Basis for Thevenin and Norton Equivalent Circuits
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1 Basis for Thevenin and Norton Equivalent Circuits
2 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
3 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.
4 imitations of eal oltage ource / eal oltage ource
5 oltage ource imitations (con t) 0Ω Ω max 0 0W / min 0W 0A
6 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.
7 imitations of eal Current ource Appear as the resistance of the load on the source approaches s. eal Current ource
8 Current ource imitations (con t) 0Ω Ω min 0 0W max 0A 0W
9 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?
10 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.
11 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
12 Example #1 Find an equivalent current source to replace s and s in the circuit below.
13 Example #1 (con t) Find and. 6kΩ 18 6kΩ 3kΩ 12 / 12 / 6kΩ 2 s s s s 12 (2) (18 36mW 12 )(2)
14 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
15 Example #1 (con t) f 20kΩ mw k k k s s s s s s s 32.0 ) 2 ( ) ( Ω Ω Ω
16 Example #1 (con t) f 6kΩ mw k k k s s s s s s s 48 ) 2 (4 12 ) ( Ω Ω Ω
17 Example #1 (con t) f 3kΩ mw k k k s s s s s s s 72 ) 2 (6 12 ) ( Ω Ω Ω
18 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.
19 Example #2 Find an equivalent voltage source to replace s and s in the circuit below.
20 Example #2 (con t) Find and. 50Ω 300Ω 50Ω (300Ω) s s s s (0.714) ( ) 1.07mW
21 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 / mW mW / 300Ω (0.714)
22 Example #2 (con t) f 50Ω mw A s s s s s s s ) )( (0.25 ) ( / Ω Ω Ω
23 Example #2 (con t) f 300Ω mw A s s s s s s s ) )( (0.418 ) ( / Ω Ω Ω
24 Example #2 (con t) f 1kΩ mw A k s s s s s s s ) )( (0.927 ) ( / Ω Ω Ω
25 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.
26 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.
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