Network Theorems FIRST DRAFT PDF

NETWORK THEOREMS  LEARNING OUTCOMES. 1. Thevenin's theorem. Understand why Thevenin’s theorem was developed. Understand and master Thevenin’s equivalent circuit. 2.SUPPER  THEVENIN’S THEOREM. It often occurs in practice that a particular element in a circuit is...

NETWORK THEOREMS  LEARNING OUTCOMES. 1. Thevenin's theorem. Understand why Thevenin’s theorem was developed. Understand and master Thevenin’s equivalent circuit. 2.SUPPER  THEVENIN’S THEOREM. It often occurs in practice that a particular element in a circuit is variable (usually called the load) while other elements are fixed. As a typical example, a household outlet terminal may be connected to different appliances constituting a variable load. Each time the variable element is changed, the entire circuit has to be analyzed all over again.  THEVENIN’S THEOREM To avoid this problem, Thevenin’s theorem provides a technique by which the fixed part of the circuit is replaced by an equivalent circuit THEVENIN’S EQUIVALENT CIRCUIT  CASE 1 If the network has no dependent sources, we turn off all independent sources. RTh is the input resistance of the network looking between terminals a and b. CASE 1 CIRCUIT  CASE 2 If the network has dependent sources, we turn off all independent sources. As with superposition, dependent sources are not to be turned off because they are controlled by circuit variables.  CASE 2 We apply a voltage source vo at terminals a and b and determine the resulting current io. Then𝑅𝑡ℎ=𝑣0 , as shown 𝑖0 in Fig.. Alternatively, we may insert a current source 𝑖0 at terminals a-b as shown in Fig and find the terminal voltage 𝑣0. CASE 2 CIRCUIT  IMPORTANT FORMULARS. 𝑉𝑇𝐻 𝐼𝐿 = 𝑅 𝑇𝐻 +𝑅𝐿 𝑅𝐿 𝑉𝐿 = 𝐼𝐿 𝑅𝐿 = 𝑉 𝑅𝑇ℎ + 𝑅𝐿 𝑡ℎ  EXAMPLE 1. Find the Thevenin equivalent circuit of the circuit shown in Fig. , to the left of the terminals a-b. Then find the current through RL = 6, 16, and 36 Ω.  SOLUTION We find 𝑅𝑡ℎ by turning off the 32-V voltage source (replacing it with a short circuit) and the 2-A current source (replacing it with an open circuit). Remove the 𝑅𝐿 load resistor from the circuit as well. The circuit becomes what is shown below. SOLUTION  SOLUTION We can calculate the 𝑅𝑡ℎ ,by calculating the total resistance in the circuit. 1.4Ω // 12Ω 12×4 =3Ω. 12+4 1.The sum of 4Ω and 12Ω resistor 3Ω is in series with 1Ω Hence 𝑅𝑡ℎ =4Ω.  SOLUTION We now calculate for 𝑉𝑡ℎ , redraw the circuit as follows.  SOLUTION 1.Find the voltage across the 12Ω, resistor. 2.𝑉𝑡ℎ = 12 𝑖1 − 𝑖2. 3.We need to find 𝑖1 𝑎𝑛𝑑𝑖2. 4.𝑖2 = -2A.  SOLUTION 1. We apply kvl to mesh 1 in order to find 𝑖1. −32 + 4𝑖1 + 12 𝑖1 − 𝑖2 = 0 16𝑖1 − 12𝑖2 = 32 16𝑖1 = 32 + 12 −2 16𝑖1 = 32 − 24 8 𝑖1 = = 0.5𝐴 16  SOLUTION Hence 𝑉𝑡ℎ = 12 0.5 + 2 = 30𝑉. Redraw the circuit as an equivalent Thevenin circuit.  SOLUTION When 𝑅𝐿 =6Ω 𝑉𝑡ℎ 𝐼𝐿 = 𝑅𝑡ℎ + 𝑅𝐿 30 𝐼𝐿 = = 3𝐴. 4+6 When 𝑅𝐿 =16Ω 𝑉𝑡ℎ 𝐼𝐿 = 𝑅𝑡ℎ + 𝑅𝐿 30 𝐼𝐿 = = 1.5𝐴. 4 + 16  SOLUTION. When 𝑅𝐿 =36Ω. 𝑉𝑡ℎ 𝐼𝐿 = 𝑅𝑡ℎ + 𝑅𝐿 30 𝐼𝐿 = = 0.75𝐴. 4 + 36  EXERCISE Using Thevenin’s theorem, find the equivalent circuit to the left of the terminals in the circuit of Fig. 4.30. Then find I.(𝑅𝑡ℎ = 45Ω, 𝑉𝑡ℎ = 90𝑉,I=1.5A).  