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Calculating Thévenin Circuits with Dependent & Independent Sources using Norton & Thévenin, Slides of Fundamentals of Electronics

Examples on how to determine thévenin equivalent circuits for circuits containing dependent and independent sources. It explains the steps to calculate thévenin resistance (rth) and voltage (vth) for circuits with only dependent sources and for circuits with both dependent and independent sources.

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2011/2012

Uploaded on 07/19/2012

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Download Calculating Thévenin Circuits with Dependent & Independent Sources using Norton & Thévenin and more Slides Fundamentals of Electronics in PDF only on Docsity! Lecture Eighteen  Norton and Thévenin Theorems  DEPENDENT SOURCES  If circuit contains only dependent sources, an independent voltage (V0 = 1 V) or current source  (I0 = 1 mA if resistances are in kΩ) is applied at the open terminals and corresponding current  (I0) or voltage (V0) at these terminals is calculated. RTh = V0/I0. The VTh in this case is essentially  zero because there is no energy source.   Example: Determine the Thévenin equivalent circuit    Since it contain only dependent source so VOC = 0. To calculate RTh, connect a 1 V supply to the  open terminal and designate currents as shown in figure    From figure  1 11 1 1 1 2 1 0 5 2 2 1 2 1 x x x V V V VV V V V k k k + = − − + + = ⇒ − =   Solving simultaneously Vx = 3/7 V  The currents are  docsity.com 1 2 3 0 1 2 3 3 mA 1k 7 1 2 1 mA 1k 7 1 1 mA 2k 2 15 mA 14 x x VI VI I I I I I = = − = = = = = + + =   Therefore       0 1 14 k 15Th R I = = Ω   CIRCUITS WITH BOTH DEPENDENT AND INDEPENDENT SOURCES  If the circuit contains both  independent and dependent sources, the open circuited terminals  are short circuited and ISC is calculated. Then RTh = VTh/ISC.  Example: Find V0 by using Thévenin theorem  1 k -+ 2 k 12 V 2 k 1 k 1 k2000 Ix Ix B A RL   Here designate the load as RL = (1k + 1k) Ω at terminal AB. After removing the load circuitry, the  circuit contains a dependent source 2000Ix’ because it is different from the original circuit.  From figure,  12ThV V′ − =   And  2 Th x VI k ′ =   Applying KCL equations  2000 122000 122 6V 1 2 1 2 2 Th Th x Th Th x Th V VI V V VV kI V k k k k k − − −′ ′ ′− − +′= + ⇒ = + ⇒ = −   docsity.com
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