PspiceTutorial_2.doc - JEE2300 PSpice Tutorial 2 Problem 1 Use a Y-to transformation to find(a io(b i1(c i2 and(d the power delivered by the ideal

# PspiceTutorial_2.doc - JEE2300 PSpice Tutorial 2 Problem 1...

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JEE2300 PSpice Tutorial 2 Problem 1. Use a Y-to- transformation to find (a) i o ; (b) i 1 ; (c) i 2 ; and (d) the power delivered by the ideal current source in the circuit show below. Use PSpice to confirm your work. 1A i o 20 100 320 50 600 240 i 2 i 1 Figure 1. (a) Need to mark Y network to perform Y-to- transformation. That is done below: 1A i o R 1 = 20 R 3 =100 320 R 2 = 50 600 240 i 2 i 1 a b c i 20 i 50 i 240 Figure 2. 1 Subscribe to view the full document.

Now we perform Y-to- transformation between nodes a,b, and c: 400 20 8000 20 2000 5000 1000 1 1 3 3 2 2 1 R R R R R R R R a 160 50 8000 50 2000 5000 1000 2 1 3 3 2 2 1 R R R R R R R R b 80 100 8000 100 2000 5000 1000 3 1 3 3 2 2 1 R R R R R R R R c Now the circuit looks like this: 1A i o R b = 160 320 R c = 80 600 240 i ac i 1 a b c R a = 400 i bc i ab i 240 Figure 3. Using KVL from Figure 3 we can see that: 160 240 240 ac i i (1) 600 400 o bc i i (2) 80 320 1 ab i i (3) We can simplify this circuit doing the following: R eq1 = (240 ) || R b = 96 and i 96 = i 240 + i ac (4) R eq2 = (600 ) || R a = 240  and 2 i 64 = i bc + i o (5) R eq3 = (320 ) || R c = 64  and i 64 = i ab + i 1 (6) New circuit looks like this: 1A R eq1 = 96 R eq3 = 64 a b c R eq2 = 240 i 96 i 64 Figure 4. From Figure 4 using KCL: 96 64 1 i i A Also using KVL for the Figure 4: 96 240 64 96 64 i i From the last two equations we get A i 24 . 0 64 A i 76 . 0 96 Using (2) and (5) we get: mA i bc 144 mA i o 96 (b) From (3) and (6) we get: 3 Subscribe to view the full document.

mA i ab 192 mA i 48 1 (c) From (1) and (4) we get mA i ac 456 mA i 304 240 Using KCL from Figure 2 we have: 1A = i 240 + i 20 + i 1 i 20 = 648mA. Also using KCL from Figure 2 we have: i 50 = i o - i 1 i 50 = 48mA ,and using KCL again from Figure 2 we have: i 20 = i 50 + i 2 i 2 = 600mA. (d) From Figure (2) we have that P 1A = - (1A)v 240 P 1A = - (1A)(240 )(i 240 ) = - (1A)(240 )(304mA) = - 72.96W – power delivered by the 1A ideal current source P 240 = (i 240 ) 2 (240 ) = (304mA) 2 (240 ) = 22.17984W P 20  = (i 20 ) 2 (20 ) = (648mA) 2 (20 ) = 8.39808W P 320  = (i 1 ) 2 (320 ) = (48mA) 2 (320 ) = 0.73728W + P 100 = (i 2 ) 2 (100 ) = (600mA) 2 (100 ) = 36W P 50 = (i 50 ) 2 (50 ) = (48mA) 2 (50 ) = 0.1152W P 600  • Fall '09

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