SUPERPOSITION THEOREM If a circuit has two or more independent sources, one way to determine the value of a specific variable (voltage or current) is to use nodal or mesh analysis. Another way is to determine the contribution of each independent source to the variable and then add them up. The latter approach is known as the superposition principle. The idea of superposition rests on the linearity property. The superposition principle states that the voltage across (or current through) an element in a linear circuit is the algebraic sum of the voltages across (or currents through) that element due to each independent source acting alone.  STEPS TO TAKE WHEN APPLYING THE SUPERPOSITION THEOREM 1.Turn off all independent sources except one source. Find the output (voltage or current) due to that active source using the techniques covered in Chapters 2 and 3. 2. Repeat step 1 for each of the other independent sources. 3.Find the total contribution by adding algebraically all the contributions due to the independent sources.  EXAMPLE Use the superposition theorem to find v in the circuit of Fig  SOLUTION Since there are two sources, let 𝑣 = 𝑣1 + 𝑣2 …………………………………………………………………….(1) where 𝑣1 and 𝑣2 are the contributions due to the 6V voltage source and the 3A current source, respectively.  OBTAINING V1 set the current source to zero and redraw the circuit as follows.  OBTAINING V1 now we need to find 𝑣1 by using voltage division or KVL KVL-METHOD. −6 + 8𝑖 + 4𝑖 = 0 12𝑖 = 6 𝑖 = 0.5𝐴 Now we apply ohms law. 𝑣1 = 4 × 0.5 = 2𝑉.  OBTAINING V1 VOLTAGE DIVISION METHOD. 𝑅1 𝑣1 = ( )𝑣 𝑅1 +𝑅2 𝑜 4 𝑣1 = 6 8+4 𝑣1 = 2𝑉  OBTAINING V2 Turn off the voltage source and redraw the circuit as follows.  OBTAINING V2 Use current division to find current 𝑖3. 8 𝑖3 = ×3 8+4 𝑖3 = 2𝐴. We now use ohms law to find 𝑣2 𝑣2 = 𝑖3 × 𝑅 = 2 × 4 = 8𝑉  OBTAINING V2 By superposition theorem. 𝑣 = 𝑣1 + 𝑣2 =2+8=10V.  EXERCISE Using the superposition theorem, find Vo in the circuit.  NORTON’S THEOREM Norton’s theorem states that a linear two-terminal circuit can be replaced by an equivalent circuit consisting of a current source 𝐼𝑁 in parallel with a resistor 𝑅𝑁 , where 𝐼𝑁 is the short- circuit current through the terminals and is 𝑅𝑁 the input or equivalent resistance at the terminals when the independent sources are turned off. NORTONS EQUIVALENT CIRCUIT.  NORTONS THEOREM We find 𝑅𝑁 in the same way we find 𝑅𝑇ℎ. In f act, from what we know about source transformation, the Thevenin and Norton resistances are equal; that is, RN = RTh To find the Norton current 𝐼𝑁 , we determine the short-circuit current flowing from terminal a to b.  EXAMPLE. Find the Norton equivalent circuit of the circuit in Fig. below at terminals a-b  SOLUTION we find 𝑅𝑁 in the same way we find 𝑅𝑇ℎ in the Thevenin equivalent circuit. Set the independent sources equal to zero. This leads to the circuit below.  8Ω,4Ω and 8Ω resistor are in series their sum is 20Ω. 5 Ω // 20Ω 5×20 𝑅𝑁 = =4Ω 5+20  FINDING 𝑰𝑵 To find IN, we short-circuit terminals a and b, as shown in Fig below. We ignore the 5-Ω resistor because it has been short-circuited. Applying mesh analysis, we obtain  Mesh 𝑖1 𝑖1 = 2𝐴 Mesh 2 20𝑖2 -4𝑖1 -12=0 𝑖2 =1 A 𝐼𝑁 =1A  EXERCISE Find the Norton equivalent circuit for the circuit in Fig below terminals a- b(𝐼𝑁 =4.5A,𝑅𝑁 =90Ω)