VIEWS: 1 PAGES: 4 CATEGORY: Physics POSTED ON: 10/17/2011 Public Domain
PHY301 – Circuit Theory Virtual University PHY 301 LECTURE 8 Example: Write KCL equations for all nodes. Solution: The KCL equations for nodes 1 through 4 follow For node 1 I +I –I =0 1 2 5 For node 2 -I +I -50I = 0 2 3 2 For node 3 -I + 50I + I = 0 1 2 4 For node 4 I –I –I =0 5 3 4 Example Write KCL equations for all nodes. Solution: For node A 10mA is entering the node and the source current I is leaving and 60mA is also entering t so, For node A I -60mA – 10mA= 0 t For node B 60mA – 40mA – 20mA = 0 Example Write KCL equation for node A. Virtual University of Pakistan Page 44 PHY301 – Circuit Theory Solution: By KCL the equation for node A -12mA + 4mA + I =0 Example Write KCL equation for node A and node B. Solution: The equation of currents at node A and B by KCL For node A -I +I + 3mA = 0 1 2 For node B -12mA + 4mA +I =0 1 Example Write KCL equation for node A . Writing equation for node A . For node A -10 I + I +44 mA – 12 mA = 0 x x Example Write KCL equation for node A . Solution: For node A equation will be by KCL I + 10I – 44mA = 0 x x Example Calculate the values of I and I 1 2. Solution: Virtual University of Pakistan Page 45 PHY301 – Circuit Theory We want to calculate the values of I and I 1 2 For node A 4mA + 8mA- I = 0 1 I = 12mA 1 Now for I 2 For node B -8mA + 2 mA + I =0 2 I = 6mA 2 Example Calculate the values of I , I , and I 1 2 3. Solution: We want to calculate the values of I , I , and I so we will use node analysis. 1 2 3, For node A -I – I +8mA = 0 1 2 -I – I = -8mA 1 2 For node B I + I +4mA = 0 2 3 I + I = - 4mA 2 3 For node C -I + 2mA – 8mA = 0 3 I = - 6mA 3 Putting the value of I in equation of node B 3 I - 6mA = -4mA 2 I = 2mA 2 Putting the value of I in equation of node A 2 -I – 2mA = -8mA 1 I = 6mA 1 Example: Find the KCL equtations for node A,node B, node C and node D. Solution: Virtual University of Pakistan Page 46 PHY301 – Circuit Theory We want to write the equations for nodes A, B, C and D. For node A -5mA + 8mA + 4mA =0 For node B I – I + 5mA =0 1 2 For node C -I – 2mA + 3mA – 8mA =0 1 For node D -4mA – 3mA + I = 00 3 Example Calculate the current I . o Solution: At node 1 (V1/12k) + ((V1 – V2)/10k) = 6mA 5V1 +6V1 – 6V2 = (60k)(6mA) 11V1– 6V2 = 360 At node 2 (V2/3k) +(V2/6k) + ((V2- V1)/10k) =0 10V2 +5V2+3V2-3V1=0 18V2 -3V1 =0 Equating equation of node1 and node 2 33V1 -18V2 =1080 -3V1 + 18V2 = 0 30V1 =1080 V1 = 36 volts 18V2 -3V1 =0 6V2 – V1=0 6V2 -36 =0 V2 = 36/6 V2 = 6volts I =V2/6k o = 6/6k = 1mA Virtual University of Pakistan Page 47