Kiel Benedick M. Gianan
12 – Canary
Chapter 1 – Shocking!
Practice Exercises
Practice Exercise 1.1
1. A piece of nylon cloth is used to clean the lenses of a pair of eyeglasses. In doing so, which becomes
positively charged? negatively charged? Assume that the lenses are made of glass.
Glass has a lower electron affinity than nylon. As a result, the nylon charges negatively and the
glass charges positively.
2. In the process of rubbing the lenses of the eyeglasses, 6.28x1010 electrons were transferred. (a) What is
the charge of the lenses and the nylon cloth? (b) What is the change in their masses?
Practice Exercise 1.3
3. Determine the ratio of the electric force to the gravitational force between an electron and a proton
when separated by a distance r.
Practice Exercise 1.6
4. A proton is released from rest in a uniform horizontal electric field. It travels 3.25 m for 5 us. Find the
acceleration of the proton and the magnitude of the electric field.
5. What electric field (magnitude and direction) is needed for an electron to remain suspended in air?
Kiel Benedick M. Gianan
12 – Canary
Chapter 2 – Aware of our Potentials
Assessment (Sentence Completion)
The electric potential at any point in an electric field is electric potential energy per unit charge at
that point. The SI unit of electric potentials is volt (V). It is named after the physicist Alessandro Volta.
One volt is equal to 1 joule/coulomb (J/C). Potential is also defined as the work done to move a unit charge
from infinity to one arbitrary point B.
Practice Exercises
Practice Exercise 2.1
A point particle of charge 2.5 nC and mass 3.25x10-3 kg is in a uniform electric field directed to the
right. It is released from rest and moves to the right. After it has traveled 12.0 cm, its speed is 25 m/s. Find
the (a) work done on the particle, (b) change in the electric potential energy of the particle, and (c)
magnitude of the electric field.
Practice Exercise 2.5
The parallel plates of an air capacitor are separated by 2.25 mm. Each plate carries a charge of 6.50
nC. The magnitude of the electric field of the plates is 4.75×105 V/m. Find the (a) potential difference
between the plates, (b) capacitance, and (c) area of a plate.
Practice Exercise 2.7
A parallel plate capacitor consists of two rectangular plates, each with an area of 4.5 cm² and
separated from each other by a 2.00 mm thick dielectric with a dielectric constant of 5.26. The capacitor is
connected to a 12.0 V battery. How much energy is stored in the capacitor?
Kiel Benedick M. Gianan
12 – Canary
Chapter 3 – Three in One
Pre-test
1.
F
2.
T
3.
T
4.
T
5.
F
Practice Exercise
Practice Exercise 3.1
An electrician experienced a mild shock when he accidentally touched a wire carrying 5.0 mA for
approximately 1.0 s. How many electrons constitute the given current?
Given:
𝐼 = 5.0 𝑚𝐴
𝑡 = 1.0 𝑠
Solution:
𝑄 = 𝑐ℎ𝑎𝑛𝑔𝑒 𝑓𝑙𝑜𝑤𝑠 𝑡ℎ𝑟𝑜𝑢𝑔ℎ 𝑎 𝑤𝑖𝑟𝑒 𝑖𝑛 𝑔𝑖𝑣𝑒𝑛 𝑡𝑖𝑚𝑒
𝑄 = 𝐼𝑡
∴ 𝑄 = 5 × 10−3 𝐶
𝑄 = 𝑡𝑜𝑡𝑎𝑙 𝑛𝑜. 𝑜𝑓 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛𝑠 × 𝑐ℎ𝑎𝑟𝑔𝑒 𝑜𝑛 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛
𝑄 = 𝑛𝑒
𝑒 = 1.6 × 10−19
5 × 10−3 = 𝑛(1.6 × 10−19 )
∴ 𝑛 = 3.125 × 1016
Practice Exercise 3.3
What color code corresponds to a resistance of 234×10³±10% with a temperature coefficient of
resistance equal to 25x10-6/C°?
GIVEN:

The resistance of the resistor is given as 243 × 105 ± 10%

The temperature coefficient of resistance is given as, 𝛼 = 25 × 10−6 /∁° or

Where, ppm is parts per million, which is 1/106 𝑜𝑟 10−6
𝛼 = 25 𝑝𝑝𝑚/∁°
TO DETERMINE:

