = 24 Ω
(b) The same current exists in all resistors in a series combination.
I=
ΔV
R eq
=
24 V
= 1.0 A
24 Ω
(c) If the three resistors were connected in parallel,
⎛ 1
1
1⎞
+
+ ⎟
⎝ R1 R 2 R 3 ⎠
R eq = ⎜
−1
1
1 ⎞
⎛ 1
=⎜
+
+
⎝ 4.0 Ω 8.0 Ω 12 Ω ⎟⎠
−1
= 2.18 Ω
Resistors in parallel have the same potential difference across them, so
I4 =
ΔV
R4
=
24 V
24 V
24 V
= 3.0 A , and I12 =
= 2.0 A
= 6.0 A , I8 =
4.0 Ω
8.0 Ω
12 Ω
Problem 19. A 9.0-Ω resistor and a 6.0-Ω resistor are connected in series with a power
supply. (a) The voltage drop across the 6.0-Ω resistor is measured to be 12 V. Find the
voltage output of the power supply. (b) The two resistors are connected in parallel across a
power supply, and the current through the 9.0-Ω resistor is found to be 0.25 A. Find the
voltage setting of the power supply.
Solution: (a)
I=
The current through this series combination isa
( ΔV )bc
R bc
b
9.0 Ω
6.0 Ω
12 V
=
= 2.0 A
6.0 Ω
Therefore, the terminal potential difference of the
power supply is
ΔV
ΔV = IR eq = ( 2.0 A ) ( 9.0 Ω + 6.0 Ω ) = 30 V
(b) When connected in parallel, the potential difference across either resistor
is the voltage setting of the power supply. Thus,
ΔV = I9 R 9 = ( 0.25 A )( 9.0 Ω ) = 2.3 V
c
Problem 20.
(a) Find the equivalent resistance between points a and b in the following
figure (b) Calculate the current in each resistor if a potential difference of 34.0 V is applied
between points a and b.
Solution: (a)
The equivalent resistance of the two parallel
4.00 Ω
resistors is
R p = ⎛⎜
1
1 ⎞
+
⎝ 7.00 Ω 10.0 Ω ⎟⎠
= 4.12 Ω
10.0 Ω
a
R ab = R 4 + R p + R 9 = ( 4.00 + 4.12 + 9.00) Ω = 17.1 Ω
( ΔV ) ab 34.0 V
R ab
=
17.1 Ω
= 1.99 A , so I4 = I9 = 1.99 A
Also,
( ΔV ) p = IabR p = (1.99 A )( 4.12 Ω) = 8.18 V
Then,
I7 =
and
I10 =
( ΔV ) p
R7
( ΔV ) p
R10
9.00 Ω
−1
Thus,
(b) Iab =
7.00 Ω
=
8.18 V
= 1.17 A
7.00 Ω
=
8.18 V
= 0.818 A
10.0 Ω
b
Problem 21. Find the equivalent resistance of the circuit in figure shown.
Solution: The equivalent resistance of the parallel combination of three resistors is
R p = ⎛⎜
1
1
1 ⎞
+
+
⎝ 18 Ω 9.0 Ω 6.0 Ω ⎟⎠
−1
= 3.0 Ω
Hence, the equivalent resistance of the circuit connected to the 30 V source is
R eq = R12 + R p = 12 Ω + 3.0 Ω = 15 Ω
Problem 22. Find the equivalent resistance of the circuit in the figure shown
Solution: The rules for combining resistors in series and parallel are used to
reduce the circuit to an equivalent resistance in the stages shown below. The
result is R eq = 9.8 Ω .
5.0 Ω
5.0 Ω
5.0 Ω
5.0 Ω
1.5 Ω
5.0 Ω
5.0 Ω
5.0 Ω
1.5 Ω
10 Ω
10
—
3
Ω
1.5 Ω
9.8 Ω
Problem 23. Consider the circuit shown in the following figure . Find (a) the current in
the 20.0-Ω resistor and (b) the potential difference between points a and b.
