Unit 8 Combination Circuits Objectives Define a combination

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. 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.

Unit 8 Combination Circuits A simple combination circuit.

Unit 8 Combination Circuits Solving Combination Circuits E=? V I=1 A R=? Ω E 1 = ? V I 1 = ? A R 1 = 325 Ω E 3 = ? V I 3 = ? A R 3 = 150 Ω E 2 = ? V I 2 = ? A R 2 = 275 Ω E 4 = ? V I 4 = ? A R 4 = 250 Ω

Unit 8 Combination Circuits Series Circuit Rules 1. The current is the same at any point in the circuit. 2. The total resistance is the sum of the individual resistances. 3. The sum of the voltage drops or the individual resistors must equal the applied (source) voltage.

Unit 8 Combination Circuits Parallel Circuit Rules 1. The voltage across any circuit branch is the same as the applied (source) voltage. 2. The total current is the sum of the current through all of the circuit branches. 3. The total resistance is equal to the reciprocal of the sum of the reciprocals of the branch resistances.

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.

Unit 8 Combination Circuits Reducing Combination Circuits Combine R 1 & R 2, and R 3 & R 4. R 1 = 325 Ω R 3 = 150 Ω R 2 = 275 Ω R 4 = 250 Ω R=? Ω

Unit 8 Combination Circuits Reducing Combination Circuits Redraw simplified circuit. R 1 + R 2 = R 1&2 = 600 ohms R 3 + R 4 = R 3&4 = 400 ohms R=? Ω R 1&2 = 600 Ω R 3&4 = 400 Ω

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 R 1&2 = 600 Ω R 3&4 = 400 Ω

Unit 8 Combination Circuits Solving Combination Circuits Solve for the branch currents using Ohm’s law. E(source) = E 1&2 = E 3&4 I 1&2 = E 1&2 / R 1&2 = 240/600 = 0. 4 A E = 240 V I=1 A R = 240 Ω E = 240 V I = 0. 4 A R 1&2 = 600 Ω R 3&4 = 400 Ω

Unit 8 Combination Circuits Solving Combination Circuits Solve for the branch currents using Ohm’s law. E(source) = E 1&2 = E 3&4 I 3&4 = E 3&4 / R 3&4 = 240/400 = 0. 6 A E = 240 V I=1 A R = 240 Ω E 1&2 = 240 V I = 0. 4 A R 1&2 = 600 Ω E 3&4 = 240 V I = 0. 6 A R 3&4 = 400 Ω

Unit 8 Combination Circuits Solving Combination Circuits Expand the circuit back to the original circuit. Branch currents remain the same. E = 240 V I=1 A R = 240 Ω E 1 = ? V I 1 = 0. 4 A R 1 = 240 Ω E 3 = ? V I 3 = 0. 6 A R 3 = 240 Ω E 2 = ? V I 2 = 0. 4 A R 2 = 240 Ω E 4 = ? V I 4 = 0. 6 A R 4 = 240 Ω

Unit 8 Combination Circuits Solving Combination Circuits Solve for each voltage drop using Ohm’s law. E 1 = I 1 x R 1 = 0. 4 x 325 = 130 V E = 240 V I=1 A R = 240 Ω E 1 = 130 V I 1 = 0. 4 A R 1 = 325 Ω E 3 = ? V I 3 = 0. 6 A R 3 = 150 Ω E 2 = ? V I 2 = 0. 4 A R 2 = 275 Ω E 4 = ? V I 4 = 0. 6 A R 4 = 250 Ω

Unit 8 Combination Circuits Solving Combination Circuits Solve for each voltage drop using Ohm’s law. E 2 = I 2 x R 2 = 0. 4 x 275 = 110 V E = 240 V I=1 A R = 240 Ω E 1 = 130 V I 1 = 0. 4 A R 1 = 325 Ω E 3 = ? V I 3 = 0. 6 A R 3 = 150 Ω E 2 = 110 V I 2 = 0. 4 A R 2 = 275 Ω E 4 = ? V I 4 = 0. 6 A R 4 = 250 Ω

Unit 8 Combination Circuits Solving Combination Circuits Solve for each voltage drop using Ohm’s law. E 3 = I 3 x R 3 = 0. 6 x 150 = 90 V E = 240 V I=1 A R = 240 Ω E 1 = 130 V I 1 = 0. 4 A R 1 = 325 Ω E 3 = 90 V I 3 = 0. 6 A R 3 = 150 Ω E 2 = 110 V I 2 = 0. 4 A R 2 = 275 Ω E 4 = ? V I 4 = 0. 6 A R 4 = 250 Ω

Unit 8 Combination Circuits Solving Combination Circuits Solve for each voltage drop using Ohm’s law. E 4 = I 4 x R 4 = 0. 6 x 250 = 150 V E = 240 V I=1 A R = 240 Ω E 1 = 130 V I 1 = 0. 4 A R 1 = 325 Ω E 3 = 90 V I 3= 0. 6 A R 3 = 150 Ω E 2 = 110 V I 2 = 0. 4 A R 2 = 275 Ω E 4= 150 V I 4 = 0. 6 A R 4 = 250 Ω

Unit 8 Combination Circuits Kirchhoff’s Laws 1. 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. 2. 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.