The color code corresponding to the resistance.
RESISTANCE TABLE FOR 6-BAND RESISTOR
Color
Band Code
Decimal Multiplier Ω
TOLERANCE %
TEMPERATURE
COEFFICIENT
OF
RESISTANCE
𝑝𝑝𝑚/∁°
BLACK
0
100
BROWN
1
101
1
100
RED
2
102
2
50
ORANGE
3
103
15
YELLOW
4
104
25
GREEN
5
105
0.5
BLUE
6
106
0.25
10
VIOLET
7
107
0.1
5
GREY
8
108
WHITE
9
109
GOLD
10−1
5
SILVER
10−2
10
NONE
20
Using this table, we can find the color code of the resistance
SOLUTION:
From the table, the color code of the resistor is found as:
2 ⟶ 𝑅𝐸𝐷
3 ⟶ 𝑂𝑅𝐴𝑁𝐺𝐸
4 ⟶ 𝑌𝐸𝐿𝐿𝑂𝑊
5
10 Ω 𝑜𝑟 100 𝑘Ω ⟶ 𝐺𝑅𝐸𝐸𝑁
10% ⟶ 𝑆𝐼𝐿𝑉𝐸𝑅
25 𝑝𝑝𝑚/∁∘ ⟶ 𝑌𝐸𝐿𝐿𝑂𝑊
∴ 𝑡ℎ𝑒 𝑐𝑜𝑙𝑜𝑟 𝑐𝑜𝑑𝑒 𝑐𝑜𝑟𝑟𝑒𝑠𝑝𝑜𝑛𝑑𝑖𝑛𝑔 𝑡𝑜 𝑡ℎ𝑒 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑖𝑠 𝑓𝑜𝑢𝑛𝑑 𝑎𝑠:
𝑹𝑬𝑫 → 𝑶𝑹𝑨𝑵𝑮𝑬 → 𝒀𝑬𝑳𝑳𝑶𝑾 → 𝑮𝑹𝑬𝑬𝑵 → 𝑺𝑰𝑳𝑽𝑬𝑹 → 𝒀𝑬𝑳𝑳𝑶𝑾
Practice Exercise 3.11
A compact fluorescent lamp (CFL) uses less energy than its equivalent incandescent bulb. How
much will a person save if he/she uses a 20 W CFL instead of a 100 W incandescent lamp for 6 h in 30
days? Electricity is priced at ₱8.17/kWh.
Given:
P CFL = 20 W
Equation:
P Incandescent = 100 W
ECFL = PCFL t
t = (6 h/day) x 30 days
E Incandescent = P Incandescent t
cost = ₱8.17/kWh
ΔE = (E Incandescent - ECFL)
Priced Saved = (cost) (ΔE)
Solution:
ECFL = PCFL
= (20 W) (6 h/day x 30 days)
= 3600 W ⋅ h
ECFL = 3.6 kW ⋅ h
E Incandescent = P Incandescent t
= (100 W) (6 h/day x 30 days)
= 18000 W ⋅ h
E Incandescent = 18 kW ⋅ h
Priced Saved = (cost) (ΔE)
= (₱8.17/kWh) (E Incandescent - ECFL)
= (₱8.17/kWh) (18 kW ⋅ h - 3.6 kW ⋅ h)
Priced Saved = ₱ 117.65
Therefore, the cost saved by the person be ₱ 117.65.
Practice Exercise 3.12
A person plugged a 1250 W hair dryer and a 2500 W electric heater into the same 120 V outlet
protected by a 7.5 A circuit breaker. Will the breaker trip?
Given:
Power of hair dryer = 1250 W
Power of electric heater = 2500 W
Voltage (V) = 120V
Current (I) = 7.5 A
Solution:
P = VI
= (120V) (7.5A)
P = 900 W
Hence, the power consumption by a heater and hair dryer is higher the maximum power. So, the
breaker will trip.
Kiel Benedick M. Gianan
12 – Canary
Test Yourself
Chapter 1
1. C
6. D
11. A
2. C
7. A
12. D
3. A
8. A
13. A
4. B
9. B
14. A
5. A
10. B
15. B
1. B
6. B
11. D
2. D
7. C
12. B
3. A
8. C
13. D
4. D
9. C
14. D
5. A
10. C
15. D
1. C
6. A
11. A
2. B
7. C
12. A
3. A
8. C
13. C
4. B
9. D
14. B
5. A
10. D
15. C
Chapter 2
Chapter 3