0195136047.pdf

22 CIRCUIT CONCEPTS

10 V

B

A

I = 10/2 = 5 A

2 Ω

+

−

(c) B

A

5 A 2 ΩV = 5 × 2 = 10 V

+

−

(d)

(e)

A

B B

2 Ω

1 Ω

5 A^1 Ω

5 A

2.5 A

2.5 A

2.5 V

1.25 A 1.25 A

5 V

10 V 5 V

2 Ω

D^2 Ω

C

+

+

−

+

−

2.5 V

+

2.5 V −

+

−

+

−

−

Maximum Power Transfer

In order to investigate the power transfer between a practical source and a load connected to it,

let us consider Figure 1.2.3, in which a constant voltage sourcevwith a known internal resistance

RSis connected to a variable load resistanceRL. Note that whenRLis equal to zero, it is called

ashort circuit, in which casevLbecomes zero andiLis equal tov/RS. WhenRLapproaches

infinity, it is called anopen circuit, in which caseiLbecomes zero andvLis equal tov. One is

generally interested to find the value of the load resistance that will absorb maximum power from

the source.

The powerPLabsorbed by the load is given by

PL=i^2 LRL (1.2.8)

where the load currentiLis given by

iL=

v^2

RS+RL

(1.2.9)

Substituting Equation (1.2.9) in Equation (1.2.8), one gets

PL=

v^2

(RS+RL)^2

RL (1.2.10)

For given fixed values ofvandRS, in order to find the value ofRLthat maximizes the power

absorbed by the load, one sets the first derivativedPL/dRLequal to zero,

dPL

dRL

=

v^2 (RL+RS)^2 − 2 v^2 RL(RL+RS)

(RL+RS)^4

= 0 (1.2.11)