《电路》（英文版）2-4 Thevenin's and Norton's theorems

s2-4 Thevenin's and Norton's theorems One of the main uses of Thevenin's and norton's theorems is the replacement of a large part of network very simple equivalent. 7 4A 126 36 R (b) ④8 R 8 (c) (d)

§2-4 Thevenin's and Norton's theorems One of the main uses of Thevenin' s and Norton's theorems is the replacement of a large part of network very simple equivalent. (a) − + 12V 3 6 RL 7 (b) 3 6 RL 7 4A ( c ) − + 8V 2 RL 7 ( d ) − + 8V 9 RL

Thevenin's theorem. Given any linear circuit, rearrange it in the form of two networks A and B that are connected together by two resistance-ess conductors. f either network contains a dependent source, its control variable must be in that same network. Define a voltage v as the open-circuit voltage, which would appear across the terminals of A, if B were disconnected so that no current is drawn from a Then all currents and voltages in B will remain uncharged if A is killed(all independent voltage sources and independent current sources in A replaced by short circuits and open circuits, respectively) and an independent voltage source is connected, with proper polarity, in series with the dead (inactived networ k

Thevenin's theorem: Given any linear circuit, rearrange it in the form of two networks A and B that are connected together by two resistance-less conductors. If either network contains a dependent source, its control variable must be in that same network. Define a voltage voc as the open-circuit voltage, which would appear across the terminals of A, if B were disconnected so that no current is drawn from A. Then all currents and voltages in B will remain uncharged if A is killed (all independent voltage sources and independent current sources in A replaced by short circuits and open circuits, respectively) and an independent voltage source is connected, with proper polarity, in series with t h e d e a d ( i n a c t i v e ) A n e t w o r k

B → B oc voc---open-voltage. Rth---equivalent-resistance Proving: Thevenin th equivalent circuit

A B voc---open-voltage. Rth---equivalent-resistance. Proving: A circuit equivalent Thevenin   oc Rth − + B − +  oc Rth

B B i=0 R B 十 十 i=i A Thevenin-eguivalent-circuit

i B  oc Rth A  oc Rth i B A A  oc  oc + −− + B  oc Rth A i'= 0 + −  oc B  oc Rth A i'' − +  oc − +  oc Rth Thevenin− equivalent − circuit B i = i

4.000R1 R2 E996 2k 3k R3 2mA 10000k 0 R14000uA600mAR2 3k 1600mA 400.0uA R3 2mA 0.000001 2000mA 0

R 1 2k 4.000V 7.998V 0V 2mA R 2 3k 4V 7.996V R 3 10000k 0 R 3 0.000001 1.600mA R 1 2k 400.0uA 2mA 2.000mA R 2 3k 1.600mA 0 4V 400.0uA

Examp!e2: Find Uoc andrin·「 2K 3k 十 Solution 1. Find v 4000 U.=4+2000× 4000∴D=Ux=8 2. Find rth (A): Method of short-circuit current(电源保留).Rm 2k 3 R U 4000 8 U,=U=0, =0.8mA. Rh=. 10kg2 000 sk 8×10

Solution: 1.Find voc. (A): Method of short-circuit current(电源保留). oc x V x x , 8 4000 = 4+ 2000   = =   0.8 . 5 4 0, 4000 0, mA k i s c x x = =  =  = =    =   = = − k i R sc oc th 10 0.8 10 8 3  Example 2: Find  OC and . Rth 2. Find Rth.  sc i 4000  x − +  x 2k 3k − + 4V 4000  x − +  x 2k 3k − + 4V  oc Rth sc i − +  − + sc oc th i R  =

(B): We apply a la source externally, measure the resultant voltage, and then set r=v/1(独立源为零) U=3k×1+(1+ 2k=5k+ 2k 3k十 4000 2 D=lOkv 4000 h 10k9 (C): We could also apply a iv source externally, determine i=1/ Rth, and then set rth=1/(独立源为零) 2k 3k 1=3+(i+)2k=5ki+ +D 400 4000 U=1∴i=1/10kR==10 10 The Thevenin-equivalent circuit 下8

（ B): We apply a 1A source externally, measure the resultant voltage, and then set .Rth =v/1(独立源为零) kV k k k 10 2 )2 5 4000 3 1 (1 = =  + + = +     Rth = = 10k 1  （C）:We could also apply a 1V source externally, determine i=1/Rth, and then set Rth =1/i.(独立源为零) i k ki i k ki x x x 1 1/10 2 )2 5 4000 1 3 ( =  = = + + = +    = = k i Rth 10 1 The Thevenin-equivalent circuit: 4000  x − +  x 2k 3k 1A = 4000  x − +  x 2k 3k 1V i − + 8V − + 10k

4.000V 7999 999∨ R1 R2 2k 3k G1 R3 4V 100000k G 0 0.0UAR1 800.0 R2 0.0 2k G1 R3 0.00000001 800.0uA G 0

R 2 3k 800.0uA + - G1 G R 3 0.00000001 800.0uA 4V 800.0uA 0 R 1 2k 800.0uA 7.999V 7.999V 0 R 1 2k 4V R 2 3k + - G1 G 0V R 3 100000k 4.000V

Norton's theorem. Given any linear circuit, rearrange it in the form of two networks A and b that are connected together by two resistance less conductors. If either network contains a dependent source. its control variable must be in that same network. Define a current isc as the short-circuit current, which would appear at the terminals of a, if B were short-circuited so that no voltage is provide by a. Then all voltages and currents in b will remain uncharged if a is killed(all independent voltage sources and independent current sources in A replaced by short circuits and open circuits, respectively) and an independent current source sc is connected, with proper polarity, in parallel with the dead (inactive)A network

Norton's theorem: Given any linear circuit, rearrange it in the form of two networks A and B that are connected together by two resistance￾less conductors. If either network contains a dependent source, its control variable must be in that same network. Define a current i sc as the short-circuit current, which would appear at the terminals of A, if B were short-circuited so that no voltage is provide by A. Then all voltages and currents in B will remain uncharged if A is killed (all independent voltage sources and independent current sources in A replaced by short circuits and open circuits, respectively) and an independent current source i sc is connected, with proper polarity, in parallel with the dead (inactive) A network