V x 4 V x. 2k = 5

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Quentin Nash
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1 Review Problem: d) Dependent sources R3 V V R Vx - R2 Vx V2 ) Determine the voltage V5 when VV Need to find voltage Vx then multiply by dependent source multiplier () Node analysis 2 V x V x R R 2 V x V x V x V x : V x V x R 3 V x V x V x V 5V : V V 5V 6 V of 9

2 Silly Bridge Circuit VL VR How do you make it so NO CURRENT goes through Rbridge? KEY: The voltages must be equal at points VL and VR V V2 R6 R8 R7 Redesign We have to find a way to get voltage across R. The resistors are not in series so you can't use a voltage divider (stupid R3 made it harder). Meh...mesh analysis, why not? 3 meshes On right side, it should be an easy equation to find VR since we have two resistors in parallel that we can combine. We can use a voltage divider equation if we get the voltage from VL! 2 of 9

3 Mesh loop I k I 2 k I I 3 I 6k I 2 k I 3 Mesh loop 2 I 2 k I k I 2 k I 2 I 3 I k I 2 k I 3 Mesh loop 3 I 3 I I 3 I 2 I 3 k I I 2 I 3 8k M : X : M X I a : I 2a : I 3a : 2.97mA.6mA.9mA Use mesh loop current I3 to get voltage across R (ohms law) I 3a kω.376 V R R R 7 k 7 :.778kΩ 3 of 9

4 Source Transformations, Thevenin Circuits, Power Transfer b). Find Vopencircuit Vthevenin Combine resistors then use the voltage divider equation to find the voltage across the load. R b : R 2b : R 6b : R 7b : Ω 3kΩ Ω kω R 5b : 8kΩ V b : V Add R2 and R7 in series R 72b : R 7b R 2b R 72b 7 kω Add R72 and R6 in parallel R 726b : R 72b R 6b R 72b R 6b R 726b.556 kω of 9

5 Voltage divider V 726b : V b R 726b R b R 726b V 726b.75V The is the voltage across R6 and R27 so use voltage divider again to find R7 V Tb : V 726b R 7b R 2b R 7b 2. Find Ishortcircuit or INorton V Tb V Short circuit across R7 which makes it go away R 2b R 6b R 62b : R 2b R 6b R 62b.2 kω V b R 62b V R62b : R 62b R b V R62b.5V V R62b I R3 : I R Nb : I R3 2b I Nb.5 ma This is the norton current 3. Find Rthevenin Long way 5 of 9

6 R R2 3k R3 R k Vtest R b R 6b Find Req R 6b : R b R 6b R 6b kω R 6b R 2b kω Two k ohms in parallel R eq : Ω. Is it possible to change R7 such that maximum power is deliver to R5? RT RL You need RT to be 8k but you have a k resistor in parallel Any resistor you put in parallel with it will result in a Req of <k 6 of 9

7 Thevenin Equivalance - Dependent source c). Find Vopen circuit VT Do nodal analysis at points Va, Vb and Vc () V a V x At node Va V a k V a k V a V b ( 2) V a V k k b k At node Vb V b V a V b.2 V k a ( 3) V a.2 V b k 7 of 9

8 At node Vc ( ) ( V a.2 ) V c k M c : C c : 3 3 X c : M c Cc X c 5 3 V a :.5V V b : 5V V c : 3V V Tc : V c 2) Find Ishortcircuit IN If you short Rload, then the current through R8 is zero so R8 goes away. The current will be I ( ).2 V a 3 3 V units here don't matter I sc : 3mA I N : I sc V c 3 Ω I sc This is the expected thevenin resitance value. 8 of 9

9 3) Find RThevenin using a test voltage sourceof V Short circuit voltage sources, open circuit INDEPENDENT current sources Find current through VTest Need Vx, what is is when Vtest V a2 V x V a2 V a2 V b2 ( ) V a2 V b2 V b2 V a2 V b2.2 V k a2 ( 2) V a2.2 V b2 k This equation below is really all we need....2 V a2 k note Vc Vtest (3) V a2.2 k V a2 : 3.2 V x :.5 V a of 9

10 Current I ma Current through R8 is also ma R eq2 : V ma Anytime you have a resistor in parallel with a voltage source it is as if the voltage is dropping across that resistor...the node above the resistor must be VTest. Therefore, it would make sense that the current through the resistor is VTest/R8... V/k ma Making Req kω R eq2 kω of 9

11 ) a) Find Vthevenin using the open circuit method VA Ix R2 5k V VB R k R8 R R3 I 3Ix VC R7 3k Voc - k KCL at node A V A 5k V A V B 3 I k x ( ) V A V 5k k B k 3 I x of 9

12 KCL at node B V B V A k V B V C V B ( 2) V A V k B V k C KCL at node C V C V B 3 I x V C k notice two resistors in series! ( 3) V B V C k Dependent source V A 5k I x ( ) V A 5k I x 5k equations unknowns VA, VB, VC, and Ix M d : of 9

13 C d : write variables as V d V A V B V C I x X d : M d C d X d V A :.66V V B : V V C : 5.33V I x :.33mA Use voltage divider now to get Voc, be careful with resistors 3 kω V ocd : V C 3kΩ kω V ocd 3.998V 3 of 9

14 b) Find INorton using the Short circuit method Ix R2 5k V i R k R8 i2 R R3 i3 I 3Ix Isc - k Source constraint i 2 : 3I x KVL on loop ( i i 2 ) ( ) 2 k i i 3 k i 5k KVL on loop 3 ( i 3 i 2 ) 2 ( ) 2 k i 3 k i 3 i k Dependent source i : I x M d2 : of 9

15 write down varialbes, i, i2, i3, and Ix V d2 C d2 : i i 2 i 3 I x X d2 : M d2 C d2 X d i.95ma i mA i 3.52mA Isc i.95ma I scd :.52mA 5 of 9

16 c) Find RThevenin using the test voltage/current method Ix R2 5k i R k R8 i2 R i3 R3 I 3Ix R7 3k i It V k Source constraint: i 2a : 3I x KVL on loop ( i i 2 ) ( ) 2 k i i 3 k i 5k KVL on loop 3: ( i 3 i 2 ) 2 ( ) 3 ( ) 2 k i 3 i k i 3 k i 3 i k KVL on loop : ( ) 3 i i 3 k Dependent source i I x 6 of 9

17 M d3 : variables i, i2, i3, i, Ix C d3 : X d3 : M d3 C d3 X d i d3 :.95mA i 2d3 :.285mA i 3d3 :.7mA i d3 :.38mA I test : i d3 I xd3 :.956mA R th : V I test R th kω Check Vth/Isc V ocd I scd 2.63 kω 7 of 9

Chapter 5. Department of Mechanical Engineering

Chapter 5. Department of Mechanical Engineering Source Transformation By KVL: V s =ir s + v By KCL: i s =i + v/r p is=v s /R s R s =R p V s /R s =i + v/r s i s =i + v/r p Two circuits have the same terminal voltage and current Source Transformation