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.

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.

A simple combination circuit.

Solving Combination Circuits

E = ? VI = 1 AR = ? Ω

E1 = ? VI1 = ? A R1 = 325 Ω

E2 = ? VI2 = ? A R2 = 275 Ω

E3 = ? VI3 = ? A R3 = 150 Ω

E4 = ? VI4 = ? A R4 = 250 Ω

Series Circuit Rules1. 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 CircuitsParallel 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.

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 CircuitsReducing Combination Circuits

Combine R1 & R2, and R3 & R4.

R = ? Ω

R1 = 325 Ω

R2 = 275 Ω

R3 = 150 Ω

R4 = 250 Ω

Reducing Combination CircuitsRedraw simplified circuit.

R1 + R2 = R1&2 = 600 ohmsR3 + R4 = R3&4 = 400 ohms

R = ? Ω R1&2 = 600 Ω R3&4 = 400 Ω

Solving Combination CircuitsSolve 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 VI = 1 AR = 240 Ω

R1&2 = 600 Ω R3&4 = 400 Ω

Solving Combination CircuitsSolve 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 VI = 1 AR = 240 Ω

E = 240 VI = 0.4 AR1&2 = 600 Ω

R3&4 = 400 Ω

Solving Combination CircuitsSolve 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 VI = 1 AR = 240 Ω

E1&2 = 240 VI = 0.4 AR1&2 = 600 Ω

E3&4 = 240 V I = 0.6 AR3&4 = 400 Ω

Solving Combination CircuitsExpand the circuit back to the original circuit.

Branch currents remain the same.

E = 240 VI = 1 AR = 240 Ω

E1 = ? VI1 = 0.4 AR1 = 240 Ω

E2 = ? VI2 = 0.4 AR2 = 240 Ω

E3 = ? VI3 = 0.6 AR3 = 240 Ω

E4 = ? VI4 = 0.6 AR4 = 240 Ω

Solving Combination CircuitsSolve for each voltage drop using Ohm’s law.

E1 = I1 x R1 = 0.4 x 325 = 130 V

E = 240 VI = 1 AR = 240 Ω

E1 = 130 VI1 = 0.4 AR1 = 325 Ω

E2 = ? VI2 = 0.4 AR2 = 275 Ω

E3 = ? VI3 = 0.6 AR3 = 150 Ω

E4 = ? VI4 = 0.6 AR4 = 250 Ω

E2 = I2 x R2 = 0.4 x 275 = 110 V

E = 240 VI = 1 AR = 240 Ω

E1 = 130 VI1 = 0.4 AR1 = 325 Ω

E2 = 110 VI2 = 0.4 AR2 = 275 Ω

E3 = ? VI3 = 0.6 AR3 = 150 Ω

E4 = ? VI4 = 0.6 AR4 = 250 Ω

E3 = I3 x R3 = 0.6 x 150 = 90 V

E = 240 VI = 1 AR = 240 Ω

E1 = 130 VI1 = 0.4 AR1 = 325 Ω

E2 = 110 VI2 = 0.4 AR2 = 275 Ω

E3 = 90 VI3 = 0.6 AR3 = 150 Ω

E4 = ? VI4 = 0.6 AR4 = 250 Ω

E4 = I4 x R4 = 0.6 x 250 = 150 V

E = 240 VI = 1 AR = 240 Ω

E1 = 130 VI1 = 0.4 AR1 = 325 Ω

E2 = 110 VI2 = 0.4 AR2 = 275 Ω

E3 = 90 VI 3= 0.6 AR3 = 150 Ω

E4= 150 VI4 = 0.6 AR4 = 250 Ω

Unit 8 Combination CircuitsKirchhoff’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.

Solving Combination Circuits Review

E = ? VI = 1 AR = ? Ω

E1 = ? VI1 = ? A R1 = 325 Ω

E2 = ? VI2 = ? A R2 = 275 Ω

E3 = ? VI3 = ? A R3 = 150 Ω

E4 = ? VI4 = ? A R4 = 250 Ω

Unit 8 Combination CircuitsSolving Combination Circuits Review:

Combine R1 & R2, and R3 & R4

R = ? Ω

R1 = 325 Ω

R2 = 275 Ω

R3 = 150 Ω

R4 = 250 Ω

Redraw simplified circuit.R1 + R2 = R1&2 = 600 ohmsR3 + R4 = R3&4 = 400 ohms

R = ? Ω R1&2 = 600 Ω R3&4 = 400 Ω

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 VI = 1 AR = 240 Ω

R1&2 = 600 Ω R3&4 = 400 Ω

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 VI = 1 AR = 240 Ω

E = 240 VI = 0.4 AR1&2 = 600 Ω

R3&4 = 400 Ω

Solving Combination Circuits Review:Solve for the branch currents using Ohm’s law.

E(source) = E1&2 = E3&4I3&4 = E3&4 / R3&4 = 240/400 = 0.6 A

E = 240 VI = 1 AR = 240 Ω

E1&2 = 240 VI = 0.4 AR1&2 = 600 Ω

E3&4 = 240 V I = 0.6 AR3&4 = 400 Ω

Expand the circuit back to the original circuit.Branch currents remain the same.

E = 240 VI = 1 AR = 240 Ω

E1 = ? VI1 = 0.4 AR1 = 240 Ω

E2 = ? VI2 = 0.4 AR2 = 240 Ω

E3 = ? VI3 = 0.6 AR3 = 240 Ω

E4 = ? VI4 = 0.6 AR4 = 240 Ω

Solving Combination Circuits Review:Solve for each voltage drop using Ohm’s law.

E1 = I1 x R1 = 0.4 x 325 = 130 V

E = 240 VI = 1 AR = 240 Ω

E1 = 130 VI1= 0.4 AR1 = 325 Ω

E2 = ? VI2 = 0.4 AR2 = 275 Ω

E3 = ? VI3 = 0.6 AR3 = 150 Ω

E4 = ? VI4 = 0.6 AR4 = 250 Ω

E2 = I2 x R2 = 0.4 x 275 = 110 V

E = 240 VI = 1 AR = 240 Ω

E1 = 130 VI1 = 0.4 AR1 = 325 Ω

E2 = 110 VI2 = 0.4 AR2 = 275 Ω

E3 = ? VI3 = 0.6 AR3 = 150 Ω

E4 = ? VI4 = 0.6 AR4 = 250 Ω

Solve for each voltage drop using Ohm’s law.E3 = I3 x R3 = 0.6 x 150 = 90 V

E = 240 VI = 1 AR = 240 Ω

E1 = 130 VI1= 0.4 AR1 = 325 Ω

E2 = 110 VI2 = 0.4 AR2 = 275 Ω

E3 = 90 VI3 = 0.6 AR3 = 150 Ω

E4 = ? VI4 = 0.6 AR4 = 250 Ω

E4 = I4 x R4 = 0.6 x 250 = 150 V

E = 240 VI = 1 AR = 240 Ω

E1 = 130 VI1= 0.4 AR1 = 325 Ω

E2 = 110 VI2 = 0.4 AR2 = 275 Ω

E3 = 90 VI3 = 0.6 AR3 = 150 Ω

E4 = 150 VI4 = 0.6 AR4 = 250 Ω

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.

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.

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.

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.

Unit 8 Combination Circuits

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