1. Network Theorems

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

3. An active network having two terminals A and B can be replaced by a constant-voltage source having an e.m.f V th and an internal resistance R th. The value of V th is equal to the open-circuited p.d between A and B. The value of R th 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) (c) (d)

5. Since no current in R 2, thus Refer to network (b), in R 2 there is not complete circuit, thus no current, thus current in R 3 And p.d across R 3 is Thus current in R (refer network (d)) Refer to network (c) the resistance at AB

6. Calculate the current through R 3 Solution With R 3 disconnected as in figure below p.d across CD is E 1 -I 1 R 1

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, R 3, thus the current

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

9. Continue…… Multiplying (b) by 3 and (c) by 4and adding the two expressions, thus Since the I 3 is positive then the direction in the figure is correct. Substitute I 1 in (a)

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, Substitute the voltage, resistance r and 10W as in figure below 10  parallel to 30  20  parallel to 15  Total

12. Another of expressing the current I L Where I S =E/R S is the current would flow in a short circuit across the source terminal( i.e when R L is replaced by short circuit) Then we can represent the voltage source as equivalent current source

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

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

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

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

17. Calculate the potential difference across the 2.0  resistor in the following circuit ………( c ) I2I2 First short-circuiting the branch containing 2.0  resistor I1I1 IsIs

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

19. Calculate the current in the 5.0  resistor in the following circuit Short-circuiting the branch that containing the 5.0  resistor Since the circuit is short-circuited across the 6.0  and 4.0  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) Hence the current in the 5  is Redraw the equivalent constant current circuit with the load 5.0  I

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

23. Dividing (e) by (f) Similarly Delta to star transformation Similarly, dividing (e) by (g) therefore We have (i) (j) Substitude R 2 and R 3 into (e) (k) (l) (m) (n) Similarly

24. Find the effective resistance at terminal between A and B of the network on the right side Solution  R = R 2 + R 4 + R 5 = 40  R a = R 2 x R 5 /  R = 4  R b = R 4 x R 5 /  R = 6  R c = R 2 x R 4 /  R = 2.4 

25. Substitute R 2, R 5 and R 4 with R a, R b dan R c : R 1 +R a 20 R 3 +R b 12 A B R3R3 6 R1R1 16 Rc 2.4 RaRa 4 RbRb 6 A B RcRc R AB = [(20x12)/(20+12)] + 2.4 = 9.9 