1. 20.1 Series and Parallel Circuits
In series circuits, current can only take one path.
The amount of current is the same at all points in a series circuit.

3. 20.1 Adding resistances in series
Each resistance in a series circuit adds to the total resistance of the circuit.
Rtotal = R1 + R2 + R3...
Total resistance
(ohms)
Individual resistances (W)

6. 20.1 Series and Parallel Circuits
In parallel circuits the current can take more than one path.
Because there are multiple branches, the current is not the same at all points in a parallel circuit.

8. Voltage and Current in a Parallel Circuit
In a parallel circuit the voltage is the same across each branch because each branch has a low resistance path back to the battery.
The amount of current in each branch in a parallel circuit is not necessarily the same.
The resistance in each branch determines the current in that branch.

9. 20.1 Advantages of parallel circuits
Parallel circuits have two big advantages over series circuits: 1. Each device in the circuit sees the full battery voltage. 2. Each device in the circuit may be turned off independently without stopping the current flowing to other devices in the circuit.

10. 20.1 Short circuit
A short circuit is a parallel path in a circuit with zero or very low resistance.
Short circuits can be made accidentally by connecting a wire between two other wires at different voltages.
Short circuits are dangerous because they can draw huge amounts of current.

12. 20.1 Adding resistance in parallel circuits
1) You are asked for the resistance.
2) You are given the circuit diagram and resistances.
3) Use the rule for parallel resistances.
4) Solve:
1/R total = 1/2 Ω + 1/4 Ω = 2/4 Ω +1/4 Ω = 3/4 Ω
R= 4/3 Ω = Ω
A circuit contains a 2 ohm resistor and a 4 ohm resistor in parallel.
Calculate the total resistance of the circuit.

13. 20.2 Three circuit laws

14. 20.2 Voltage divider V0 = R1 Vi R1 + R2 Input Output voltage
A circuit divides any supplied voltage by a ratio of the resistors.
V0 = R Vi
R1 + R2
Input
voltage
(volts)
Output
voltage
(volts)
resistor ratio
(W)

15. 20.2 Solving circuit problems
Identify what the problem is asking you to find. Assign variables to the unknown quantities.
Make a large clear diagram of the circuit. Label all of the known resistances, currents, and voltages. Use the variables you defined to label the unknowns.
You may need to combine resistances to find the total circuit resistance. Use multiple steps to combine series and parallel resistors.

16. 20.3 Alternating and direct current
The current from a battery is always in the same direction.
One end of the battery is positive and the other end is negative.
The direction of current flows from positive to negative.
This is called direct current, or DC.

17. 20.3 Alternating and direct current
If voltage alternates, so does current.
When the voltage is positive, the current in the circuit is clockwise.
When the voltage is negative the current is the opposite direction.
This type of current is called alternating current, or AC.

18. 20.2 Solving circuit problems
If you know the total resistance and current, use Ohm’s law as V = IR to calculate voltages or voltage drops. If you know the resistance and voltage, use Ohm’s law as I = V ÷ R to calculate the current.
An unknown resistance can be found using Ohm’s law as R = V ÷ I, if you know the current and the voltage drop through the resistor.
Use Kirchhoff’s current and voltage laws as necessary.

19. 20.2 Solving circuit problems
A bulb with a resistance of 1Ω is to be used in a circuit with a 6-volt battery.
The bulb requires 1 amp of current.
If the bulb were connected directly to the battery, it would draw 6 amps and burn out instantly.
To limit the current, a resistor is added in series with the bulb.
What size resistor is needed to make the current 1 amp?
1) You are asked to calculate the resistance.
2) You are told it is a series circuit and given the voltage, total current, and one resistance.
3) Use Ohm’s law, R = V ÷ I, and add the resistance in series.
4) Solve:
Total resistance = 6V ÷ 1A = 6Ω.
SInce the bulb is 1Ω, the additional resistor must be 5Ω to get a total 6Ω of resistance.

20. 20.2 Network circuits
In many circuits, resistors are connected both in series and in parallel.
Such a circuit is called a network circuit.
There is no single formula for adding resistors in a network circuit.
For very complex circuits, electrical engineers use computer programs that can rapidly solve equations for the circuit using Kirchhoff’s laws.

21. 20.2 Calculate using network circuits
Three bulbs, each with a resistance of 3Ω, are combined in the circuit in the diagram
Three volts are applied to the circuit.
Calculate the current in each of the bulbs.
From your calculations, do you think all three bulbs will be equally bright?
1) You are asked to calculate the currents.
2) You are given the circuit diagram, voltages, and resistances.
3) Use Ohm’s law, R = V ÷ I, and the series and parallel resistance formulas.
4) First, reduce the circuit by combining the two parallel resistances.
1/R total = 1/3 Ω + 1/3 Ω = 2/3 Ω
R= 3/2 Ω = 1.5 Ω
5) Calculate the total resistance of 4.5Ω by adding up the remaining series resistances.
Calculate the total current using Ohm’s law: I = 3V ÷ 4.5Ω = 0.67A.
The two bulbs in parallel have the same resistance, so they divide the current equally;
each one gets 0.33 amps.
The single bulb in series gets the full current of 0.67 amps, but the other two bulbs
get only 0.33 amps each. That means the bulbs in parallel will be much dimmer since
they only get half the current.

22. 20.3 Power
Voltage (volts)
P = VI
Power (watts)
Current (amps)

23. 20.3 Calculate power
A light bulb with a resistance of 3Ω is connected to a 1.5- volt battery in the circuit shown at right.
Calculate the power used by the light bulb.
1) You are asked to find the power used by the light bulb.
2) You are given the voltage of the battery and the bulb’s resistance.
3) Use Ohm’s law, I = V/R, to calculate the current; then use the power equation, P=VI, to calculate the power.
4) Solve: I = 1.5V ÷ 1.5Ω = 1A
P = 1.5V × 1A = 1.5 W; the bulb uses 1.5 watts of electric power.

24. 20.3 Paying for electricity
Electric companies charge for the number of kilowatt-hours used during a set period of time, often a month.
One kilowatt-hour (kWh) means that a kilowatt of power has been used for one hour.
Since power multiplied by time is energy, a kilowatt-hour is a unit of energy.
One kilowatt-hour is 3.6 x 106 joules.

25. 20.3 Calculate power
Your electric company charges 14 cents per kilowatt-hour. Your coffee maker has a power rating of 1,050 watts.
How much does it cost to use the coffee maker one hour per day for a month?
1) You are asked to find the cost of using the coffee maker.
2) You are given the power in watts and the time.
3) Use the power formula P = VI and the fact that 1 kWh = 1kW x 1h.
4) Solve: Find the number of kilowatts of power that the coffee maker uses.
1,050 W × 1 kW/1,000 W = 1.05 kW
Find the kilowatt-hours used by the coffee maker each month.
1.05 kW × 1 hr/day x 30 days/month = 31.5 kWh per month.
Find the cost of using the coffee maker.
31.5 kWh/month × $0.14/kWh = $4.41 per month.

26. 20.3 Alternating and direct current
AC current is used for almost all high-power applications because it is easier to generate and to transmit over long distances.
The 120 volt AC (VAC) electricity used in homes and businesses alternates between peak values of +170 V and V at a frequency of 60 Hz.
AC electricity is usually identified by the average voltage, (120 VAC) not the peak voltage.

27. Application: Wiring in Homes and Buildings

28. Application: Wiring in Homes and Buildings