Problem set #3 EE 221, 09/05/2002 – 09/12/2002
Chapter 3, Problem 4.
In Fig. 3.42,
(a) How many nodes are there?
(b) How many branches are there?
(c) If we move from B to F to E to C, have we
formed a path? A loop?
Chapter 3, Solution 4.
(a) Five nodes; (b) seven branches; (c) path, yes – loop, no.
Chapter 3, Problem 6.
Referring to Fig. 3.44,
(a ) Find ix if iy =2 A and iz =0 A.
(b) Find iy if ix = 2 A and iz =2 iy
(c) Find iz if ix=iy= iz .
Chapter 3, Solution 6.
KCL, we may write: 5 + iy + iz = 3 + ix
(a) ix = 2 + iy + iz = 2 + 2 + 0 = 4 A
(b) iy = 3 + ix – 5 – iz
iy = –2 + 2 – 2 iy
Thus, we find that iy = 0.
(c) 5 + iy + iz = 3 + ix (let ix = iy = iz = i)
5+i+i = 3+i
i = -2 A
Chapter 3, Problem 22.
Find the power absorbed by each of the
six circuit elements in Fig. 3.56.
1
2
Problem set #3 EE 221, 09/05/2002 – 09/12/2002
Chapter 3, Solution 22.
Applying KVL about this simple loop circuit (the dependent sources are still linear
elements, by the way, as they depend only upon a sum of voltages)
-40 + (5 + 25 + 20)i – (2v3 + v2) + (4v1 – v2) = 0
where we have defined i to be flowing in the clockwise direction, and
v1 = 5i, v2 = 25i, and v3 = 20i.
Performing the necessary substition, Eq. [1] becomes
50i - (40i + 25i) + (20i – 25i) = 40
so that i = 40 V/-20 Ω = -2 A
Computing the absorbed power is now a straightforward matter:
p40V
p5Ω
p25Ω
p20Ω
pdepsrc1
pdepsrc2
= (40)(-i)
= 5i2
= 25i2
= 20i2
= (2v3 + v2)(-i) = (40i + 25i)
= (4v1 – v2)(-i) = (20i - 25i)
= 80 W
= 20 W
= 100 W
= 80 W
= -260 W
= -20 W
and we can easily verify that these quantities indeed sum to zero as expected.
Chapter 3, Problem 32.
(a) Apply the techniques of single-node-pair analysis to
the upper right node in
Fig. 3.66 and find ix .(b) Now work with the upper left
node and find v8.(c)
How much power is the 5-A source generating?
Chapter 3, Solution 32.
Define a voltage v9 across the 9-W resistor, with the “+” reference at the top node.
(a) Summing the currents into the right-hand node and applying KCL,
5 + 7 = v9 / 3 + v9 / 9
Solving, we find that v9 = 27 V. Since ix = v9 / 9,
ix = 3 A.
(b) Again, we apply KCL, this time to the top left node:
2 – v8 / 8 + 2ix – 5 = 0
Since we know from part (a) that ix = 3 A, we may calculate v8 = 24 V.
(c) p5A = (v9 – v8) · 5 = 15 W.
(generating +15 W!)
[1]
Problem set #3 EE 221, 09/05/2002 – 09/12/2002
3
Chapter 3, Problem 54.
Use source- and resistor-combination techniques as a help in finding vx and ix in the circuit of Fig.
3.85.
Chapter 3, Solution 54.
To begin with, the 10 Ω and 15 V resistors are in parallel ( = 6 Ω), and so are the
20 Ω and 5 Ω resistors (= 4 Ω).
Also, the 4-A, 1-A and 6-A current sources are in parallel, so they can be combined into a
single 4 + 6 – 1 = 9 A current source.
Next, we note that (14 Ω + 6 Ω) || (4 Ω + 6 Ω) = 6.667 Ω
so that
vx = 9(6.667) = 60 V
and
ix = -60/10 = -6 A.
Chapter 3, Problem 65.
Use current and voltage division on the circuit
of Fig. 3.95 to find an expression
for (a) v2; (b) v1; (c) i4.
Problem set #3 EE 221, 09/05/2002 – 09/12/2002
4
Chapter 3, Solution 65.
R 2 || (R 3 + R 4 )
R1 + [R 2 || (R 3 + R 4 )]
R 2 (R 3 + R 4 ) (R 2 + R 3 + R 4 )
= VS
R1 + R 2 (R 3 + R 4 ) (R 2 + R 3 + R 4 )
R 2 (R 3 + R 4 )
= VS
R1 (R 2 + R 3 + R 4 ) + R 2 (R 3 + R 4 )
(a) v2 = VS
R1
R 1 + [R 2 || (R 3 + R 4 )]
R1
= VS
R 1 + R 2 (R 3 + R 4 ) (R 2 + R 3 + R 4 )
R 1 (R 2 + R 3 + R 4 )
= VS
R 1 (R 2 + R 3 + R 4 ) + R 2 (R 3 + R 4 )
(b) v1 = VS
 v 
R

2

(c) i4 =  1  
 R1   R 2 + R 3 + R 4 
R1 (R 2 + R 3 + R 4 ) R 2
R1 [R1 (R 2 + R 3 + R 4 ) + R 2 (R 3 + R 4 )(R 2 + R 3 + R 4 )]
R2
= VS
R 1 (R 2 + R 3 + R 4 ) + R 2 (R 3 + R 4 )
= VS
Chapter 3, Problem 70.
With reference to the circuit of Fig.
3.100, find (a) Ix if I1=12 mA; (b) I1 if
Ix=12 mA; (c) Ix if I2=15 mA; (d) Ix if
Is=60 mA.
Chapter 3, Solution 70.
Define R1 = 10 + 15 || 30 = 20 Ω and R2 = 5 + 25 = 30 Ω.
(a) Ix = I1 . 15 / (15 + 30) = 4 mA
(b) I1 = Ix . 45/15 = 36 mA
(c) I2 = IS R1 / (R1 + R2) and I1 = IS R2 / (R1 + R2)
So I1/I2 = R2/R1
Therefore
I1 = R2I2/R1 = 30(15)/20 = 22.5 mA
Thus, Ix = I1 . 15/ 45 = 7.5 mA
(d) I1 = IS R2/ (R1 + R2) = IS (30)/ 50 = 36 mA
Thus, Ix = I1 15/ 45 = 12 mA.