Basis for Thevenin and Norton Equivalent Circuits
Objective of ecture Describe the differences between ideal and real voltage and current sources Chapter 8.1 and 8.2 rinciples of Electric Circuits Demonstrate how a real voltage source and real current source are equivalent so one source can be replaced by the other in a circuit. Chapter 6.6 Electric Circuit Fundamentals Chapter 8.3 rinciples of Electric Circuits Chapter 4.4 Fundamentals of Electric Circuits Chapter 2.6 Electrical Engineering: rinciples and Applications
oltage ources deal An ideal voltage source has no internal resistance. t can produce as much current as is needed to provide power to the rest of the circuit. eal A real voltage sources is modeled as an ideal voltage source in series with a resistor. There are limits to the current and output voltage from the source.
imitations of eal oltage ource / eal oltage ource
oltage ource imitations (con t) 0Ω Ω max 0 0W / min 0W 0A
Current ources deal An ideal current source has no internal resistance. t can produce as much voltage as is needed to provide power to the rest of the circuit. eal A real current sources is modeled as an ideal current source in parallel with a resistor. imitations on the maximum voltage and current.
imitations of eal Current ource Appear as the resistance of the load on the source approaches s. eal Current ource
Current ource imitations (con t) 0Ω Ω min 0 0W max 0A 0W
Electronic esponse For a real voltage source, what is the voltage across the load resistor when s? For a real current source, what is the current through the load resistor when s?
Equivalence An equivalent circuit is one in which the i-v characteristics are identical to that of the original circuit. The magnitude and sign of the voltage and current at a particular measurement point are the same in the two circuits.
Equivalent Circuits in both circuits must be identical. and in the left circuit and on the left 1 2 eal oltage ource eal Current ource
Example #1 Find an equivalent current source to replace s and s in the circuit below.
Example #1 (con t) Find and. 6kΩ 18 6kΩ 3kΩ 12 / 12 / 6kΩ 2 s s s s 12 (2) (18 36mW 12 )(2)
Example #1 (con t) There are an infinite number of equivalent circuits that contain a current source. f, in parallel with the current source, s Ω s is an open circuit, which means that the current source is ideal. 2(6kΩ) s 12 (2) 24mW 12 24mW
Example #1 (con t) f 20kΩ mw k k k s s s s s s s 32.0 ) 2 (2.67 12 ) (2 12 12 2.67 2 20 20 6 Ω Ω Ω
Example #1 (con t) f 6kΩ mw k k k s s s s s s s 48 ) 2 (4 12 ) (2 12 12 4 2 6 6 6 Ω Ω Ω
Example #1 (con t) f 3kΩ mw k k k s s s s s s s 72 ) 2 (6 12 ) (2 12 12 6 2 3 3 6 Ω Ω Ω
Example #1 (con t) Current and power that the ideal current source needs to generate in order to supply the same current and voltage to a load increases as decreases. Note: s can not be equal to 0Ω. The power dissipated by is 50% of the power generated by the ideal current source when.
Example #2 Find an equivalent voltage source to replace s and s in the circuit below.
Example #2 (con t) Find and. 50Ω 300Ω 50Ω 0.714 0.714(300Ω) 0.214 s s s s 0.214 (0.714) 0.214 (5 0.714) 1.07mW
Example #2 (con t) There are an infinite number of equivalent circuits that contain a voltage source. f, in series with the voltage source, s 0Ω s is a short circuit, which means that the voltage source is ideal. s / 0.214 0.714 0.153mW 0.214 0.153mW / 300Ω 0.214 (0.714)
Example #2 (con t) f 50Ω mw A s s s s s s s 0.179 ) )(0.714 0.214 (0.25 ) (0.714 0.214 0.714 / 0.25 0.214 300 50 300 Ω Ω Ω
Example #2 (con t) f 300Ω mw A s s s s s s s 0.306 ) )(0.714 0.214 (0.418 ) (0.714 0.214 0.714 / 0.418 0.214 300 300 300 Ω Ω Ω
Example #2 (con t) f 1kΩ mw A k s s s s s s s 0.662 ) )(0.714 0.214 (0.927 ) (0.714 0.214 0.714 / 0.927 0.214 300 1 300 Ω Ω Ω
Example #2 (con t) oltage and power that the ideal voltage source needs to supply to the circuit increases as increases. s can not be equal to Ω. The power dissipated by is 50% of the power generated by the ideal voltage source when.
ummary An equivalent circuit is a circuit where the voltage across and the current flowing through a load are identical. As the shunt resistor in a real current source decreases in magnitude, the current produced by the ideal current source must increase. As the series resistor in a real voltage source increases in magnitude, the voltage produced by the ideal voltage source must increase. The power dissipated by is 50% of the power produced by the ideal source when.