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PHY301 – Circuit Theory Virtual University PHY 301 LECTURE 18 Example: Calculate the voltage V o. Solution: We want to calculate the voltage V The circuit can be redrawn as o. Now Vx = (I - I ) 6k -------------(A) 2 3 I = Vx/2k ---------------(B) 1 Put Vx from A into B we have I = (I – I )6k/2k 1 2 3 = 3(I – I ) 2 3 Now I = 2mA 2 I = 6mA - 3I 1 3 Now KVL equation for mesh 3 6kI +6k(I – I ) +2k(I -I ) =0 3 3 2 3 1 6kI + 6kI – 12 +2kI -2k(6mA -3I )=0 3 3 3 3 14kI – 12 -12 + 6kI = 0 3 3 20kI = 24 3 I = 1.2mA 3 V = 6kI 0 3 = 6k(1.2mA) =7.2 volts Virtual University of Pakistan Page 75 PHY301 – Circuit Theory Example: Calculate the voltage V a. Solution: We want to calculate the voltage V The circuit can be redrawn as a. Va can also be found by super node technique but it will take a lengthy calculation lets see how it becomes very easy with loop analysis. Here I =1A 4 KVL for mesh 1 1(I – I ) +3(I -3I )-5 =0 1 2 1 3 I – I +3I -3I -5 = 0 1 2 1 3 4I – 3I – I = 5 --------(A) 1 3 2 KVL for mesh 2 2I2 +3I2 -2Va+I2-I1 = 0 6I2 –I1 = 2Va------(B) Va = 3I2 Put this value in (a) 6I2 –I1 = 3I2 I1 = 0 KVL for mesh 3 Virtual University of Pakistan Page 76 PHY301 – Circuit Theory 2Va + 5(I – I ) +3(I –I ) =0 3 4 3 1 Putting the values of I ,I and Va 1 4 6I + 3I -0 +5I – 5 = 0 2 3 3 8I + 6I = 5 3 2 Solving equation for mesh 1 and mesh 3 - 18I - 6I = 30 3 2 8I + 6I = 5 3 2 -10I = 35 3 I = - 3.5A 3 Putting values of I and I in eq of mesh 1 i.e in (A) we have 3 1 10.5– I =5 2 I = 5.5 A 2 As Va =3I 2 Therefore Va =3 * 5.5 Va = 16.5 V Example: Find Current through all meshes. Solution: Incorrect approach I =-5A 2 I =+5A 3 We drive all these KVL equation for mesh 1 -5+I -I +2I =0 -------------------- (1) 1 3 1 Virtual University of Pakistan Page 77 PHY301 – Circuit Theory KVL equation for super mesh 1(I -I )+4I +3I +2+4I - 4I +6I =0 ------------ (2) KVL 3 1 2 2 3 4 3 equation for Mesh 4 4I – 4I +2I -3=0 ------------ (3) 4 3 4 Simplifying Equation (1) 3I - I = 5 1 3 Equation (2) -I +7I +11I -4I =-2 1 2 3 4 Equation (3) - 4I + 6I =3 3 4 Also -I +I =5 2 3 Therefore I =2.481A 1 I =-2.556A 2 I =2.444A 3 I =2.130A 4 Example: Calculate current I . 0 Solution: We want to calculate the current I The circuit can be redrawn as 0. Here I = -1mA 1 I = -2mA 2 KVL equation for loop 3 1k( I + I ) + 2kI + 1k(I + I ) = 2 + 4 3 1 3 3 2 1kI + 1kI + 2kI + 1kI + 1kI = 6 3 1 3 3 2 4kI – 1 – 2 = 6 3 4kI – 3 = 6 3 I = 9/4 mA 3 I = 2.25 mA 3 I = I = 2.25 mA 3 0 Virtual University of Pakistan Page 78 PHY301 – Circuit Theory Example: Calculate current I . 0 Solution: We want to calculate the current I The circuit can be redrawn as 0. Here I = 2mA 1 I = - 4mA 2 KVL for loop 3 2kI + 2k(I + I ) = 12 3 3 2 2kI + 2kI + 2kI = 12 3 3 2 4kI – 8 =12 3 4kI = 20 3 I = 5mA 3 Now I =I +I 0 3 1 I = 5mA + 2mA 0 = 7mA Virtual University of Pakistan Page 79

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posted: | 10/17/2011 |

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