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Unit 8 Combination Circuits

Unit 8 Combination Circuits. Objectives: Define a combination circuit. List the rules for parallel circuits. List the rules for series circuits. Solve for combination circuit values. Unit 8 Combination Circuits. Characteristics There are multiple current paths.

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Unit 8 Combination Circuits

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  1. Unit 8 Combination Circuits Objectives: • Define a combination circuit. • List the rules for parallel circuits. • List the rules for series circuits. • Solve for combination circuit values.

  2. Unit 8 Combination Circuits Characteristics • There are multiple current paths. • Resistors may be in series or parallel with other resistors. • A node is where three or more paths come together. • The total power is the sum of the resistors’ power.

  3. Unit 8 Combination Circuits A simple combination circuit.

  4. Unit 8 Combination Circuits Solving Combination Circuits E1 = ? V I1 = ? A R1 = 325 Ω E3 = ? V I3 = ? A R3 = 150 Ω E = ? V I = 1 A R = ? Ω E2 = ? V I2 = ? A R2 = 275 Ω E4 = ? V I4 = ? A R4 = 250 Ω

  5. Unit 8 Combination Circuits Series Circuit Rules • The current is the same at any point in the circuit. • The total resistance is the sum of the individual resistances. • The sum of the voltage drops or the individual resistors must equal the applied (source) voltage.

  6. Unit 8 Combination Circuits Parallel Circuit Rules • The voltage across any circuit branch is the same as the applied (source) voltage. • The total current is the sum of the current through all of the circuit branches. • The total resistance is equal to the reciprocal of the sum of the reciprocals of the branch resistances.

  7. Unit 8 Combination Circuits Simplifying the Circuit • Resistors in series can be combined to form an equivalent resistance. • Resistors in parallel can be combined to form an equivalent resistance. • The equivalent resistances are used to draw simplified equivalent circuits.

  8. Unit 8 Combination Circuits Reducing Combination Circuits Combine R1 & R2, and R3 & R4. R3 = 150 Ω R1 = 325 Ω R = ? Ω R4 = 250 Ω R2 = 275 Ω

  9. Unit 8 Combination Circuits Reducing Combination Circuits Redraw simplified circuit. R1 + R2 = R1&2 = 600 ohms R3 + R4 = R3&4 = 400 ohms R1&2 = 600 Ω R3&4 = 400 Ω R = ? Ω

  10. Unit 8 Combination Circuits Solving Combination Circuits Solve for the applied voltage using Ohm’s law. Note that the I(total) was given data. E(source) = I(total) x R(total) = 1 x 240 = 240 V E = 240 V I = 1 A R = 240 Ω R1&2 = 600 Ω R3&4 = 400 Ω

  11. Unit 8 Combination Circuits Solving Combination Circuits Solve for the branch currents using Ohm’s law. E(source) = E1&2 = E3&4 I1&2 = E1&2 / R1&2 = 240/600 = 0.4 A E = 240 V I = 0.4 A R1&2 = 600 Ω E = 240 V I = 1 A R = 240 Ω R3&4 = 400 Ω

  12. Unit 8 Combination Circuits Solving Combination Circuits Solve for the branch currents using Ohm’s law. E(source) = E1&2 = E3&4 I3&4 = E3&4 / R3&4 = 240/400 = 0.6 A E1&2 = 240 V I = 0.4 A R1&2 = 600 Ω E3&4 = 240 V I = 0.6 A R3&4 = 400 Ω E = 240 V I = 1 A R = 240 Ω

  13. Unit 8 Combination Circuits Solving Combination Circuits Expand the circuit back to the original circuit. Branch currents remain the same. E1 = ? V I1 = 0.4 A R1 = 240 Ω E3 = ? V I3 = 0.6 A R3 = 240 Ω E = 240 V I = 1 A R = 240 Ω E4 = ? V I4 = 0.6 A R4 = 240 Ω E2 = ? V I2 = 0.4 A R2 = 240 Ω

