Unit 7 Parallel Circuits Objectives Discuss the characteristics

Unit 7 Parallel Circuits Objectives: • Discuss the characteristics of parallel circuits. • State 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.

Unit 7 Parallel Circuits • Objectives: 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) = 3 A + 2 A + 1 A = 6 A

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. E = 120 V E 3 = 120 V E 1 = 120 V E 2 = 120 V

Unit 7 Parallel Circuits Parallel Resistance Formulas The Reciprocal Formula: 1/R(total) = 1/R 1 + 1/R 2 + 1/R 3 … 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) = (R 1 x R 2) / (R 1 + R 2)

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/R 1 + 1/R 2 + 1/R 3 … 1/R(number) R(total) R 1 R 2 R 3

Unit 7 Parallel Circuits Reciprocal Formula Example 1/R(total) = 1/R 1 + 1/R 2 + 1/R 3 … 1/R(number) 1/R(total) = 1/50 = 1/150 + 1/300 + 1/100 R(total) = 50 ohms R(total) 50 Ω R 1 150 Ω R 2 300 Ω R 3 100 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) R 1 R 2 R 3

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 R 1 24 Ω R 2 24 Ω R 3 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) = (R 1 x R 2) / (R 1 + R 2) R(total) R 1 R 2 R 3

Unit 7 Parallel Circuits Product Over Sum Formula Example Step One: R(2 & 3) = (R 2 x R 3) / (R 2 + R 3) R(2 & 3) = (30 x 60) / (30 + 60) = 1800 / 90 R(2 & 3) = 20 ohms R(total) 10 Ω R 1 20 Ω R 2 30 Ω R 3 60 Ω

Unit 7 Parallel Circuits Product-Over-Sum Formula Example Step Two: R(1 & 2 & 3) = R 1 x R(2 & 3) / R 1 + R(2 & 3) R(1 & 2 & 3) = (20 x 20) / (20 + 20) = 400 / 40 = 10 R(1 & 2 & 3) = 10 ohms = R(total) 10 Ω R 1 20 Ω R 2 30 Ω R 3 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) E 3 = 120 V I 3 = 4 A R 3 = 30 Ω P 3 = 360 W E = 120 V I = 24 A R=5Ω P = 2880 W E 1 = 120 V I 1 = 8 A R 1 = 15 Ω P 1 = 960 W E 2 = 120 V I 2 = 12 A R 2 = 10 Ω P 2 = 120 W

Unit 7 Parallel Circuits Current Divider Formula Example I(unknown) = I(total) x R(total)/R(unknown) Find I 1, I 2, and I 3. E 3 = 160 V I 3 = ? A R 3 = 120 Ω P 3 = 360 W E = 160 V I=2 A R = 80 Ω P = 320 W E 1 = 160 V I 1 = ? A R 1 = 1200 Ω P 1 = 960 W E 2 = 160 V I 2 = ? A R 2 = 300 Ω P 2 = 120 W

Unit 7 Parallel Circuits Current Divider Formula Example I(unknown) = I(total) x R(total)/R(unknown) I 1 = 2 x (80/1200) =. 133 amps E = 160 V I=2 A R = 80 Ω P = 320 W E 3 = 160 V I 3 = ? A R 3 = 120 Ω P 3 = 360 W E 1 = 160 V I 1 =. 133 A R 1 = 1200 Ω P 1 = 960 W E 2 = 160 V I 2 = ? A R 2 = 300 Ω P 2 = 120 W

Unit 7 Parallel Circuits Current Divider Formula Example I(unknown) = I(total) x R(total)/R(unknown) I 2 = 2 x (80/300) =. 533 amps E 3 = 160 V I 3 = ? A R 3 = 120 Ω P 3 = 360 W E = 160 V I=2 A R = 80 Ω P = 320 W E 1 = 160 V I 1 =. 133 A R 1 = 1200 Ω P 1 = 960 W E 2 = 160 V I 2 =. 533 A R 2 = 300 Ω P 2 = 120 W

Unit 7 Parallel Circuits Current Divider Formula Example I(unknown) = I(total) x R(total)/R(unknown) I 3 = 2 x (80/120) = 1. 33 amps E 3 = 160 V I 3 = 1. 33 A R 3 = 120 Ω P 3 = 360 W E = 160 V I=2 A R = 80 Ω P = 320 W E 1 = 160 V I 1 =. 133 A R 1 = 1200 Ω P 1 = 960 W E 2 = 160 V I 2 =. 5 A R 2 = 300 Ω P 2 = 120 W

Unit 7 Parallel Circuits 1. 2. 3. 4. 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.

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.