This preview shows pages 1–2. Sign up to view the full content.
Physics 21
Fall, 2007
Solution, Hour Exam #1
The graders for the problems were:
1 Belony, 2 McGeehan, 3 Sweeney, 4 Kanofsky, 5 Lyu
For questions about the grading, see the grader by Oct. 19.
Problem 1.
2
Ω
3
Ω
4
Ω
I
1
I
3
I
2
10.0 V
8.0 V
a
b
loop 1
loop 2
loop 3
(a) (10 pts.) Write the loop and junction equations needed
to determine the currents
I
1
,
I
2
,and
I
3
in the circuit
shown. Indicate clearly the loop used to determine each
loop equation.
The junction and loop equations are:
junction:
I
2
=
I
1
+
I
3
loop 1:
8
−
2
I
1
−
3
I
2
=0
loop 2:
10
−
3
I
2
−
4
I
3
loop 3:
8
−
2
I
1
−
10 + 4
I
3
(b) (6 pts.) Determine the currents
I
1
,
I
2
I
3
(including
the correct sign) by explicit solution of the equations.
You must show your work.
We can use any two loop equations; the di±erence be
tween the loop 1 and 2 equations is the loop 3 equation.
Let’s use 1 and 2. First rearrange them and then elim
inate
I
3
using the junction equation.
loop 1:
2
I
1
+3
I
2
=8
loop 2:
3
I
2
+4
I
3
=10
3
I
2
+4(
I
2
−
I
1
)=10
loop 2
±
:
−
4
I
1
+7
I
2
Multiply loop 1 by 2 and add to loop 2
±
:
loop 1
±
:
4
I
1
+6
I
2
=16
loop 2
±
:
−
4
I
1
I
2
add:
13
I
2
=26
⇒
I
2
=2A
Now back substitute to get all the currents:
I
1
=1A
,I
2
3
(c) (4 pts.) What is the potential di±erence
V
b
−
V
a
between
the points marked
a
and
b
on the diagram? Use the
calculated values of the currents and show how you get
your answer.
The potential di±erence can be calculated by any path.
There are three ways to get the answer:
V
b
−
V
a
=4
I
3
+8=10+2
I
1
−
3
I
2
=12V
.
Problem 2.
The thin rod shown in the diagram is extends
along the
x
axis from
x
=0to
x
=
a
. A total charge
Q
is
spread uniformly along the rod.
x
y
P
= (0,
y
,0)
+a
x'
dQ
r

r
'
(a) (3 pts.) Write an expression for the linear charge density
λ
on the rod.
λ
=
Q/a
.
(b) (4 pts.) Give the components of the vector (
r
−
r
±
)from
the charge
dQ
shownontherodat
x
=
x
±
to the point
P
on the
y
axis.
The ²eld point
r
is
y
ˆ
j
, and the charge point
r
±
is
x
±
ˆ
i
.
Then
r
−
r
±
=
−
x
±
ˆ
i
+
y
ˆ
j
.
(c) (4 pts.) Determine the electric ²eld
d
E
at the point
P
due to the element
dQ
of charge at the point
x
±
on the
rod. Express
d
E
in terms of
Q
,
a
,
x
±
y
.
Start with the general expression (from equation sheet)
d
E
(at
r
)=
1
4
π±
0
dQ
(
r
−
r
±
)

r
−
r
±

3
,
where
dQ
=
λdx
±
=(
Q/a
)
dx
±
and

r
−
r
±

=
p
(
x
±
)
2
+
y
2
.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview. Sign up
to
access the rest of the document.
 Spring '08
 Hickman
 Physics

Click to edit the document details