BROCK UNIVERSITY
PHYS 1P22/1P92
Test 2 SOLUTIONS
27 June 2013
1. [2 marks] A 3.0 V battery powers a flashlight bulb that has a resistance of 6.0 Ω.
Determine the amount of charge that flows from the battery in 20 minutes.
Solution: Using Ohm’s law, the current from the battery is
I=
V
3.0 V
=
= 0.5 A
R
6.0 Ω
The amount of charge that flows from the battery in 20 minutes is
Q = I∆t = (0.5 A) (20 min) = (0.5 A) (1200 s) = 600 C
2. [4 marks] A 9.0 V battery can supply a total charge of 450 mA·h. Determine the total
energy that the battery can supply.
Solution: The total charge in coulombs that can flow from the battery is
450 mA · h = (0.450 A) × (3600 s) = 1620 C
Thus, the total amount of energy that the battery can supply is
E = QV = (1620 C) (9 V) = 14580 J = 15 kJ
3. [4 marks] Determine the equivalent resistance between points a and b in the figure.
Solution: The equivalent resistance of the 30 Ω and 45 Ω resistors in parallel is
−1 −1 −1
1
1
3
2
5
90
R=
+
=
+
=
=
= 18 Ω
30 45
90 90
90
5
The total of this 18 Ω resistance with the 42 Ω resistance in series is 18 + 42 = 60 Ω.
The total equivalent resistance is therefore
−1 −1 −1
1
1
2
3
5
120
R=
+
=
+
=
=
= 24 Ω
60 40
120 120
120
5
4. [4 marks] Determine the current through each resistor and the potential difference
across each resistor.
Solution: First determine the equivalent resistance of the circuit so that the current
from the battery can be determined. The equivalent resistance of the circuit is
−1
−1
−1
2
5
30
1
3
1
=6+
=6+
=6+
+
+
= 12 Ω
R=6+
15 6 + 4
30 30
30
5
Thus, the current from the battery is
I=
V
24
=
= 2.0 A
R
12
Thus, the current through the 6.0 Ω resistor at the left of the figure is 2.0 A, and so
the voltage drop across it is 2 × 6 = 12 V.
This means that the voltage drop across the 15 Ω resistor is 24 − 12 = 12 V, and so
the current through it is 12/15 = 0.80 A.
The voltage drop across the remaining two resistors in series (the ones furthest to the
right in the figure) is also 12 V, so the current through this branch of the circuit is
12/(6 + 4) = 1.2 A. (As a check, note that 2 = 0.80 + 1.2.) The voltage drop across
the 6 Ω resistor at the right is (1.2)(6) = 7.2 V, and the voltage drop across the 4 Ω
resistor at the right is (1.2)(4) = 4.8 V.
The voltage drops across each resistor can be checked by using Kirchoff’s loop law.
5. [6 marks] Clearly indicate whether each statement is TRUE or FALSE. Then provide
a BRIEF explanation (i.e. one or two sentences). Also include a clear and complete
correction of the statement if it is false. Your explanation may include formulas or
diagrams, if you wish. Remember, brevity and clarity are courtesy.
(a) For several resistors connected in series to a battery, the voltage across each
resistor is the same as the voltage of the battery.
Solution: FALSE. The current through each resistor is the same as the current
from the battery in a series circuit. The sum of the voltage drops across each
resistor is equal to the battery voltage in a series circuit.
(b) For several resistors connected in parallel to a battery, the power dissipated in
each resistor is the same as the power output of the battery.
Solution: FALSE. The voltage drop across each resistor is equal to the battery
voltage in a parallel circuit. The sum of the power dissipated in each resistor is
equal to the power delivered by the battery.
(c) Light bulbs connected with a short circuit are brighter because the electricity
doesn’t have as far to go.
Solution: FALSE. In a short circuit more current flows through the “short”
wire, so less current flows through the light bulbs. As a result, the light bulbs
are dimmer (and possible unlit, depending on how complicated the circuit is and
where the short is) in the presence of a short.