Unit 8 Combination Circuits Solving Combination Circuits Review E=? V I=1 A R=? Ω E 1 = ? V I 1 = ? A R 1 = 325 Ω E 3 = ? V I 3 = ? A R 3 = 150 Ω E 2 = ? V I 2 = ? A R 2 = 275 Ω E 4 = ? V I 4 = ? A R 4 = 250 Ω

Unit 8 Combination Circuits Solving Combination Circuits Review: Combine R 1 & R 2, and R 3 & R 4 R 1 = 325 Ω R 3 = 150 Ω R 2 = 275 Ω R 4 = 250 Ω R=? Ω

Unit 8 Combination Circuits Solving Combination Circuits Review: Redraw simplified circuit. R 1 + R 2 = R 1&2 = 600 ohms R 3 + R 4 = R 3&4 = 400 ohms R=? Ω R 1&2 = 600 Ω R 3&4 = 400 Ω

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 R 1&2 = 600 Ω R 3&4 = 400 Ω

Unit 8 Combination Circuits Solving Combination Circuits Review: Solve for the branch currents using Ohm’s law. E(source) = E 1&2 = E 3&4 I 1&2 = E 1&2 / R 1&2 = 240/600 = 0. 4 A E = 240 V I=1 A R = 240 Ω E = 240 V I = 0. 4 A R 1&2 = 600 Ω R 3&4 = 400 Ω

Unit 8 Combination Circuits Solving Combination Circuits Review: Solve for the branch currents using Ohm’s law. E(source) = E 1&2 = E 3&4 I 3&4 = E 3&4 / R 3&4 = 240/400 = 0. 6 A E = 240 V I=1 A R = 240 Ω E 1&2 = 240 V I = 0. 4 A R 1&2 = 600 Ω E 3&4 = 240 V I = 0. 6 A R 3&4 = 400 Ω

Unit 8 Combination Circuits Solving Combination Circuits Review: Expand the circuit back to the original circuit. Branch currents remain the same. E = 240 V I=1 A R = 240 Ω E 1 = ? V I 1 = 0. 4 A R 1 = 240 Ω E 3 = ? V I 3 = 0. 6 A R 3 = 240 Ω E 2 = ? V I 2 = 0. 4 A R 2 = 240 Ω E 4 = ? V I 4 = 0. 6 A R 4 = 240 Ω

Unit 8 Combination Circuits Solving Combination Circuits Review: Solve for each voltage drop using Ohm’s law. E 1 = I 1 x R 1 = 0. 4 x 325 = 130 V E = 240 V I=1 A R = 240 Ω E 1 = 130 V I 1= 0. 4 A R 1 = 325 Ω E 3 = ? V I 3 = 0. 6 A R 3 = 150 Ω E 2 = ? V I 2 = 0. 4 A R 2 = 275 Ω E 4 = ? V I 4 = 0. 6 A R 4 = 250 Ω

Unit 8 Combination Circuits Solving Combination Circuits Review: Solve for each voltage drop using Ohm’s law. E 2 = I 2 x R 2 = 0. 4 x 275 = 110 V E = 240 V I=1 A R = 240 Ω E 1 = 130 V I 1 = 0. 4 A R 1 = 325 Ω E 3 = ? V I 3 = 0. 6 A R 3 = 150 Ω E 2 = 110 V I 2 = 0. 4 A R 2 = 275 Ω E 4 = ? V I 4 = 0. 6 A R 4 = 250 Ω

Unit 8 Combination Circuits Solving Combination Circuits Review: Solve for each voltage drop using Ohm’s law. E 3 = I 3 x R 3 = 0. 6 x 150 = 90 V E = 240 V I=1 A R = 240 Ω E 1 = 130 V I 1= 0. 4 A R 1 = 325 Ω E 3 = 90 V I 3 = 0. 6 A R 3 = 150 Ω E 2 = 110 V I 2 = 0. 4 A R 2 = 275 Ω E 4 = ? V I 4 = 0. 6 A R 4 = 250 Ω

Unit 8 Combination Circuits Solving Combination Circuits Review: Solve for each voltage drop using Ohm’s law. E 4 = I 4 x R 4 = 0. 6 x 250 = 150 V E = 240 V I=1 A R = 240 Ω E 1 = 130 V I 1= 0. 4 A R 1 = 325 Ω E 3 = 90 V I 3 = 0. 6 A R 3 = 150 Ω E 2 = 110 V I 2 = 0. 4 A R 2 = 275 Ω E 4 = 150 V I 4 = 0. 6 A R 4 = 250 Ω

Unit 8 Combination Circuits Review: 1. The three rules for series circuits are: a. The current is the same at any point in the circuit. b. The total resistance is the sum of the individual resistances. c. The applied voltage is equal to the sum of the voltage drops across the individual components.

Unit 8 Combination Circuits Review: 2. The three rules for parallel circuits are: a. The total voltage is the same as the voltage across any branch. b. The total current is the sum of the individual currents. c. The total resistance is the reciprocal of the sum of the reciprocals of the branch resistances.

Unit 8 Combination Circuits Review: 3. Combination circuits are circuits that contain both series and parallel branches. 4. A node is where three or more paths come together. 5. The total power is the sum of all the circuit resistors’ power.

Unit 8 Combination Circuits Review: 6. When solving combination circuits, simplify, reduce, and redraw equivalent value circuits. 7. Apply the series rules and the parallel rules selectively to various parts of the combination circuit.