Unit 7 Parallel Circuits. Objectives: Discuss the characteristics of parallel circuits. State the three rules for solving electrical values of resistance

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.


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.


Parallel circuits are circuits that have more than one path for current to flow.

I (total current) = 3A + 2A + 1A = 6A


Lights and receptacles are connected in parallel. Each light or receptacle needs 120 volts.

Panel


The voltage drop across any branch of a parallel circuit is the same as the applied (source) voltage.

Panel


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


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)


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


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


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


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


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


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


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


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.


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


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



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



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


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.


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.


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.

Documents

Unit 7 Parallel Circuits. Objectives: Discuss the characteristics of parallel circuits. State the three rules for solving electrical values of resistance