  14. Unit 8 Combination Circuits Solving Combination Circuits Solve for each voltage drop using Ohm’s law. E1 = I1 x R1 = 0.4 x 325 = 130 V E1 = 130 V I1 = 0.4 A R1 = 325 Ω E3 = ? V I3 = 0.6 A R3 = 150 Ω E = 240 V I = 1 A R = 240 Ω E4 = ? V I4 = 0.6 A R4 = 250 Ω E2 = ? V I2 = 0.4 A R2 = 275 Ω

  15. Unit 8 Combination Circuits Solving Combination Circuits Solve for each voltage drop using Ohm’s law. E2 = I2 x R2 = 0.4 x 275 = 110 V E1 = 130 V I1 = 0.4 A R1 = 325 Ω E3 = ? V I3 = 0.6 A R3 = 150 Ω E = 240 V I = 1 A R = 240 Ω E4 = ? V I4 = 0.6 A R4 = 250 Ω E2 = 110 V I2 = 0.4 A R2 = 275 Ω

  16. Unit 8 Combination Circuits Solving Combination Circuits Solve for each voltage drop using Ohm’s law. E3 = I3 x R3 = 0.6 x 150 = 90 V E1 = 130 V I1 = 0.4 A R1 = 325 Ω E3 = 90 V I3 = 0.6 A R3 = 150 Ω E = 240 V I = 1 A R = 240 Ω E4 = ? V I4 = 0.6 A R4 = 250 Ω E2 = 110 V I2 = 0.4 A R2 = 275 Ω

  17. Unit 8 Combination Circuits Solving Combination Circuits Solve for each voltage drop using Ohm’s law. E4 = I4 x R4 = 0.6 x 250 = 150 V E1 = 130 V I1 = 0.4 A R1 = 325 Ω E3 = 90 V I 3= 0.6 A R3 = 150 Ω E = 240 V I = 1 A R = 240 Ω E4= 150 V I4 = 0.6 A R4 = 250 Ω E2 = 110 V I2 = 0.4 A R2 = 275 Ω

  18. Unit 8 Combination Circuits Kirchhoff’s Laws • The algebraic sum of the voltage sources and voltage drops in a closed circuit must equal zero. This law states that the sum of the voltage drops in a series circuit must equal the applied voltage. • The algebraic sum of the current entering and leaving a point must equal zero. The second law is for parallel circuits and states that the total current is the sum of all the branch currents.

  19. Unit 8 Combination Circuits Solving Combination Circuits Review E1 = ? V I1 = ? A R1 = 325 Ω E3 = ? V I3 = ? A R3 = 150 Ω E = ? V I = 1 A R = ? Ω E2 = ? V I2 = ? A R2 = 275 Ω E4 = ? V I4 = ? A R4 = 250 Ω

  20. Unit 8 Combination Circuits Solving Combination Circuits Review: Combine R1 & R2, and R3 & R4 R3 = 150 Ω R1 = 325 Ω R = ? Ω R4 = 250 Ω R2 = 275 Ω

  21. Unit 8 Combination Circuits Solving Combination Circuits Review: Redraw simplified circuit. R1 + R2 = R1&2 = 600 ohms R3 + R4 = R3&4 = 400 ohms R1&2 = 600 Ω R3&4 = 400 Ω R = ? Ω

  22. Unit 8 Combination Circuits Solving Combination Circuits Review: Solve for the applied voltage using Ohm’s Law. Note that the I(total) was given data. E(source) = I(total) x R(total) = 1 x 240 = 240 V E = 240 V I = 1 A R = 240 Ω R1&2 = 600 Ω R3&4 = 400 Ω

  23. Unit 8 Combination Circuits Solving Combination Circuits Review: Solve for the branch currents using Ohm’s law. E(source) = E1&2 = E3&4 I1&2 = E1&2 / R1&2 = 240/600 = 0.4 A E = 240 V I = 0.4 A R1&2 = 600 Ω E = 240 V I = 1 A R = 240 Ω R3&4 = 400 Ω

