This preview shows pages 1–3. Sign up to view the full content.
EE40 Spring 2008 Homework 2 Solutions
1)
a)
The circuit can be redrawn as
And then the portion:
can be replaced by a (2+4)//6 = 6//6 = 3
Ω
resistor as shown in the drawing. Then we get,
by similar reasoning (3+3)//3 = 6//3 = 2
Ω
for the larger parallel block.
Therefore the total resistance is R
eq
= 2+2+8+2+2 = 16
Ω
.
b)
Shorting c to d means no voltage drop between those two points, so no current on the 8 or
2
Ω
resistors, and we can just ignore them. Hence the total resistance is R
eq
= 2+2+0+2 =
6
Ω
.
2)
Hambley P 2.17:
Combining the 20
Ω
resistor with the 30
Ω
gives 20*30/50 = 60/5=12
Ω
,
then we note as in the drawing above that 60 // (8+12) = 60 // 20 =
120/8=15, for a total
resistance of 7+15=22
Ω
.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document 3)
Hambley P 2.24
Label the blacked nodes in the text by d,a,b,c going counterclockwise from top left.
The total current flowing into and out of the parallel combination of the 6
Ω
and 12
Ω
resistors on the right is 2A. By the current divider formula, i
2
= 2*(1/12) /( 1/12 + 1/6 ) =
2/3A.
The total current flowing into and out of the parallel combination of the 12
Ω
and 24
Ω
resistors on the left is 2A. By the current divider formula,
A
i
3
4
2
24
/
1
12
/
1
12
/
1
1
=
+
=
.
Thus
v
a
v
b
= 12*i
2
= 8V
v
b
–
v
c
= 8*2 = 16V
v
c
–
v
d
= 12*i
1
= 16V
By KVL,
v
=
v
a

v
d
= 8+16+16 = 40V
4)
Hambley P 2.32
With the switch open, by the voltage division formula
v
2
= 10*R
2
/ (R
2
+ 6)
V.
Since
This is the end of the preview. Sign up
to
access the rest of the document.
This note was uploaded on 04/07/2008 for the course EE 40 taught by Professor Changhasnain during the Spring '07 term at University of California, Berkeley.
 Spring '07
 ChangHasnain

Click to edit the document details