1250
HOMEWORK #3-1 prob 5 solution
F14
EX:
R3
v2
αv1 +
–
R2
v1
ia
va +
–
R1
Use the node-voltage method to find v1 and v2.
SOL'N:
We first observe that our dependent variable, v1, is a node voltage. Thus,
this variable is already defined in terms of node voltages.
We have only a voltage source between the nodes labeled v1 and v2. Thus,
we have a supernode. We draw a bubble around these two nodes, (with
the dependent source inside), and we set the sum of currents flowing out
of the bubble equal to zero:
v
v −v
v −v
v −v
−ia + 1 + 1 2 + 2 1 + 2 a = 0 A
R1
R2
R2
R3
Note that the 3rd and 4th terms cancel out. It turns out that R2 has no
impact on this problem.
We put this equation in standard form:
v1
1
1
v
+ v2
= ia + a
R1
R3
R3
We obtain a second equation for the supernode from the voltage
relationship between the nodes.
v 2 − v1 = αv1
NOTE:
The node next to the plus sign of the dependent source, v2, is
added in this equation and the node next to the minus sign of
the dependent source, v1, is subtracted in this equation.
We solve these two equations in two unknowns. Here, we eliminate v2
first.
v 2 = v1 + αv1 = (1+ α)v1
Substituting into the first equation gives one equation with one unknown.
v1
1
1
+ (1+ α)v1
= ia +
R1
R3
va
R3
or
⎛1
1⎞
v
v1⎜ + (1+ α) ⎟ = ia + a
R3 ⎠
R3
⎝ R1
or
v1 =
v
ia + a
R3
1
1
+ (1+ α)
R1
R3
or
R (R i + v a )
v1 = 1 3 a
R3 + (1+ α)R1
Using the second of the original equations, we find v2.
v 2 = (1+ α)v1 =
(1+ α)R1 (R3ia + v a )
R3 + (1+ α)R1