0195136047.pdf

46 CIRCUIT CONCEPTS

Power absorbed by resistorR 2 =I 22 R 2 =( 3 )^24 =36 W

Power absorbed by resistorR 3 =I 32 R 3 =( 1 )^212 =12 W

Power delivered by sources= 144 − 72 =72 W

Power absorbed by resistorsR 1 ,R 2 , andR 3 = 24 + 36 + 12 =72 W

The conservation of power is satisfied by the circuit.

EXAMPLE 1.3.6

Consider the network shown in Figure E1.3.6 containing a voltage-controlled source producing

the controlled currentic=gv, wheregis a constant with units of conductance, and the control

voltage happens to be the terminal voltage in this case.

(a) Obtain an expression forReq=v/i.

(b) For (i)gR=^1 / 2 , (ii)gR=1, and (iii)gR=2, findReqand interpret what it means in

each case.

viR iR c = gv

i

Node 2

Node 1

−

+

Figure E1.3.6

Solution

(a) Applying the KCL at node 1, we get

i+ic=iR=

v

R

Therefore,

i=

v

R

−ic=

v

R

−gv=

1 −gR

R

v

Then,

Req=

v

i

=

R

1 −gR

(b) (i) ForgR=^1 / 2 ,Req= 2 R. The equivalent resistance is greater thanR; the internal

controlled source provides part of the current throughR, thereby reducing the input

currentifor a given value ofv. Wheni<v/R, the equivalent resistance is greater

thanR.