Download Section 14.1 1. series circuit 2. Kirchhoff`s voltage law 3. voltage

Section 14.1
1. series circuit
2. Kirchhoff’s voltage law
3. voltage drop
Section 14.2
4. Kirchhoff’s current law
5. short circuit
6. parallel circuit
Section 14.3
7. kilowatt
8. electrical power
9. kilowatt-hour
10. horsepower
11. direct
12. alternating
13. transformer
Reviewing Concepts
Section 14.1
1.
2. The current is the same at every point.
3. If bulbs are wired in series, when the circuit is opened, such as when a
bulb burns out, the current stops flowing everywhere in the circuit
and ALL the bulbs go out.
4. The total resistance in a series circuit is the sum of the resistances in
the circuit.
5. As more bulbs are added, the total resistance in a series circuit
increases and the brightness of the bulbs decreases.
6. Kirchhoff’s voltage law is an application of the law of conservation of
energy. The power used by the bulbs in the circuit is equal to the
power supplied by the battery. The sum of the voltage drops across
the bulbs is the same as the voltage of the battery.
Section 14.2
7. A parallel circuit is a circuit with more than one path over which the
current may flow.
8.
9. The current entering a point in a circuit is equal to the current flowing
out of the point.
10. voltage
11. Each device in the circuit has the same voltage drop as the voltage
source and each device in the circuit may be turned off independently
without stopping the current in the other devices in the circuit.
12. Appliances in a home are connected together in parallel circuits. If
one appliance is turned off, the others will still operate.
13. The total resistance decreases. There are more paths for the current to
follow and more current flows for the same voltage drop.
14. RT = R1 × R2
R1 + R2
The reciprocal of the total resistance is the sum of the reciprocals of
the individual resistances.
15. A short circuit is a parallel path in which there is little resistance.
16. Short circuits allow large currents to flow which can generate enough
heat in a wire to melt the wire or start a fire.
Section 14.3
17. 60 joules of energy is used in the light bulb every second.
18. To calculate the power of an appliance, multiply current flowing
through the appliance by the voltage drop across the appliance.
P = IV
19. The kilowatt-hour is a measure of the energy. One kilowatt of power
has been used for one hour.
20. Direct current is current that flows in one direction while alternating
current switches direction constantly.
21. Alternating current used in the United States reverses direction 60
times each second.
22. Thinner wires have more resistance than thicker wires.
23. Longer wires have more resistance than shorter wires.
24. Too long wires will cause a voltage drop and appliances may not
work properly.
25. Many appliances are designed to run on direct current of low voltage.
The box on the end of some appliance cords is used to change 120 V
alternating current, available at wall receptacles, into low voltage
direct current.
Solving Problems
Section 14.1
1. Rt = R1 + R2 + R3 = 5 Ω + 3 Ω + 8 Ω
Rt = 16 Ω
2. Vt = V1 + V2; V1 = V2
Vt = 2V1
9 volts = 2V1
V1 = 4.5 volts
3.
Rt = R1 + R2 = 3 Ω + 3 Ω
Rt = 6 Ω
Since the current is the same everywhere in a series circuit:
It = Vt ÷ Rt
It = (12 V) ÷ (6 Ω)
It = 2 amps
Vt = V1 + V2; V1 = V2
12 volts = 2V1
V = 6 volts
4.
Rt = R1 + R2 + R3 = 1 Ω + 1 Ω + 1 Ω
Rt = 3 Ω
It = Vt ÷ Rt
It = (12 V) ÷ (3 Ω)
It = 4 amps
5. Answers are:
a. Rt = R1 + R2 = 1 Ω + 1 Ω
Rt = 2 Ω
It = Vt ÷ Rt
It = (6 V) ÷ (2 Ω)
It = 3 amps
b. Rt = R1 + R2 = 3 Ω + 3 Ω
Rt = 6 Ω
It = Vt ÷ Rt
It = (6 V) ÷ (6 Ω)
It = 1 amps
c. Rt = R1 + R2 + R3 = 1 Ω + 2 Ω + 3 Ω
Rt = 6 Ω
It = Vt ÷ Rt
It = (12 V) ÷ (6 Ω)
It = 2 amps
6. Rt = R1 + R2 = 1 Ω + 1 Ω
Rt = 2 Ω
V = It × Rt
V = (1.5 A) × (2 Ω)
V = 3 volts
7. Rt = V/I
Rt = (24 V)/(3 A) = 8 Ω
R1 = R2 = Rt/2 = 8 Ω/2 = 4 Ω
Section 14.2
8. Answers are:
a. 2 amps flows through point P from left to right. The current
flowing from the source is 4 amps. 2 amps flows through the
upper branch of the circuit and 2 amps flows through the center
branch of the circuit.
b. 4 amps flowing through point P from bottom to top. The sum of
the current in the branches is 4 amps.
c. 2 amps are flowing through point P from left to right. The sum of
the current flowing back to the source is 3 amps, 1 amp from the
upper branch and 2 amps from the lower branch.
