Unit 7 Parallel Circuits

1. 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.

2. Unit 7 Parallel Circuits Objectives: • Calculate current values using the current divider formula.

3. 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.

4. 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

5. Unit 7 Parallel Circuits Lights and receptacles are connected in parallel. Each light or receptacle needs 120 volts. Panel

6. Unit 7 Parallel Circuits The voltage drop across any branch of a parallel circuit is the same as the applied (source) voltage. Panel

7. Unit 7 Parallel Circuits The voltage drop across any branch of a parallel circuit is the same as the source voltage. E = 120 V E3 = 120 V E1 = 120 V E2 = 120 V

8. 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)

9. 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) R2 R1 R3

10. 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 R3 100 Ω R1 150 Ω R2 300 Ω

11. 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) R2 R1 R3

12. 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) 8 Ω R2 24 Ω R1 24 Ω R3 24 Ω

13. 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) R2 R1 R3

14. 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 R2 30 Ω R3 60 Ω R1 20 Ω

15. 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 R2 30 Ω R3 60 Ω R1 20 Ω

16. 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.

17. 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 = 5 Ω 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

18. Unit 7 Parallel Circuits Current Divider Formula Example I(unknown) = I(total) x R(total)/R(unknown) Find I1, I2, and I3. E3 = 160 V I3 = ? A R3 = 120 Ω P3 = 360 W E = 160 V I = 2 A R = 80 Ω P = 320 W E1 = 160 V I1 = ? A R1 = 1200 Ω P1 = 960 W E2 = 160 V I2 = ? A R2 = 300 Ω P2 = 120 W

19. Unit 7 Parallel Circuits Current Divider Formula Example I(unknown) = I(total) x R(total)/R(unknown) I1 = 2 x (80/1200) = .133 amps E = 160 V I = 2 A 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 = ? A R2 = 300 Ω P2 = 120 W

20. Unit 7 ParallelCircuits Current Divider Formula Example I(unknown) = I(total) x R(total)/R(unknown) I2 = 2 x (80/300) = .533 amps E3 = 160 V I3 = ? A R3 = 120 Ω P3 = 360 W E = 160 V I = 2 A R = 80 Ω P = 320 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

21. 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 E3 = 160 V I3 = 1.33 A R3 = 120 Ω P3 = 360 W E = 160 V I = 2 A R = 80 Ω P = 320 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

22. Unit 7 Parallel Circuits Review: • Parallel circuits have more than one circuit path or branch. • The total current is equal to the sum of the branch currents. • The voltage drop across any branch is equal to the source voltage. • The total resistance is less than any branch resistance.

23. Unit 7 Parallel Circuits Review: • The total resistance can be found using the reciprocal formula. • The product-over-sum formula and the resistors of equal value formula are special formulas. • Circuits in homes are connected in parallel.

24. Unit 7 Parallel Circuits Review: • The total power is equal to the sum of the resistors’ power. • Parallel circuits are current dividers. • The amount of current flow through each branch of a parallel circuit is inversely proportional to its resistance.