Download Electricity Practice Test -- SOLUTIONS

ELECTRICITY PRACTICE TEST
SPH3U
Date:__________
Knowledge & Understanding
Short Answer:
1. How many electrons flow past a spot in a conductor in 5 seconds if a current of 3A is present?
Q
t
Ne
I
t
It
Therefore there are 9.375 x 1019 electrons that flow past the conductor.
N
e
(3)(5)
N
1.6  10 19
9.375  1019  N
I
2. A total of 3.5 x 1016 electrons flow past a load in a circuit delivering 0.07 J of energy. Determine the
voltage of the load?
Ee
Q
E
V  e
Ne
V 
The voltage of the load is 12.5 V
0.07
(3.5  10 )(1.6  10 19 )
V  12.5V
V 
16
3. A load in a circuit consumes 200 W of electrical power. If the voltage of the load is 30V, determine the total
amount of charge that passes in 5 min.
P  VI
Q
P V 
t 
Pt
2000 C of charge passes through the load in 5 minutes.
Q
V
(200)(300)
Q
30
2000C  Q
SPH3U: Physics (Grade 11 University)
Practice Test: Electricity
Inquiry
1. Completely solve each circuit below. Include your reasoning.
a)
1
2
3
4
5
6
7
8
9
T
1
2
T ..
3
4
V
40
50
70
30
40
30
10
10
40
200
I
10
10
2
8
2
6
4
2
10
10
R
4
5
35
3.75
20
5
2.5
5
4
20
P
400
500
140
240
80
180
40
20
400
2000
Check:
5
400+500+140+240+80+180+40+20+400 = 2000
6
7
8
9
V3 = V4 + V5
(parallel)
V4 = 30V
R4 = V4/I4
P4 = V4I4
V7 = V8 (parallel)
V5 = V6 + V7
(parallel)
V6 = 30 V
I6 = V6/R6
P6 = V6I6
I4 = I5 + I6 (KCL)
I5 = 2A
I6 = I7 + I8 (KCL)
I8 = 4A
R5=V5/I5; P5 = V5I5
R7=V7/I7; P7 = V7I7
R8 = V8/I8
P8 = V8I8
V9 = P9/I9
R9 = V9/I9
I9=IT=I1=I2
(series)
VT = ITRT
PT = VTIT
V1 = P1/I1
R1 = V1/I1
V2 = I2R2
P2 = V2I2
I2 = I3 + I4 (KCL)
I3 = 2A
V3 = I3R3
P3 = V3I3
SPH3U: Physics (Grade 11 University)
Practice Test: Electricity
b)
..
1
..
..
T ..
I1=V1/R1
P1=V1R1
I1=IT (series)
VT=PT/IT
RT=VT/IT
1
2
3
4
T
2
3
I1=I2+I3 (KCL)
I3= 2A
R3=V3/I3
P3=V3I3
I4=I2 (series)
..
4
R4=V4/I4
P4=V4I4
V3=V2+V4 (parallel)
V2=40V
R2=V2/I2
P2=V2I2
V
20
40
60
20
80
I
5
3
2
3
5
R
4
13.3
30
6.7
16
P
100
120
120
60
400
Check:
100+120+120+60 = 400
Application [LEVEL]
2. Describe a situation in which a series circuit is a better design than a parallel one.




Series circuits are ones in which there is only one path for electricity to flow.
Parallel circuits are ones in which there are multiple paths for electricity to flow.
If there are multiple loads in a series circuit and one fails, they all fail (no complete path for electricity)
Therefore, the only time a series circuit is preferable is when there is only one load present (i.e handheld electronic devices).
3. Boosting a car using someone else’s battery requires the combination of two batteries. Are these batteries
connected in series or in parallel? Explain.



In order for a car to be boosted properly, the boosting battery must be connected to the dead battery in
a way that the voltages are the same.
This will give the dead battery circuit the energy it needs to get started again.
Therefore, the boosting battery must be connected in parallel with the dead battery in order for their
voltages to be equal.