# TTsu10 - 7.5 7.5 Ω a b V 7.5 Ω 1A a b I N = 7.5 V / 7.5...

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http://people.clarkson.edu/~svoboda/eta/dcWorkout/EquivR.html DCCA - THEVENIN'S THEOREM Any circuit can be converted into an equivalent circuit (Thevenins circuit) consisting of a constant voltage source in series with a resistor. V Th is equal to the voltage across terminals a & b when they are open (the resistor in question is removed from the circuit). R Th is the resistance between a & b when the voltage source is deactivated (power supply is shorted). (Current supplies are removed from the circuit) Example V Th = ( 29 ( 29 30 10 + 30 10 7 5 = . v (voltage division) R Th = 10 || 30 = (10)(30)/(10+30) = 7.5 Norton's Theorem TT1 a b V TH TH R a b 10 30 7.5 7.5 10 V a b b a V 70

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Any network can be replaced by an equivalent circuit consisting of a current source in parallel with an equivalent resistance. R N a b I N I N is equal to the current flowing between a & b when they are short- circuited. R N

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Unformatted text preview: 7.5 7.5 Ω a b V 7.5 Ω 1A a b I N = 7.5 V / 7.5 Ω = 1 A Converting Thevenin's Into Norton's R N = 7.5 Ω R N = R Th I N = V Th / R Th Best to simply determine Thevenins and then convert to Norton. Examples TT2 1. What resistor draws a current of 5 A when connected across terminals a and b of the circuit shown below? 100 V 5 20 6 5A Ω Ω Ω R = ? a b Solution: Find the Thevenin equivalent circuit. V Th = V ab (when open) = ( 29 ( 29 20 5 20 100 + V = 80 V R Th = 6 + 5 || 20 = 6 + 1/ (1/ 5 +1 / 20) = 10 Ω V / R t = 5 A ⇒ R t = 80 V/5 A = 16 Ω so R = 16 - 10 = 6 Ω R 80 V 5A 10 Ω Thevenin Equivalent 2. Find the Thevenin equivalent circuit at terminals a & b of the circuit below: TT3 Solution: 3. Find the Thevenin equivalent circuit at terminals a & b for the circuit below: TT4...
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## This note was uploaded on 05/21/2011 for the course CGN 3710 taught by Professor Bloomquist during the Summer '10 term at University of Florida.

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TTsu10 - 7.5 7.5 Ω a b V 7.5 Ω 1A a b I N = 7.5 V / 7.5...

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