BROCK UNIVERSITY
Test 2: February 2015
Course: PHYS 1P22/1P92
Examination date: 7 Febuary 2015
Time of Examination: 11:00–11:50
Number of pages: 6 + formula sheet
Number of students: 106
Instructor: S. D’Agostino
No aids are permitted except for a non-programmable, non-graphics calculator.
Solve all problems in the space provided.
Total number of marks: 24
SOLUTIONS
1. [5 points] The voltage drop across the terminals of a light bulb is 120 V, and the light
bulb dissipates 60 W of power.
(a) Determine the current flowing through the bulb.
(b) Determine the resistance of the bulb.
(c) Determine the amount of energy dissipated by the bulb in 24 h.
(d) Determine how much charge flows through the bulb in 24 h.
(e) Determine the cost to run the light bulb continuously for 30 days if electricity
costs 0.12 dollars per kW·h.
Solution:
(a)
I=
P
60
=
= 0.5 A
∆V
120
R=
∆V
120
=
= 240 Ω
I
0.5
(b)
(c)
∆E = P ∆t
∆E = (60 W) (24 h)
J
3600 s
∆E = 60
24 h ×
s
h
6
∆E = 5.2 × 10 J
If you prefer the result in kW·h, which is a more practical energy unit, calculate as
follows:
∆E
∆E
∆E
∆E
= P ∆t
= (60 W) (24 h)
= 1440 W · h
= 1.44 kW · h
(d)
∆Q = I∆t
∆Q = 0.5 A × 24 h
∆Q = 0.5 A × 24 h ×
3600 s
h
∆Q = 43, 200 C
∆Q = 43.2 kC
(e)
Cost = 0.12
kW·h
$
× 1.44
× 30 days = $5.18
kW·h
day
2. [3 points] If a copper wire is replaced by another copper wire that has twice the length
and one-third the diameter, how does the resistance change?
Solution: The resistance increases by a factor of 18, as the following calculation shows:
R2
R1
R2
R1
R2
R1
R2
R1
R2
R1
R2
R1
R2
R1
ρ 2 L2 ρ 1 L1
÷
A2
A1
ρ 2 L2 ρ 1 L1
=
÷
πr22
πr12
ρ 2 L2
πr12
=
×
πr22
ρ 1 L1
ρ 2 L2
r2
= 2 × 1
r2
ρ 1 L1
ρ2 L2 r12
=
×
×
ρ1 L1 r22
2
ρ 2 L2
r1
=
×
×
ρ 1 L1
r2
2
1
= (1) × (2) ×
1/3
=
R2
= (1) × (2) × (3)2
R1
R2
=2×9
R1
R2
= 18
R1
3. [4 points] Determine the equivalent resistance of the circuit.
10 Ω
10 Ω
0
12
20
Ω
50 V
40 Ω
25 Ω
Ω
10 Ω
30 Ω
Solution: First identify simple series or simple parallel combinations of resistors. In
the following diagram I’ve identified two resistors that are in series by printing their
labels in red:
10 Ω
12
20
0
10 Ω
Ω
50 V
25 Ω
40 Ω
Ω
10 Ω
30 Ω
Combining the two red resistors in series results in 10+10 = 20 Ω, so these two resistors
can be replaced by a single resistor of resistance 20 Ω, as in the following diagram:
10 Ω
20 Ω
0
12
20
Ω
50 V
40 Ω
25 Ω
Ω
30 Ω
Now combine the two 20-Ω resistors in parallel and replace them by a single resistor
of resistance
−1 −1
1
1
2
20
+
= 10 Ω
=
=
20 20
20
2
10 Ω
0
12
10
Ω
50 V
40 Ω
25 Ω
Ω
30 Ω
Now the 10-Ω resistor and the 30-Ω resistor, which are in series, can be combined and
replaced by a single resistor of resistance
10 Ω + 30 Ω = 40 Ω
10 Ω
0
12
40
Ω
50 V
40 Ω
Ω
25 Ω
Next, combine the two 40-Ω resistors, which are in parallel, and replace them by a
single resistor of resistance
−1 −1
1
1
2
40
+
=
=
= 20 Ω
40 40
40
2
10 Ω
0
12
Ω
50 V
20 Ω
25 Ω
Now the 10-Ω resistor and the 20-Ω resistor, which are in series, can be combined and
replaced by a single resistor of resistance
10 Ω + 20 Ω = 30 Ω
30 Ω
0
12
Ω
50 V
25 Ω
Next, combine the 30-Ω resistor and the 120-Ω resistor, which are in parallel, and
replace them by a single resistor of resistance
−1 −1 −1
4
5
120
1
1
1
=
=
=
+
+
= 24 Ω
30 120
120 120
120
5
24
Ω
50 V
25 Ω
The equivalent resistance of the circuit can now be determined by combining the remaining two resistors, which are in series, to obtain
24 Ω + 25 Ω = 49 Ω
4. [4 points] Determine the current through each resistor.
10 Ω
10 Ω
50 V
50 V
10 Ω
30 Ω
10 Ω
Solution: I have labelled currents as in the following diagram:
I1
10 Ω
I3
10 Ω
I2
50 V
50 V
I1
10 Ω
I3
30 Ω
10 Ω
I3
Using Kirchhoff’s junction law at either of the two labelled junction points, we obtain
I2 = I1 + I3
(1)
Using Kirchhoff’s loop law in the left loop, we obtain
50 − 10I1 + 50 − 10I1 = 0
which is equivalent to
20I1 = 100
(2)
Using Kirchhoff’s loop law in the right loop, we obtain
50 − 10I3 − 30I3 − 10I3 = 0
which is equivalent to
50I3 = 50
(3)
Solving equation (2), we obtain
I1 = 5 A
Solving equation (3), we obtain
I3 = 1 A
Substituting these results into equation (1), we obtain
I2 = I1 + I3
I2 = 5 + 1
I2 = 6 A
Clearly state whether each of the following statements is true or false. Then, whether
the statement is true or false, explain the situation briefly and clearly, in at most
a few sentences. Your explanation may include formulas or diagrams, if you wish.
Remember, brevity and clarity are courtesy.
5. [2 points] A bird can safely perch on an uninsulated 10,000-V power line.
Solution: TRUE. Because the resistance of the wire is extremely low, the voltage drop
between two points on the wire that are very close together is small. The small voltage
difference between the bird’s feet drives a very small current through the bird’s body,
because the bird’s own resistance is considerably greater than the wire’s resistance.
6. [2 points] According to Ohm’s law, when current flows through a resistor, charge is
consumed and converted into thermal energy, which is then dissipated by the resistor.
Solution: FALSE. Charge is not consumed; think about the principle of conservation
of charge. What actually happens in the resistor is that electric potential energy carried
by the electrons is converted to thermal energy, which is then dissipated by the resistor.
7. [2 points] In a series circuit with two resistors connected to a battery, the current in
the resistor closer to the negative battery terminal is greater, because some current is
consumed as it passes through the first resistor.
Solution: FALSE. Current is not consumed; think about the principle of conservation
of charge, which is equivalent to Kirchhoff’s junction law. In a series circuit, the current
is the same in all parts of the circuit.
8. [2 points] Household circuits are wired in series because in this way less wire is used
and so it’s less expensive.
Solution: FALSE. Household circuits are wired in parallel, NOT in series. Parallel
circuits are used so that the current to each appliance or light bulb is not decreased
when several of them are switched on. Furthermore, if your home were wired in series
you would not be able to separately control each item included in a circuit; they would
have to either all be switched on or all be switched off.