Phys 122
Introductory Physics II
TUTORIAL 3
Electric
Circuits.
Key
DC Current and Ohm’s Law.
1.
The filament of a light bulb has a resistance of 580
Ω
. A voltage of 120 V is connected across the filament.
How much current is in the filament?
We know that
V
=
IR
.
Therefore,
I
=
V
R
=
120 V
580
Ω
=
0.21 A
2.
A battery charger is connected to a dead battery and delivers a current of 6.0 A for 5.0 hours, keeping the
voltage across the battery terminals at 12 V in the process. How much energy is delivered to the battery?
First determine the total charge delivered to the battery using Equation 20.1:
∆
q
=
I
∆
t
= (6.0 A)(5.0 h)[(3600 s)/(1 h)] = 1.1
×
10
5
C
To find the energy delivered to the battery, multiply this charge by the energy per unit charge
(i.e., the voltage) to get
Energy = (
∆
q
)
V
= (1.1
×
10
5
C)(12 V) =
1.3
×
10
6
J
3.
In the Arctic, electric socks are useful. A pair of socks uses a 9.0V battery pack for each sock. A current of
0.11 A is drawn from each battery pack by wire woven into the socks. Find the resistance of the wire in one sock.
Ohm's law (Equation 20.2),
V
=
IR
, gives the result directly:
R
=
V
I
=
9.0 V
0.11 A
=
82
Ω
4.
The resistance of a bagel toaster is 14
Ω
. To prepare a bagel, the toaster is operated for one minute from a
120V outlet. How much energy is delivered to the toaster?
Energy =
V
∆
q
Solving Equation 20.1 for
∆
q
gives
∆
q
=
I
∆
t
, which can be substituted in the previous result to give
Energy =
V
∆
q
=
VI
∆
t
According to Ohm’s law (Equation 20.2), the current is
I
=
V
/
R
, which can be substituted in the
energy expression to show that
1
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Energy
=
VI
∆
t
=
V
V
R
∆
t
=
V
2
∆
t
R
=
120 V
(
29
2
60 s
(
29
14
Ω
=
6.2
×
10
4
J
5.
A beam of protons is moving toward a target in a particle accelerator. This beam constitutes a current whose
value is 0.50
μ
A. (a) How many protons strike the target in 15 seconds? (b) Each proton has a kinetic energy of 4.9
×
10
–12
J. Suppose the target is a 15gram block of aluminum, and all the kinetic energy of the protons goes into heating
it up. What is the change in temperature of the block at the end of 15 s?
The number
N
of protons that strike the target is equal to the amount of electric charge
∆
q
striking the
target divided by the charge
e
of a proton,
N =
(
∆
q
)
/e
.
a.
The number
N
of protons that strike the target is
6
13
19
(0.50
10
A)(15 s)
4 7
10
1.6
10
C
q
I
t
N
e
e


∆
∆
×
=
=
=
=
×
×
.
b.
The amount of heat
Q
provided by the kinetic energy of the protons is
13
12
(4.7
10
protons)(4.9
10
J/proton)
230 J
Q

=
×
×
=
Since
Q
cm T
=
∆
and since Table 12.2 gives the specific heat of aluminum as
c
= 9.00
×
10
2
J/
(kg
.
C°), the change in temperature of the block is
2
3
230 J
17 C
J
9.00
10
(15
10
kg)
kg C
Q
T
cm

∆
=
=
=
°
×
×
⋅
°
6.
Two wires are identical, except that one is aluminum and one is copper. The aluminum wire has a resistance
of 0.20
Ω
. What is the resistance of the copper wire?
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 Winter '08
 Loughlin
 Resistor, ohm, Electrical resistance, equivalent resistance

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