  24. Unit 8 Combination Circuits Solving Combination Circuits Review: Solve for the branch currents using Ohm’s law. E(source) = E1&2 = E3&4 I3&4 = E3&4 / R3&4 = 240/400 = 0.6 A E = 240 V I = 1 A R = 240 Ω E1&2 = 240 V I = 0.4 A R1&2 = 600 Ω E3&4 = 240 V I = 0.6 A R3&4 = 400 Ω

  25. Unit 8 Combination Circuits Solving Combination Circuits Review: Expand the circuit back to the original circuit. Branch currents remain the same. E1 = ? V I1 = 0.4 A R1 = 240 Ω E3 = ? V I3 = 0.6 A R3 = 240 Ω E = 240 V I = 1 A R = 240 Ω E4 = ? V I4 = 0.6 A R4 = 240 Ω E2 = ? V I2 = 0.4 A R2 = 240 Ω

  26. Unit 8 Combination Circuits Solving Combination Circuits Review: Solve for each voltage drop using Ohm’s law. E1 = I1 x R1 = 0.4 x 325 = 130 V E1 = 130 V I1= 0.4 A R1 = 325 Ω E3 = ? V I3 = 0.6 A R3 = 150 Ω E = 240 V I = 1 A R = 240 Ω E4 = ? V I4 = 0.6 A R4 = 250 Ω E2 = ? V I2 = 0.4 A R2 = 275 Ω

  27. Unit 8 Combination Circuits Solving Combination Circuits Review: Solve for each voltage drop using Ohm’s law. E2 = I2 x R2 = 0.4 x 275 = 110 V E1 = 130 V I1 = 0.4 A R1 = 325 Ω E3 = ? V I3 = 0.6 A R3 = 150 Ω E = 240 V I = 1 A R = 240 Ω E4 = ? V I4 = 0.6 A R4 = 250 Ω E2 = 110 V I2 = 0.4 A R2 = 275 Ω

  28. Unit 8 Combination Circuits Solving Combination Circuits Review: Solve for each voltage drop using Ohm’s law. E3 = I3 x R3 = 0.6 x 150 = 90 V E1 = 130 V I1= 0.4 A R1 = 325 Ω E3 = 90 V I3 = 0.6 A R3 = 150 Ω E = 240 V I = 1 A R = 240 Ω E4 = ? V I4 = 0.6 A R4 = 250 Ω E2 = 110 V I2 = 0.4 A R2 = 275 Ω

  29. Unit 8 Combination Circuits Solving Combination Circuits Review: Solve for each voltage drop using Ohm’s law. E4 = I4 x R4 = 0.6 x 250 = 150 V E1 = 130 V I1= 0.4 A R1 = 325 Ω E3 = 90 V I3 = 0.6 A R3 = 150 Ω E = 240 V I = 1 A R = 240 Ω E4 = 150 V I4 = 0.6 A R4 = 250 Ω E2 = 110 V I2 = 0.4 A R2 = 275 Ω

  30. Unit 8 Combination Circuits Review: • The three rules for series circuits are: • The current is the same at any point in the circuit. • The total resistance is the sum of the individual resistances. • The applied voltage is equal to the sum of the voltage drops across the individual components.

  31. Unit 8 Combination Circuits Review: • The three rules for parallel circuits are: • The total voltage is the same as the voltage across any branch. • The total current is the sum of the individual currents. • The total resistance is the reciprocal of the sum of the reciprocals of the branch resistances.

  32. Unit 8 Combination Circuits Review: • Combination circuits are circuits that contain both series and parallel branches. • A node is where three or more paths come together. • The total power is the sum of all the circuit resistors’ power.

  33. Unit 8 Combination Circuits Review: • When solving combination circuits, simplify, reduce, and redraw equivalent value circuits. • Apply the series rules and the parallel rules selectively to various parts of the combination circuit.

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