Unit 7 Parallel Circuits
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Transcript of Unit 7 Parallel Circuits
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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.
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Unit 7 Parallel Circuits
Objectives:• Calculate current values using the
current divider formula.
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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.
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Unit 7 Parallel CircuitsParallel circuits are circuits that have more than one
path for current to flow.I (total current) = 3A + 2A + 1A = 6A
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Unit 7 Parallel Circuits
Lights and receptacles are connected in parallel. Each light or receptacle needs 120 volts.
Panel
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Unit 7 Parallel CircuitsThe voltage drop across any branch of a
parallel circuit is the same as the applied (source) voltage.
Panel
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Unit 7 Parallel CircuitsThe 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
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Unit 7 Parallel CircuitsParallel 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)
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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
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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 + 1/100R(total) = 50 ohms
R(total)50 Ω
R1150 Ω
R2300 Ω
R3100 Ω
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Unit 7 Parallel CircuitsResistors 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
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Unit 7 Parallel CircuitsResistors 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
R324 Ω
R124 Ω
R224 Ω
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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
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Unit 7 Parallel CircuitsProduct Over Sum Formula Example
Step One:R(2 & 3) = (R2 x R3) / (R2 + R3)
R(2 & 3) = (30 x 60) / (30 + 60) = 1800 / 90R(2 & 3) = 20 ohms
R(total) 10 Ω
R120 Ω
R230 Ω
R360 Ω
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Unit 7 Parallel CircuitsProduct-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
R120 Ω
R230 Ω
R360 Ω
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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.
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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
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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
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Unit 7 Parallel CircuitsCurrent Divider Formula Example
I(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
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Unit 7 Parallel CircuitsCurrent Divider Formula Example
I(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
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Unit 7 Parallel CircuitsCurrent Divider Formula Example
I(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
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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.
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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.
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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.