sol-w3-2 - Problem 3.62 Solution Known quantities The...

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Unformatted text preview: Problem 3.62 Solution: Known quantities: The schematic of the circuit (see Figure P35). Find: The Thévenin equivalent resistance seen by resistor R3 , the Thevenin (open- circuit) voltage and the Norton (short-circuit) cmrent when R3 is the load. Analysis: (1) Remove the load, leaving the load terminals open circuited, and the voltage sources. Redraw the circuit. R1R4 R2R5 R-R R :R R- + T 1“ 4 2” 5 R1+R4 R2+R5 (2) Remove the load, leaving the load terminals open circuited. Redraw the circuit. Fornode#l: fl+fl:0 R1 R4 V9 V”) + V59 For node #2: 4 ~ : 0 R2 R5 Solving the system, R v1 = 1 V51 R1 +R4 R V? : 7 2 V32 R2 + R5 Therefore, R1 Ra v T V V ’ V + - V DC 1 2 R1+R4 51 R2+R5 5_ (3) Replace the load with a short circuit. Redraw the circuit. For mesh (a): ia(R1+R4)iR1ic = V51 For mesh (b): ib (R2 + R5)i Rzic = V52 For mesh (c): 7R1ia 7 Rgib +1’C(R1 + R2): 0 Solving the system, . (R1R2 + R11535 + R2R5)VS1+ R1R2V52 1 , R1R4(R2 + R5)+ R2R5(R1+ R4) 1 _ “ R1R4(R2 +R5)+ 122125021 + R4) Therefore, . . R1(R2+R5)V51+R2(R1+R4)Vsz 1SC _ Ic _ H ib Problem 3.63 Solution: Known quantities: The schematic of the circuit (see Figure P310). Find: The Thevenin equivalent resistance seen by resistor R5 , the Thévenin (open-circuit) voltage and the Norton (short- circuit) current When R5 is the load. Analysis: (1) Remove the load, leaving the load terminals open circuited, and the voltage sources. Redraw the circuit. RT : 25 Q N (75 (2+ 200 Q): 22.92 9 (2) Remove the load, leaving the load terminals open circuited. 50 9 Redraw the circuit. For node #1: 759 u v v 7v 71+¥ 02 200 75 For node #2: V2"’1+Q+ V2’V3 75 25 50 For node #3: v3 ’v2 ,1- 50 10V For the voltage source: +510V=0 v3+102v2 Solving the system, v1 =13_33 v, v2 = 3.33 v and v3 :7667 v. Therefore, vac = v3 = —6.67 V. (3) Replace the load with a short circuit. Redraw the circuit. Tor mesh (3): i,,(50)= 10 For mesh (b): i1,(300)7 i, (25): 40 For mesh (6): i,,(25)7 i425): 10 Solving the system, i, : 200 mA, ib :109 mA and ic :7291 mA. Therefore, iSC :i, 27291mA. ...
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