Chapter 5. Department of Mechanical Engineering


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2 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
3 Source Transformation
4 Source Transformation Example 1: Find the values of i s and R in two circuits if they are equivalent R=10 Ω i s =12/R=1.2 A
5 Source Transformation Example 2: Find current i in circuit (a) i=(5v1.2v)/(5ω+12ω)=0.224a
6 Superposition Example1:(a) A circuit containing two independent sources. (b) The circuit after the ideal ammeter has been replaced by the equivalent short circuit and a label has been added to indicate the current measured by the ammeter i m. Deactivate the current source Deactivate the voltage source i 1 =6/(3+6)=0.67 Α i 2 =[3/(3+6)]% 2=0.67 Α i m =i 1 + i 2 =1.33 Α
7 Superposition Example2: a) A circuit containing two independent sources. (b) The circuit after the ideal voltmeter has been replaced by the equivalent open circuit and a label has been added to indicate the voltage measured by the voltmeter v m. With only the voltage source With only the current source
8 Superposition Example3: (a) A circuit containing two independent sources. (b) The circuit after the ideal ammeter has been replaced by the equivalent short circuit and a label has been added to indicate the current measured by the ammeter i m. With only the voltage source With only the current source
9 Superposition Example4: (a) A circuit containing two independent sources. (b) The circuit after the ideal voltmeter has been replaced by the equivalent open circuit and a label has been added to indicate the voltage measured by the voltmeter v m. With only the current source With only the voltage source
10 Superposition Example5:Find current i. Deactivate the current source Deactivate the voltage source
11 Superposition Example5: Find current i. For circuit (a): by KVL, we have 24(3+2)i 13i 1 =0 Then i 1 =3 (A) i=i 1 +i 2 =1.25 (A) For circuit (b): we use node voltage analysis at node a i 2 7+ (v a 3i 2 )/2=0 For 3Ω resistor, we have i 2 =v a /3 Then i 2 =7/4
12 Thévenin s Theorem (a) A circuit partitioned into two parts: circuit A and circuit B. (b) Replacing circuit A by its Thévenin equivalent circuit. A: Driving circuit B: Load
13 Thévenin s Theorem The Thévenin equivalent circuit involves three parameters: (a) the opencircuit voltage, v oc, (b) the shortcircuit current i sc, and (c) the Thévenin resistance, R t. v oc =R t i sc
14 Thévenin s Theorem (a) The Thévenin resistance, R t, (b) A method for measuring or calculating the Thévenin resistance, R t. R t =v t /i t
15 Thévenin s Theorem Example1: Find current i using Thévenin s Theorem + v oc  Steps for determining the Thévenin equivalent circuit for the circuit left of the terminals R t =4+5//20=8Ω V oc = % 50 =40 V 40 i= R+8
16 Thévenin s Theorem Example2: Find the Thévenin equivalent circuit for: Keep in mind: The Thévenin equivalent circuit involves three parameters: (a) the opencircuit voltage, v oc, (b) the shortcircuit current i sc, and (c) the Thévenin resistance, R t. v oc =R t i sc
17 Thévenin s Theorem First, find R t : Circuit reduction by deactivate all ideal sources Then find the equivalent resistant R t =10//40 + 4=12Ω
18 Thévenin s Theorem Then, find v oc : + v oc Using Node voltage mothod to find v c, since 1b is open circuit, no voltage drop for 4Ω resistor, v oc = v c v c v c + 2=0 40 Solve for v c v c =8V
19 Thévenin s Theorem For circuit with dependent sources, we can not directly obtain the R t from simple circuit reduction. The procedure to get R t : Find open circuit voltage v oc, Find the shortcircuit current i sc, R t = v oc Example 3: Find the Thévenin s equivalent circuit for the following circuit: i sc
20 Thévenin s Theorem First, find open circuit voltage V oc V oc For the left loop, apply KVL: 206i+2i6i=0 i=2 (A) V oc =6i=12 (V)
21 Thévenin s Theorem Make ab a short circuit, and find i sc, By mesh current method, we have 206i 1 +2i 16(i 1 i 2 )=0 And 6(i 2 i 1 )10i 2 =0 i 2 = i sc =120/136 (A) R t = v oc i sc
22 Thévenin s Theorem The Thévenin s resistance is R t = v oc 10 = i sc 120/136 =13.6 Ω The Thévenin s equivalent circuit is
23 Norton s equivalent Circuit (a) A circuit partitioned into two parts: circuit A and circuit B. (b) Replacing circuit A by its Norton equivalent circuit. Norton equivalent is simply the source transformation of the Thévenin equivalent
24 Norton s equivalent Circuit Example1: Find the Norton Equivalent Circuit for Find R n by replacing the voltage source with a short circuit R n = 6x = 4 kω
25 Norton s equivalent Circuit Find the short circuit current i sc, Apply KVL for the large loop: i sc =0 i sc =1.25mA (Note: No current for the 6kΩ resistor it is shorted) i sc =1.25mA R n =4000Ω
26 Norton s equivalent Circuit Example2: Find the Norton Equivalent Circuit for First, find the open circuit voltage: Apply KVL for the close loop: 12+6i a 2i a =0 i a =3(A) v oc =2i a =6(V) v oc
27 Norton s equivalent Circuit Then, find the short circuit current: Apply KVL for the left loop: 12+6i a 2i a =0 i a =3(A) i sc i sc = 2i a 3 R t = =2 (A) v oc i sc = 6 V 2 A =3Ω
28 Maximum Power Transfer For a circuit A and load resistor R L Circuit A contains resistors and independent and dependent sources. The Thévenin equivalent is substituted for circuit A. Here we use v s for the Thévenin source voltage.
29 Maximum Power Transfer It can be proved that when R t =R L maximum power transferred from circuit A to the load resistor, and the power is v 2 s P max = 4R t We can also use Norton s equivalent circuit to substitute circuit A. Here we use i s as the Norton source current. Again, the maximum power occurs at R t =R L and the maximum power is P max = R t i s 2 4
30 Maximum Power Transfer Example: Find the Load R L that result in maximum power delivered to the load. Also determine P max First, we use the circuit (b) to obtain the Thévenin equivalent circuit. Find the open circuit voltage v oc. Apply KVL to the close loop: 66i+2v ab 4i=0 v oc And v ab =4i i=3a v oc = v ab = 12 (V)
31 Maximum Power Transfer Find the short circuit current for circuit (c), i sc. Since ab is short, v ab =0. Apply KVL to the close loop: 66i sc +2v ab =0 i sc =1 (A) Find the equivalent resistance: R t = v oc i sc =12 Ω R L =R t =12 Ω P max = v oc 2 4R t = (12) =3 W
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