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Network Theorems - Alternating Current examples - J. R. Lucas In the previous chapter, we have been dealing mainly with direct current resistive circuits in order to the principles of the various theorems clear. As was mentioned, these theories are equally valid for a.c. Example 1 I1 19.89 µF −j160 Ω E1 = 100∠0o V j20 Ω 63.66 mΗ I2 20 Ω I E2 = 70∠-30o V For the circuit shown in the figure, if the frequency of the supply is 50 Hz, determine using Ohm’s Law and Kirchoff’s Laws the current I in the 160 Ω capacitor. Impedance of capacitor and inductor at 50 Hz are j XL = j 2π × 50 × 63.66 × 10−3 = j 20 Ω and - j XC = 1/( j 2π × 50 × 19.89 × 10−6) = − j 160 Ω Using Kirchoff’s current law I = I1 − I2 Using Kirchoff’s voltage law 100∠0o = 20 I1 – j160 (I1 – I2) and ⇒ 70∠-30 = – j 20 I2 – j 160 (I1 – I2) ⇒ o 5 = (1-j8) I1 +j8 I2 o ...... (1) 7∠-30 = – j 16 I1+ j 14 I2 ...... (2) multiplying equation (1) by 7 and equation (2) by 4 and subtracting gives ⇒ 35 - 28∠-30o = (7 – j 56 + j 64) I1 + 0 i.e. I1 = 10.751 + j 14 = (7 + j 8)I1 17.65∠52.48 o = 1.660∠3.67o A o 10.63∠48.81 substituting in (1), j8 I2 = 5 – (1-j8) × 1.660∠3.67o i.e. 8∠90o I2 = 5 – 8.062∠-82.87o × 1.660∠3.67o = 5 – 13.383∠-79.20o = 2.492 + j13.146 i.e. 13.38∠79.27 o I2 = = 1.673∠-10.73o A o 8∠90 Thus the required current I is = 1.660∠3.67o – 1.673∠-10.73o = 1.657 + j 0.106 – 1.644 + j 0.311 = 0.013 +j 0.417 = 0.42∠88.2o A The problem could probably have been worked out with lesser steps, but I have done it in this manner so that you can get more familiarised with the solution of problems using complex numbers. Network Theorems- a c examples – Professor J R Lucas 1 November 2001 Example 2 Let us solve the same problem as earlier, but using Superposition theorem. I1 20 Ω 19.89 µF −j160 Ω E1 = 100∠0o V j20 Ω 63.66 mΗ I2 I E2 = 70∠-30o V This circuit can be broken into its two constituent components as shown. j20 Ω 63.66 mΗ I2A I1A 20 Ω E1 = 100∠0o V 19.89 µF −j160 Ω IA j20 Ω 63.66 mΗ I2B I1B 20 Ω IB IB 19.89 µF −j160 Ω + E2 = 70∠-30o V Using series parallel addition of impedances, we can obtain the supply currents as follows. Equivalent Zs1 = 20 + (-j160)//j20, = 20 + Zs2 = j20 + 20//(-j160) − j160 × j 20 , − j140 = j20 + 20 × (− j160) 20 − j160 = 20 + j 22.857, = j20 + 19.692 – j 2.462 = 19.692 + j17.538 = 30.372∠48.81o Ω, = 26.370∠41.69o Ω source current I1A = 100∠0 o , 30.372∠48.81o −I2B = = 3.293∠-48.81o, 70∠ − 30 o 26.370∠41.69 o = 2.655∠-71.69o, Using the current division rule (note directions of currents and signs), IA = 3.293∠− 48.81o × j 20 , − j140 IB = 2.655∠− 71.69 o × = 0.470∠131.19o, = 20 20 − j160 53.10∠− 71.69 o = 0.329∠11.18o o 161.25∠ − 82.87 Using superposition theorem, the total current in I = 0.470∠131.19o + 0.329∠11.18o = -0.310 + j 0.354 + 0.323 + j0.064 = 0.013 + j 0.419 = 0.42∠88.2o A which is the same answer obtained in the earlier example. Network Theorems- a c examples – Professor J R Lucas 2 November 2001 Example 3 Let us again consider the same example to illustrate Thevenin’s Theorem. I1 20 Ω P 19.89 µF −j160 Ω E1 = 100∠0o V j20 Ω 63.66 mΗ I2 I E2 = 70∠-30o V Q Consider the capacitor disconnected at P and Q. Current flowing in the circuit under this condition = 100∠0 o − 70∠ − 30 o 20 + j 20 100 − 60.62 + j 35 52.69∠41.63o = = = 1.863∠ − 3.37 o o 20 + j 20 28.28∠45 ∴Thevenin’s voltage source = 100∠0o – 20 × 1.863∠-3.37o = 62.80 + j 2.19 = 62.84∠2.00o 20 × j 20 = 14.142∠45o = 10 + j 10 20 + j 20 Also, Thevenin’s impedance across Q = 20//j20 = ∴Thevenin’s equivalent circuit is 10 + j 10 Ω Eth = 62.84∠2.00o V From this circuit, it follows that P I 19.89 µF −j160 Ω I = 62.84∠2.00 o 62.84∠2.00 o = 10 + j10 − j160 150.33∠ − 86.19 o = 0.418∠-88.2o A which is again the same result. Q Example 4 Let us again consider the same example to illustrate Norton’s Theorem. I1 20 Ω P j20 Ω 63.66 mΗ I2 I E1 = 100∠0o V 19.89 µF −j160 Ω E2 = 70∠-30o V Q Consider the capacitor short-circuited at P and Q. The Norton’s current source is