Unit 7
Parallel Circuits
Unit 7 Parallel Circuits
Objectives:
• Discuss the characteristics of parallel circuits.
• State the three rules for solving electrical values
of resistance for parallel circuits.
• Solve the missing values in a parallel circuit using
the three rules and Ohm’s law.
• Calculate current values using the current divider
formula.
Unit 7 Parallel Circuits
Three Parallel Circuit Rules
• The voltage drop across any branch is
equal to the source voltage.
• The total current is equal to the sum of the
branch currents.
• The total resistance is the reciprocal of the
sum of the reciprocals of each individual
branch.
Unit 7 Parallel Circuits
Parallel circuits are circuits that have more than
one path for current to flow.
I (total current) = 3A + 2A + 1A = 6A
Unit 7 Parallel Circuits
Lights and receptacles are connected in parallel.
Each light or receptacle needs 120 volts.
Panel
Unit 7 Parallel Circuits
The voltage drop across any branch of a parallel
circuit is the same as the applied (source) voltage.
Panel
Unit 7 Parallel Circuits
The voltage drop across any branch of a parallel
circuit is the same as the source voltage.
E3 = 120 V
E = 120 V
E1 = 120 V
E2 = 120 V
Unit 7 Parallel Circuits
Parallel Resistance Formulas
The Reciprocal Formula:
1/R(total) = 1/R1 + 1/R2 + 1/R3 …1/R(number)
The Resistors of Equal Value Formula:
R (total) = R (any resistor)/N (number of resistors)
The Product-Over-Sum Formula:
R (total) = (R1 x R2) / (R1 + R2)
Unit 7 Parallel Circuits
The Reciprocal Formula
The total resistance of a parallel circuit is the
reciprocal of the sum of the reciprocals of the
individual branches.
1/R(total) = 1/R1 + 1/R2 + 1/R3 …1/R(number)
R (total)
R1
R2
R3
Unit 7 Parallel Circuits
Reciprocal Formula Example
1/R(total) = 1/R1 + 1/R2 + 1/R3 …1/R(number)
1/R(total) = 1/50 = 1/150 + 1/300 + 1/100
R (total) = 50 ohms
R(total)
50 Ω
R1
150 Ω
R2
300 Ω
R3
100 Ω
Unit 7 Parallel Circuits
Resistors of Equal Value Formula
The total resistance of a parallel circuit is equal to the
value of one resistor, divided by the number of
resistors.
R (total) = R (any resistor) / N (number of resistors)
R (total)
R1
R2
R3
Unit 7 Parallel Circuits
Resistors of Equal Value Example
R (total) = R (any resistor)/ N (number of resistors)
R (total) = 24 (any resistor)/ 3 (number of resistors)
R (total) = 24/3 = 8 ohms
R (total)
8Ω
R1
24 Ω
R2
24 Ω
R3
24 Ω
Unit 7 Parallel Circuits
The Product-Over-Sum Formula
The total resistance of two resistors or branches is
equal to the value of the product of the resistors
divided by the sum of resistors.
R (total) = (R1 x R2) / (R1 + R2)
R (total)
R1
R2
R3
Unit 7 Parallel Circuits
Product Over Sum Formula Example
Step One:
R(2 & 3) = (R2 x R3) / (R2 + R3)
R(2 & 3) = (30 x 60) / (30 + 60) = 1800 / 90
R(2 & 3) = 20 ohms
R (total)
10 Ω
R1
20 Ω
R2
30 Ω
R3
60 Ω
Unit 7 Parallel Circuits
Product-Over-Sum Formula Example
Step Two:
R(1 & 2 & 3) = R1 x R(2 & 3) / R1 + R(2 & 3)
R(1 & 2 & 3) = (20 x 20) / (20 + 20) = 400 / 40 = 10
R(1 & 2 & 3) = 10 ohms = R (total)
R (total)
10 Ω
R1
20 Ω
R2
30 Ω
R3
60 Ω
Unit 7 Parallel Circuits
Product-Over-Sum Formula Review
• The ohm value of two branches is combined.
• This process is repeated using the combined
ohm value with the next branch.
• When all the branches are combined, this
equals the total resistance.
Unit 7 Parallel Circuits
Current Divider Formula
I (unknown) = I (total) x R (total)/ R (unknown)
E3 = 120 V
I3 = 4 A
R3 = 30 Ω
P3 = 360 W
E = 120 V
I = 24 A
R=5Ω
P = 2880 W
E1 = 120 V
I1 = 8 A
R1 = 15 Ω
P1 = 960 W
E2 = 120 V
I2 = 12 A
R2 = 10 Ω
P2 = 120 W
Unit 7 Parallel Circuits
Current Divider Formula Example
I (unknown) = I (total) x R (total)/ R (unknown)
Find I1, I2, and I3.
E = 160 V
I=2A
R = 80 Ω
P = 320 W
E3 = 160 V
I3 = ? A
R3 = 120 Ω
P3 = 360 W
E1 = 160 V
I1 = ? A
R1 = 1200 Ω
P1 = 960 W
E2 = 160 V
I2 = ? A
R2 = 300 Ω
P2 = 120 W
Unit 7 Parallel Circuits
Current Divider Formula Example
I (unknown) = I (total) x R (total)/R (unknown)
I1 = 2 x (80/1200) = .133 amps
E3 = 160 V
I3 = ? A
R3 = 120 Ω
P3 = 360 W
E = 160 V
I=2A
R = 80 Ω
P = 320 W
E1 = 160 V
I1 = .133 A
R1 = 1200 Ω
P1 = 960 W
E2 = 160 V
I2 = ? A
R2 = 300 Ω
P2 = 120 W
Unit 7 Parallel Circuits
Current Divider Formula Example
I (unknown) = I (total) x R (total)/R (unknown)
I2 = 2 x (80/300) = .533 amps
E = 160 V
I=2A
R = 80 Ω
P = 320 W
E3 = 160 V
I3 = ? A
R3 = 120 Ω
P3 = 360 W
E1 = 160 V
I1 = .133 A
R1 = 1200 Ω
P1 = 960 W
E2 = 160 V
I2 = .533 A
R2 = 300 Ω
P2 = 120 W
Unit 7 Parallel Circuits
Current Divider Formula Example
I (unknown) = I (total) x R (total)/R (unknown)
I3 = 2 x (80/120) = 1.33 amps
E = 160 V
I=2A
R = 80 Ω
P = 320 W
E3 = 160 V
I3 = 1.33 A
R3 = 120 Ω
P3 = 360 W
E1 = 160 V
I1 = .133 A
R1 = 1200 Ω
P1 = 960 W
E2 = 160 V
I2 = .5 A
R2 = 300 Ω
P2 = 120 W
Unit 7 Parallel Circuits
Review:
1. Parallel circuits have more than one
circuit path or branch.
2. The total current is equal to the sum of
the branch currents.
3. The voltage drop across any branch is
equal to the source voltage.
4. The total resistance is less than any
branch resistance.
Unit 7 Parallel Circuits
Review:
5. The total resistance can be found using
the reciprocal formula.
6. The product-over-sum formula and the
resistors of equal value formula are
special formulas.
7. Circuits in homes are connected in
parallel.
Unit 7 Parallel Circuits
Review:
8. The total power is equal to the sum of
the resistors’ power.
9. Parallel circuits are current dividers.
10. The amount of current flow through each
branch of a parallel circuit is inversely
proportional to its resistance.