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Transcript
Thevenin and Norton Equivalent Circuits Problem 18 (a) Convert the circuit to the left of terminals AB in Figure 1 (a) to an equivalent Thevenin circuit by initially converting to a Norton equivalent circuit (b) Determine the current flowing in the 1.8 resistor. (a) E1 = 12V E2 = 24V 1.8 r1 = 3 r2 = 2 Figure 1 For the branch containing the 12 V source, converting to a Norton equivalent circuit gives ISC = 12/3 = 4 A and r1 = 3 . For the branch containing the 24 V source, converting to a Norton equivalent circuit gives ISC2 = 24/2 =12 A and r2=2 . Thus Figure 1b shows a network equivalent to Figure 1a . From Figure 1b the total shortcircuit current is 4 + 12 = 16A . ISC = 4A ISC = 12A r1= 3 A And the total resistance 3 x 2 = 1.2 3+2 R2= Thus Figure 11.45(b) simplifies to Figure 1(c). The opencircuit voltage across AB of Figure 1(c), 2 B FigureA1b 16A r1= 1.2 B E = (16)(l.2) = 19.2 V, and the resistance 'looking-in' at AB is1.2. Hence the Thevenin equivalent circuit is as shown in Figure 1(d). (b) When the 1.8 resistance is connected between terminals A and B of Figure 1(d) the current I flowing is given by: Figure 1c A r1= 1.2 I = 19.2 = 6.4A 1.2 +1.8 19.2 V B Figure 1d 29 Problem 19. Determine by successive conversions between Thevenin and Norton equivalent networks a Thevenin equivalent circuit for terminals AB of Figure 2(a). Hence determine the current flowing in the 200 resistance. For the branch containing the 10 V source, converting to a Norton equivalent network gives 1 mA E1 = 10V 600 E2 = 6V 200 2k 3 k Figure 2 ISC = 6 = 2mA and r2 = 3k 3000 1 mA E1 = 10V 600 E2 = 6V 2k ISC = 10 = 5mA and r1 = 2k 2000 For the branch containing the 6 V source, converting to a Norton equivalent network gives A 3 k Thus the network of Figure 2(a) converts to Figure 2(b). Combining the 5 mA and 2 mA current sources gives the equivalent network of Figure 2(c) where the short-circuit current for the original two branches considered is 7 mA and the resistance is B Figure 2b 1 mA 600 2 x 3 = 1.2 k. 2+3 Both of the Norton equivalent networks shown in Figure 2(c) may be converted to Thevenin equivalent circuits. The open-circuit voltage across CD is (7 x l0-3)(1.2 x 10-3) = 8.4 V and the resistance 'looking-in' at CD is 1.2 k. A 7mA The open-circuit voltage across EF is (1 x 10-3)(600) = 0.6 V and the resistance 'looking-in' at EF is 0.6 k. Thus Figure 2(c) converts to Figure 2(d). Combining the two Thevenin circuits gives 1.2k B Figure 2c E = 8.4 - 0.6 = 7.8 V and the resistance r = (1.2 + 0.6) k = 1.8 k. 0.6 V 600 8.4 V A Thus the Thevenin equivalent circuit for terminals AB of Figure 2(a) is as shown in Figure 2(e). Hence the current I flowing in a 200 resistance connected between A and B is given by: 1.2k B I = 7.8 = 7.8 = 3.9 mA 1800+200 2000 Figure 2d 30 A 7.8 V 1.8k B Figure 2e 31