1. Linear Circuit Analysis Linear Circuits Superposition

2. Linear Circuits Most circuits we will study are linear Linear circuits contain linear elements – those that have a linear relationship between their voltage and their current – Resistors – Voltage and Current Sources – Dependent sources that depend on a voltage or current (but not if they depend on a product of current and voltage or some current or voltage to a power different than one)

3. Superposition Linear circuits may be analyzed by looking at the voltages and currents produced by each independent source when acting alone and then adding them together When one source acts on a circuit, the other independent sources must be set to zero – A voltage source set to zero is equivalent to a short circuit – A current source set to zero is equivalent to an open circuit

4. Sources when set to Zero Voltage Sources Current Sources + Vs=0 Ξ Short - Circuit Is Ξ Open Circuit

5. Superposition Superposition allows us to reduce the complexity of a circuit Superposition requires that more circuits be analyzed (more simpler circuits) Superposition allows us to analyze circuits that contain sources of different types – DC, AC, different frequency AC sources, triangular waves, square waves, etc.

6. Circuit Analysis using Linearity In a linear circuit with one independent source, if you double the value of the source, then every current and voltage in the circuit doubles Therefore, you can guess at any voltage or current in the circuit and see what the source would have to be in order to get that current and then apply a scale factor

7. Linear Circuit Analysis When a circuit has one independent source, you can analyze it by choosing a current or voltage somewhere in the circuit and then determine what the source had to be in order to get that value Then you can apply a scale factor to all the assumed currents and voltages in the circuit to get the actual values of the source and other voltages and currents

8. Example Let the 4 A current source act alone, by zeroing out the voltage source (effectively making it a short circuit.

9. Example Let the 4 A current source act alone, by zeroing out the voltage source (effectively making it a short circuit). Short circuit Short Circuit

10. Example Let the current source be Is, and choose a convenient value for I 4 of 10 A, then V B = 25 volts and I 3 = 6.25 A Since I 2 =I 4 +I 3, then I 2 = 16.25 A Short circuit Short Circuit Is

11. Example If I 2 = 16.25 A, and V A -V B = I 2 ∙2 Ω Since V B =25 v, then V A =57.5 v Then I 1 =V A /14Ω= 4.11 A So Is = I 1 + I 2 = 20.36 A Short circuit Short Circuit Is

12. Example But Is = 4 A, so if we multiply everything by a scale factor of 4/20.36 we will get the correct voltages and currents. So I 4 = 10 A ∙ (4 / 20.36) = 1.96 A Short circuit Short Circuit Is

13. Linear Circuit Analysis When a circuit has one independent source, you can analyze it by choosing a current or voltage somewhere in the circuit and then determine what the source had to be in order to get that value Then you can apply a scale factor to all the assumed currents and voltages in the circuit to get the actual values of the source and other voltages and currents

14. Superposition Example We found if I 4 =10 A, Is=20.36 A, V B = 25 volts, I 3 = 6.25 A, I 2 = 16.25 A, V A =57.5 v & I 1 = 4.11 A Knowing that Is = 4 A, we can multiply each voltage and current by 4/20.36, so Is=4 A, I 4 =1.96 A, V B = 4.91 v, I 3 = 1.23 A, I 2 = 3.19 A, V A = 11.30 v & I 1 =.81 A Short circuit Short Circuit Is

15. Example To finish the analysis by superposition, let the 34 v source act alone, by zeroing out the current source (effectively making it an open circuit. Open Circuit

16. Example Pretend that the source is Vs and choose a convenient value for such as I 1 = -I 2 = 1 A. So V A = 14 volts and V B = 16 volts and I 3 = 4 A and I 4 = I 2 - I 3 = -5 A Open Circuit Vs

17. Example If I 4 = -5 A, then V B -Vs = I 4 ∙2.5 Ω = -12.5 v But V B = 16 v, so Vs = 28.5 volts Vs is really 34 v, so our scale factor is 34/28.5, so I 4 = -5 A∙(34/28.5) = -5.96 A Open Circuit Vs

18. Example If I 4 = 1.96 A when the 4 A source was acting alone and if I 4 = -5.96 A when the 34 volt source was acting alone, Then I 4 = 1.96 - 5.96 = -4.00 A when both sources are working together

19. Summary Linear circuits allow us to analyze the response to each independent source and add them together (superposition) – Especially valuable for different types of sources Linear circuit analysis (where a current or voltage is assumed in order to get a scale factor to get the true voltages and currents) can be used for any circuit with a single independent source