Unit 7Parallel 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.

E1 = 120 V E2 = 120 V

E3 = 120 VE = 120 V

Unit 7 Parallel Circuits

Parallel Resistance FormulasThe 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 FormulaThe 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 Example1/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

R1150 Ω

R2300 Ω

R3100 Ω

Unit 7 Parallel Circuits

Resistors of Equal Value FormulaThe 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 ExampleR (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)8 Ω

R324 Ω

R124 Ω

R224 Ω

Unit 7 Parallel Circuits

The Product-Over-Sum FormulaThe 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 ExampleStep 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

R120 Ω

R230 Ω

R360 Ω

Unit 7 Parallel Circuits

Product-Over-Sum Formula ExampleStep 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

R120 Ω

R230 Ω

R360 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 FormulaI (unknown) = I (total) x R (total)/ R (unknown)

E1 = 120 VI1 = 8 A R1 = 15 Ω P1 = 960 W

E2 = 120 VI2 = 12 A R2 = 10 ΩP2 = 120 W

E3 = 120 VI3 = 4 A R3 = 30 Ω P3 = 360 W

E = 120 VI = 24 AR = 5 ΩP = 2880 W

Unit 7 Parallel CircuitsCurrent Divider Formula Example

I (unknown) = I (total) x R (total)/ R (unknown)

Find I1, I2, and I3.

E1 = 160 VI1 = ? A R1 = 1200 ΩP1 = 960 W

E2 = 160 VI2 = ? A R2 = 300 ΩP2 = 120 W

E3 = 160 VI3 = ? A R3 = 120 ΩP3 = 360 W

E = 160 VI = 2 AR = 80 Ω P = 320 W

Unit 7 Parallel Circuits

Current Divider Formula ExampleI (unknown) = I (total) x R (total)/R (unknown)

I1 = 2 x (80/1200) = .133 amps

E1 = 160 VI1 = .133 A R1 = 1200 ΩP1 = 960 W

E2 = 160 VI2 = ? A R2 = 300 ΩP2 = 120 W

E3 = 160 VI3 = ? A R3 = 120 ΩP3 = 360 W

E = 160 VI = 2 AR = 80 ΩP = 320 W

Unit 7 Parallel Circuits

Current Divider Formula ExampleI (unknown) = I (total) x R (total)/R (unknown)

I2 = 2 x (80/300) = .533 amps

E1 = 160 VI1 = .133 A R1 = 1200 ΩP1 = 960 W

E2 = 160 VI2 = .533 A R2 = 300 Ω P2 = 120 W

E3 = 160 VI3 = ? A R3 = 120 Ω P3 = 360 W

E = 160 VI = 2 AR = 80 ΩP = 320 W

Unit 7 Parallel Circuits

Current Divider Formula ExampleI (unknown) = I (total) x R (total)/R (unknown)

I3 = 2 x (80/120) = 1.33 amps

E1 = 160 VI1 = .133 A R1 = 1200 ΩP1 = 960 W

E2 = 160 VI2 = .5 A R2 = 300 ΩP2 = 120 W

E3 = 160 VI3 = 1.33 A R3 = 120 ΩP3 = 360 W

E = 160 VI = 2 AR = 80 ΩP = 320 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.