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Network Theorems

Network Theorems. Circuit analysis. Mesh analysis Nodal analysis Superposition Thevenin’s Theorem Norton’s Theorem Delta-star transformation. Thevenin theorem.

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Network Theorems

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  1. Network Theorems

  2. Circuit analysis • Mesh analysis • Nodal analysis • Superposition • Thevenin’s Theorem • Norton’s Theorem • Delta-star transformation

  3. Thevenin theorem • An active network having two terminals A and B can be replaced by a constant-voltage source having an e.m.f Vth and an internal resistance Rth. • The value of Vth is equal to the open-circuited p.d between A and B. • The value of Rth is the resistance of the network measured between A and B with the load disconnected and the sources of e.m.f replaced by their internal resistances.

  4. Networks to illustrate Thevenin theorem (a) (b) (d) (c)

  5. Refer to network (b), in R2 there is not complete circuit, thus no current, thus current in R3 And p.d across R3 is Since no current in R2, thus Refer to network (c) the resistance at AB Thus current in R(refer network (d))

  6. Example 13 Calculate the current through R3 Solution With R3 disconnected as in figure below p.d across CD is E1-I1R1

  7. continue To determine the internal resistance we remove the e.m.f s Replace the network with V=5.2V and r=1.2, then the at terminal CD, R3, thus the current

  8. Example 14 Solution Determine the value and direction of the current in BD, using (a) Kirchoff’s law (b) Thevenin theorem (a) Kirchoff’s law Using K.V.L in mesh ABC + the voltage E …..(a) Similarly to mesh ABDA ……(b) For mesh BDCB …..(c)

  9. Continue…… Multiplying (b) by 3 and (c) by 4and adding the two expressions, thus Substitute I1 in (a) Since the I3 is positive then the direction in the figure is correct.

  10. continue By Thevenin Theorem P.D between A and B (voltage divider) P.D between A and D (voltage divider) P.D between B and D

  11. continue For effective resistance, 10W parallel to 30 W 20W parallel to 15 W Total Substitute the voltage, resistance r and 10W as in figure below

  12. Constant current generator Another of expressing the current IL Where IS=E/RS is the current would flow in a short circuit across the source terminal( i.e when RL is replaced by short circuit) Then we can represent the voltage source as equivalent current source

  13. Example 15 Calculate the equivalent constant-voltage generator for the following constant current source Vo Current flowing in 15W is 1 A, therefore Current source is opened thus the 5 W and 15 W are in series, therefore

  14. Example 16 Analysis of circuit using constant current source From circuit above we change all the voltage sources to current sources

  15. continue At node 1 At node 2 X 120 X 30 …..(a) ……(b)

  16. continue ………( c ) (c) + (b) From (a) Hence the current in the 8 W is So the answers are same as before

  17. Example 17 Calculate the potential difference across the 2.0W resistor in the following circuit ………( c ) First short-circuiting the branch containing 2.0W resistor I2 I1 Is

  18. continue Redraw for equivalent current constant circuit Using current division method Hence the voltage different in 8 W is

  19. Example 18 Calculate the current in the 5.0W resistor in the following circuit Short-circuiting the branch that containing the 5.0 W resistor Since the circuit is short-circuited across the 6.0W and 4.0W so they have not introduced any impedance. Thus using current divider method

  20. continue The equivalent resistance is a parallel (2.0+8.0)//(6.0+4.0) Redraw the equivalent constant current circuit with the load 5.0W I Hence the current in the 5 W is

  21. Delta-star transformation

  22. Delta to star transformation From delta cct , impedance sees from AB From star cct , impedance sees from AB (a) Thus equating (b) Similarly from BC (c) and from AC (d) (b) – (c) By adding (a) and (d) ; (b) and (d) ;and (c) and (d) and then divided by two yield (g) (f) (e)

  23. Delta to star transformation (i) Dividing (e) by (f) therefore (j) Similarly, dividing (e) by (g) (j) (k) We have (l) Substitude R2 and R3 into (e) (m) Similarly (n) Similarly

  24. Example 18 Find the effective resistance at terminal between A and B of the network on the right side Solution R = R2 + R4 + R5 = 40  Ra = R2 x R5/R = 4  Rb = R4 x R5/R = 6  Rc = R2 x R4/R = 2.4 

  25. A A 16 R1+Ra 20 12 R3+Rb R1 R3 6 Rb Ra 6 4 Rc 2.4 Rc 2.4 B B Substitute R2, R5 and R4 with Ra, Rb dan Rc: RAB = [(20x12)/(20+12)] + 2.4 = 9.9 

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