given as 100∠0 o 70∠ − 30 o + = 5 –1.75 –j3.031 20 j 20 = 3.25 – j 3.031 = 4.444∠-43.00o Network Theorems- a c examples – Professor J R Lucas 3 November 2001 Norton’s admittance = 1 1 + = 0.05 − j 0.05 S 20 j 20 I 4.444∠-43.0o or same as Thevenin’s impedance 10 + j10 Ω −j160 The Norton’s equivalent circuit is as shown in the figure. The current through the capacitor can be determined using the current division rule. I = 4.444∠−43.0o × 14.142∠45o 10 + j10 = 4.444∠−43.0o × 10 + j10 − j160 150.33∠ − 86.19 o = 0.418∠-88.2o A Example 5 Using Millmann’s theorem find the current in the capacitor. I1 20 Ω 19.89 µF −j160 Ω E1 = 100∠0o V S j20 Ω 63.66 mΗ I2 I E2 = 70∠-30o V N VSN 1 1 ⋅ 100∠0 o + ⋅0 + Y.V 20 − j160 ∑ = = 1 1 ∑Y + + 20 − j160 = 1 ⋅ 70∠ − 30 o j 20 1 j 20 4.444∠− 43.00 o 5 + 0 − 1.75 − j 3.031 3.25 − j 3.031 = = 0.05 + j 0.00625 − j 0.05 0.05 − j 0.04375 0.06643∠ − 41.19 = 66.89∠-1.81o V ∴ I = 66.89∠-1.81o /(-j160) = 0.418∠88.19o A Example 6 Determine the delta equivalent of the star connected network shown. S A YAB A B B j20 Ω 20 Ω ≡ −j160 Ω YCA YBC C C Network Theorems- a c examples – Professor J R Lucas 4 November 2001 YAB 1 1 × 20 j 20 1 1 = = , = 1 1 1 j 20 − 2.5 + 20 17.5 + j 20 + + 20 − j160 j 20 YBC 1 1 × 1 1 − j160 j 20 = = , ∴ ZBC = 160 – j 140 Ω = 1 1 1 160 + j 20 − j160 160 − j140 + + 20 − j160 j 20 YCA 1 1 × 20 − j160 1 1 = = , ∴ ZCA = –140 – j 160 Ω = 1 1 1 − j160 + 20 − 160 − 140 − j160 + + 20 − j160 j 20 ∴ ZAB = 17.5 + j 20 Ω Example 7 Determine the star equivalent of the delta connected network shown. 17.5+j20 Ω A S A B ZB ZA − 140 − j160Ω 160−j140 Ω ≡ B ZC C C ZA = (17.5 + j 20)(−140 − j160) 26.575∠48.81o × 212.603∠ − 131.19 o = 17.5 + j 20 − 140 − j160 + 160 − j140 37.5 − j 280 = 5650∠ − 82.38 o = 20.00∠ − 0.01o = 20 Ω (same as original value in Ex 6). o 282.5∠ − 82.37 ZB = (17.5 + j 20)(160 − j140) 26.575∠48.81o × 212.603∠ − 41.19 o = 17.5 + j 20 − 140 − j160 + 160 − j140 37.5 − j 280 = 5650∠7.62 o = 20.00∠89.99 o = j 20 Ω (same as original value in Ex 6). o 282.5∠ − 82.37 ZC = = 212.603∠− 41.19 o × 212.603∠− 131.19 o (160 − j140)(−140 − j160) = 37.5 − j 280 17.5 + j 20 − 140 − j160 + 160 − j140 45200∠− 172.38 o = 160.00∠− 90.01o = − j160 Ω (same as original value in Ex 6). o 282.5∠− 82.37 In order to show that the working is correct, I have selected the reverse problem for this example and used the results of the previous example to find the original quantities. You can see that the answers differ only due to the cumulative calculation errors. Network Theorems- a c examples – Professor J R Lucas 5 November 2001 Example 8 Determine using compensation theorem, the current I, if the available capacitor is 20 µF, instead of the 19.89 µF already assumed in the earlier problems. I1 j20 Ω 63.66 mΗ I2 20 Ω 19.89 µF −j160 Ω E1 = 100∠0o V I E2 = 70∠-30o V Solution 20 µF corresponds to 1 = -j159.15 Ω j 20 × 10 × 2 × π × 50 −6 change of impedance ∆Z = –j159.15 – (–j160) = j0.85 Ω. from earlier calculations I = 0.418∠-88.2o A ∴ using compensation theorem, I . ∆Z = 0.418∠-88.2o× j0.85 = 0.355∠1.8o V ∴ changes in current in the network can be obtained from Note that the direction of ∆I is marked in the same direction as the original I, so that the source would in fact send a current in the opposite direction. i.e. = = ∆I1 j20 Ω 63.66 mΗ ∆I2 20 Ω 0.355∠1.8o 0.355∠1.8 o – ∆I = − j159.15 + 20 // j 20 ∆I 0.355∠1.8 o 0.355∠1.8 o = 20 × j 20 − j159.15 + 10 + j10 − j159.15 + 20 + j 20 20 µF −j159.15 Ω 0.355∠1.8 o = 0.00237∠88.0o, giving ∆I as –0.00237∠88.0o, or 0.00237∠268.0o A o 149.48∠− 86.16 i.e. correct current I = 0.418∠-88.2o + 0.00237∠268.0o = 0.013 – j 0.4177 –0.00008–j0.00237 = 0.013 – j 0.420 = 0.420∠-88.2o A Comparing result using Thevenin’s equivalent circuit derived in example 3 10 + j 10 Ω 62.84∠2.00o V From this circuit, it follows that P I 20 µF −j159.15 Ω I = 62.84∠2.00 o 62.84∠2.00 o = 10 + j10 − j159.15 149.48∠ − 86.16 o = 0.420∠-88.2o A which is the same result. Q Network Theorems- a c examples – Professor J R Lucas 6 November 2001