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CIRCUITS WITH DEPENDENT SOURCES
A CONVENTION ABOUT DEPENDENT SOURCES.
UNLESS OTHERWISE SPECIFIED THE CURRENT
AND VOLTAGE VARIABLES ARE ASSUMED IN SI
UNITS OF Amps AND Volts
DEPENDENT
VARIABLE


GENERAL STRATEGY
TREAT DEPENDENT SOURCES AS REGULAR
SOURCES AND ADD ONE MORE EQUATION FOR
THE CONTROLLING VARIABLE
 VA 
FIND VO
VD   I X
CONTROLLING
VARIABLE
FOR THIS EXAMPLE THE MULTIPLIER MUST HAVE
UNITS OF OHM
OTHER DEPENDENT SOURCES
VD  VX (  scalar)
I D  VX
( Siemens)
I D  I X
(  scalar)
A PLAN:
SINGLE LOOP CIRCUIT.
USE KVL TO DETERMINE CURRENT
KVL :  12  3k * I1  VA  5k * I1  0
ONE EQUATION, TWO UNKNOWNS. CONTROLLING
VARIABLE PROVIDES EXTRA EQUATION
AN ALTERNATIVE DESCRIPTIO N
 V  UNITS
VD  I X ,   2 
 mA 
ASSUMES CURRENT IN mA
KVL
ARE EXPLICIT
VA  2k * I1
REPLACE AND SOLVE FOR THE CURRENT
I1  2mA
USE OHM’S LAW
VO  5k * I1  10V
KCL TO THIS NODE. THE
DEPENDENT SOURCE IS JUST
ANOTHER SOURCE
FIND VO
A PLAN:
IF V_s IS KNOWN V_0 CAN BE DETERMINED USING VOLTAGE DIVIDER.
TO FIND V_s WE HAVE A SINGLE NODE-PAIR CIRCUIT
THE EQUATION FOR THE CONTROLLING
VARIABLE PROVIDES THE ADDITIONAL EQUATION
ALGEBRAICALLY, THERE ARE TWO UNKNOWNS
AND JUST ONE EQUATION
SUBSTITUTION OF I_0 YIELDS
VOLTAGE DIVIDER
* / 6k  5VS  60 VO 
NOTICE THE CLEVER WAY OF WRITING mA TO
HAVE VOLTS IN ALL NUMERATORS AND THE
SAME UNITS IN DENOMINATOR
4k
2
VS  (12)V
4k  2k
3
FIND VO
KVL TO
THIS LOOP
A PLAN:
ONE LOOP PROBLEM.
FIND THE CURRENT
THEN USE OHM’S LAW.
THE DEPENDENT SOURCE IS ONE MORE VOLTAGE
SOURCE
THE EQUATION FOR THE CONTROLLING VARIABLE
PROVIDES THE ADDITIONAL EQUATION
REPLACE AND SOLVE FOR CURRENT I
… AND FINALLY
FIND G 
vO ( t )
vi (t )
KCL
A PLAN:
ONE LOOP ON THE LEFT - KVL
ONE NODE-PAIR ON RIGHT - KCL
KVL
KVL
KCL
gm v g ( t ) 
ALSO A VOLTAGE DIVIDER
vO ( t )
0
RL