Solution: Turn the circuit given in Figure
90° counterclockwise to observe that it
is equivalent to that shown in Figure 1 below. This reduces, in stages, as
shown in the following figures.
b
b
I20
25.0
V
5.00
Ω
5.00 c
Ω
20.0
Ω
10.0
Ω
10.0
Ω
I20
25.0
V
5.00
Ω
10.0
Ω
25.0
Ω
10.0
Ω
a
Figure 1
a
Figure 2
b
I
I
25.0
V
2.94
Ω
25.0
V
12.9
Ω
10.0
Ω
a
Figure 3
Figure 4
From Figure 4,
ΔV 25.0 V
I=
=
= 1.93 A
R
12.9 Ω
(b) From Figure 3,
( ΔV )ba = IR ba
= (1.93 A ) ( 2.94 Ω ) = 5.68 V
(a) From Figures 1 and 2, the current through the 20.0 Ω resistor is
I20 =
( ΔV ) ba 5.68 V
R bca
=
25.0 Ω
= 0.227 A
Problem 24.
Two resistors, A and B, are connected in parallel across a 6.0-V battery. The
current through B is found to be 2.0 A. When the two resistors are connected in series to the
6.0-V battery, a voltmeter connected across resistor A measures a voltage of 4.0 V. Find the
resistances of A and B.
Solution: First, consider the parallel case. The resistance of resistor B is
RB =
( ΔV ) B
IB
=
6.0 V
= 3.0 Ω
2.0 A
In the series combination, the potential difference across B is given by
( ΔV )B = ( ΔV )battery − ( ΔV ) A = 6.0 V -4.0 V = 2.0 V
The current through the series combination is then
Is =
( ΔV ) B
RB
=
2.0 V 2
= A,
3.0 Ω 3
and the resistance of resistor A is R A =
( ΔV ) A
Is
=
4.0 V
= 6.0 Ω
23A
Problem 25. The resistance between terminals a and b in the following figure is 75 Ω. If
the resistors labeled R have the same value, determine R.
Solution:
The equivalent resistance is R eq = R + R p , where R p is the total
resistance of the three parallel branches;
R p = ⎛⎜
1
1
1
⎞
+
+
⎝ 120 Ω 40 Ω R + 5.0 Ω ⎟⎠
Thus, 75 Ω = R +
−1
1
⎛ 1
⎞
=⎜
+
⎝ 30 Ω R + 5.0 Ω ⎟⎠
−1
=
( 30 Ω)( R + 5.0 Ω)
( 30 Ω)( R + 5.0 Ω) R 2 + ( 65 Ω) R + 150 Ω 2
R + 35 Ω
=
R + 35 Ω
R + 35 Ω
,
which reduces to R 2 − (10 Ω ) R + 2475 Ω 2 = 0 or ( R − 55 Ω )( R + 45 Ω ) = 0 .
Only the positive solution is physically acceptable, so R = 55 Ω
Problem 26. Find the current in the 12-Ω resistor in the following figure
Solution: The resistors in the circuit can be combined in the stages shown below
to yield an equivalent resistance of R ad = ( 63 11) Ω .
6.0 Ω
3.0 Ω
a 3.0 Ω
b
I1
c
3.0 Ω
6.0 Ω
4.0 Ω
I2
I12
d
3.0 Ω
a
I1
d
e
3.0 Ω
I
e
12 Ω
b
I2
2.0 Ω
3.0 Ω
c
2.0 Ω
I
18 V
18 V
Figure 1
Figure 2
6.0 Ω
3.0 Ω I1
a
b
3Ω
a
d
b
30
—
11
Ω
d
63
—
11
a
Ω
I2
I
I
5.0 Ω
18 V
Figure 3
From Figure 5, I=
Then, from Figure 4,
( ΔV ) ad
R ad
I
18 V
18 V
Figure 4
Figure 5
18 V
=
= 3.14 A .
( 63 11) Ω
( ΔV )bd = IR bd = ( 3.14 A ) ( 30 11 Ω) = 8.57 V .
Now, look at Figure 2 and observe that
I2 =
so
( ΔV )bd
3.0 Ω + 2.0 Ω
=
8.57 V
= 1.71 A ,
5.0 Ω
( ΔV )be = I2R be = (1.71 A )( 3.0 Ω ) = 5.14 V
Finally, from Figure 1,
I12 =
( ΔV )be 5.14 V
R12
=
12 Ω
= 0.43 A
d