9. For each of the circuits:
a. The voltage is equal to the voltage of the source.
b. I = V/R for each resistor
c. It = I1 + I2
d. Rt = Vt/It
10. Answers are:
a.
b. I = V/R = (6 V)/(6 Ω) = 1 amp in each
c. It = I1 + I2 = 1 amp + 1 amp = 2 amps
d. Rt = Vt/It = (6 V)/(2 A) = 3 Ω
e.
ABC
V = 12 V V = 24 V V = 24 V
I = 6 amps
in each
I = 8 amps
in each
I(6Ω) = 4 amps
I(12Ω) = 2 amps
It = 12 amps It = 16 amps It = 6 amps
Rt = 1 Ω Rt = 2 Ω Rt = 4 Ω
RT = R1 × R2
R1 + R2
=3Ω
11. Answers are:
a.
b. I4 = V/R4 = (24 V)/(4 Ω) = 6 amps
I12 = V/R12 = (24 V)/(12 Ω) = 2 amps
c. It = I4 + I12 = 6 amps + 2 amps = 8 amps
d. Rt = Vt/It = (24 V)/(8 A) = 3 Ω
e. 8 ohms
Section 14.3
12. Answers are:
a. V = I × R = (2 amps) × (4 Ω) = 8 V
b. R1 = V/I = (12 V) / (3 amps) = 4 Ω
c. It = I1 + I2
12 A = 4 A + I2
I2 = 8 A
R = V/I = (24 V)/(4 A) = 6 Ω
13. Answers are:
a. P = IV = (10 amps)(120 V) = 1200 watts
b. P = IV = (2 amps)(120 V) = 240 watts
c. P = IV = (0.5 amps)(120 V) = 60 watts
14. Answers are:
a. I = P/V = (100 W) / (120 V) = 0.83 amps
b. I = P/V = (1200 W) / (120 V) = 10 amps
c. I = P/V = (30 W) / (120 V) = 0.25 amps
15. V = P/I = (15 W) / (1.5 amps) = 10 V
16. The 1.5 V cells will be connected in series so the total voltage
required will be divided by 1.5 V to determine the number of 1.5 V
cells needed.
Vt = P/I = (6 W) / (2 amps) = 3v
# of batteries = Vt/Volts per battery = 2
17. Answers are:
a. 1000 W = 1 kW
b. Energy = Power × Time = 1kW × 8 hours = 8 kW-hr
c. Total Cost = (cost per kW-hr) × (# of kW-hr)
Total cost = ($0.15/kW-hr)(8 kW-hr) = $1.20
18. Answers are:
a. 1000 W = 1 kW; 300 w = 0.3 kW
b. Energy = Power × Time = 0.3 kW × 2 hours = 0.6 kW-hr
c. Total Cost = (cost per kW-hr) × (# of kW-hr)
Total cost = ($0.15/kW-hr)(0.6 kW-hr) = $0.09 (9 cents)
Applying Your Knowledge
Section 14.1
1. Answers will vary. Almost any appliance will use series circuits, and
parallel circuits as well. A good example of a series circuit is a light
on a switch. Any lamp or LED that can turn on and off is in a series
circuit that includes the lamp and a switch to turn it on and off.
Section 14.2
2.
3. Answers are:
a. First, solve for the parallel resistors:
Then, add the series resistors:
Rt = 3 Ω + 1 Ω + 1 Ω = 5 Ω
b. Add the individual series resistors, then the parallel:
R=2Ω +4Ω =6Ω
R=3Ω +3Ω =6Ω
c.
R = 4 Ω + 4 Ω + 1 Ω= 9 Ω
Section 14.3
4. I = P/V
Answers will vary and may include:
900 W toaster = 7.5 amps
1800 W hair dryer = 15 amps
70 W portable TV = 0.58 amps
8 W clock radio = 0.067 amps
5. Answers will vary. For an example of a portable television:
a. 70 W / 1000 watts per kW = 0.070 kW
b. 2 hours per day
c. energy per day = (2 hrs/day)(0.070 kW) = 0.14 kW-hr/day
d. energy per yr = (365.25 days/yr)(0.14 kW-hr/day) = 51 kW-hr/yr
e. cost = $0.08175/kW-hr
f. cost per year = (51 kW-hr/yr)($0.08175/kW-hr) = $4.18 per yr