《电路》（英文版）2-5 Maximum power transfer in the DC case

82-5 Maximum power transfer in the dC case S R L RS+RI R PRL=I RI )2R1 Rs+RL B(pz),2(R、+R1)2-R1×2(Rs+Rn) R。-R 0 (R s+R) (Rs+R1)3 Equation is satisfied when Rl=r s. The maximum power is Pmax ARL

§2-5 Maximum power transfer in the DC case S L S L R R i + =  L S L S RL L L R R R p i R 2 2 ( ) + = =  4 2 2 2 ( ) ( ) ( ) ( ) S L S L L S L RL S L R R R R R R R p dR d + + −  + = Equation is satisfied when R L= R S . The maximum power is L S R P 4 2 max  = 0 3 2 = + − = ( ) ( ) S L S L S R R R R 

Example: What is the maximum power that can be delivered to an external load ri by the network 200 36 0.2A R Oc Solution:。=20×+u1=1.51u1=30(+0.2)1=24 40 36 40+0.2×30 20 +0.12 =0.2 ×30=2.4 20+3040 20+30 s2.4 0+0.12=0.18 Rm=36/0.18=200c2 36 162Ⅳ na 4×200

Example: What is the maximum power that can be delivered to an external load RL by the network. 1 1 1 1.5 40 20    oc =  + = 0.2) 24V 40 30( 1 1 1 = +  =   oc = 36V 0 12 20 30 40 30 0 2 40 1 1 . = + . + = +    sc i 30 2.4 20 30 20 1 0.2  = +  =  = + 0.12 = 0.18 = 36/ 0.18 = 200 40 2.4 s c Rt h i P 1.62W 4 200 362 max =  = Solution: sc i  oc

1 {R两端并何元件,使≠=0} 375 20 375 20 20 =0 50 50 3.75A 10 y3.75A

{ R 两端并何元件，使i=0} i

active Source controlled node pat loop reference branch clockwise counterclockwise

active source controlled node path loop reference branch clockwise counterclockwise

A mesh current may often be identified as abranch current, as i, and i, are identified above. This is not always true, however, for consideration of a square nine-mesh network soon shows that the central mesh current cannot be identified as the current in any branch

A mesh current may often be identified as abranch current,as i1 and i2 are identified above.This is not always true, however, for consideration of a square nine-mesh network soon shows that the central mesh current cannot be identified as the current in any branch

There is also no reason that an independent source must assume only its given value or a zero value in the several experiments; it is only necessary for the sum of the several values to be equal to the original value. An inactive source almost always leads to the simplest circuit, however

There is also no reason that an independent source must assume only its given value or a zero value in the several experiments; it is only necessary for the sum of the several values to be equal to the original value. An inactive source almost always leads to the simplest circuit, however