m g d
V
=
m g d
tan
θ
q
=
(0
.
000148 kg) (9
.
8 m
/
s
2
) (0
.
059 m)
2
.
7
×
10

8
C
×
tan 18
◦
=
1
.
0298 kV
.
021
(part 1 of 2) 10 points
Consider the circuit shown below.
There
are three resistors,
r
1
,
r
2
, and
R
; and two
emf’s,
E
1
and
E
2
.
The directions of the cur
rents
i
1
,
i
2
, and
i
3
are shown in the figure.
E
F
A
B
D
C
A
E
1
E
2
i
1
r
1
i
3
R
i
2
r
2
Apply Kirchhoff’s rules. Neglect the differ
ence of an overall sign, if applicable.
What
equation
does
the
loop
ABCDA
yield?
1.
E
1
+
E
2

i
2
r
2
+
i
1
r
1
= 0
2.
E
1
 E
2

i
1
r
2
+
i
2
r
1
= 0
3.
E
1
+
E
2
+
i
2
r
2
+
i
1
r
1
= 0
4.
E
1
 E
2
+
i
2
r
2
+
i
1
r
1
= 0
5.
E
1
+
E
2
+
i
2
r
2

i
1
r
1
= 0
6.
E
1
 E
2

i
1
r
2

i
2
r
1
= 0
7.
E
1
+
E
2

i
2
r
2

i
1
r
1
= 0
8.
E
1
 E
2

i
2
r
2
+
i
1
r
1
= 0
9.
E
1
 E
2

i
2
r
2

i
1
r
1
= 0
10.
E
1
 E
2
+
i
2
r
2

i
1
r
1
= 0
correct
Explanation:
Recall that Kirchhoff’s loop rule states that
the sum of the potential differences across all
the elements around a closed circuit loop is
zero. If a resistor is traversed in the direction
of the current, the change in potential is

i R .
If an emf source is traversed from the

to
+ terminals, the change in potential is +
E
.
Apply the opposite sign for traversing the
elements in the opposite direction.
Hence, by inspection up to an overall sign,
following is the correct equation
ABCDA
:
E
1
 E
2
+
i
2
r
2

i
1
r
1
= 0
.
022
(part 2 of 2) 10 points
Let
E
1
=
E
2
= 5 V, and
r
1
=
r
2
= 4
.
5 Ω, and
R
= 1
.
9 Ω.
E
F
A
B
D
C
A
5 V
5 V
i
1
4
.
5 Ω
i
3
1
.
9 Ω
i
2
4
.
5 Ω
Find the current
i
3
.
Hint:
From symmetry, one expects
i
1
=
i
2
.
Correct answer: 1
.
20482 A.
Explanation:
We are given that
E
1
=
E
2
and
r
1
=
r
2
.
This implies that
i
1
=
i
2
. Hence the junction
rule yields
i
1
+
i
2
= 2
i
2
=
i
3
i
2
=
i
3
2
.
Platt, David – Quiz 2 – Due: Oct 18 2005, 10:00 pm – Inst: Ken Shih
13
Substituting
this
into
the
loop
equation
DCFED
gives
E
2

i
3
R

i
3
2
r
2
= 0
.
Solving for
i
3
yields
i
3
=
E
2
R
+
r
2
2
=
(5 V)
(1
.
9 Ω) +
(4
.
5 Ω)
2
=
1
.
20482 A
.
023
(part 1 of 2) 10 points
Consider two conductors 1 and 2 made of the
same ohmic material;
i.e.
,
ρ
1
=
ρ
2
.
Denote
the length by
‘
, the cross sectional area by
A
.
The same voltage
V
is applied across the
ends of both conductors. The field
E
is inside
of the conductor.
V
1
~
E
1
I
1
‘
1
r
1
V
2
~
E
2
I
2
‘
2
r
2
If
A
2
= 2
A
1
, ‘
2
= 2
‘
1
and
V
2
=
V
1
, find
the ratio
E
2
E
1
of the electric fields.
1.
E
2
E
1
=
1
8
2.
E
2
E
1
= 1
3.
E
2
E
1
= 2
4.
E
2
E
1
=
1
16
5.
E
2
E
1
=
1
3
6.
E
2
E
1
= 4
7.
E
2
E
1
=
1
2
correct
8.
E
2
E
1
= 8
9.
E
2
E
1
=
1
12
10.
E
2
E
1
=
1
4
Explanation:
E
2
E
1
=
V
2
‘
2
V
1
‘
1
=
‘
1
‘
2
=
‘
1
2
‘
1
=
1
2
.
024
(part 2 of 2) 10 points
Determine the ratio
I
2
I
1
of the currents .
1.
I
2
I
1
= 8
2.
I
2
I
1
=
1
3
3.
I
2
I
1
= 2
4.
I
2
I
1
= 4
5.
I
2
I
1
=
1
2
6.
I
2
I
1
= 1
correct
7.
I
2
I
1
=
1
8
8.
I
2
I
1
=
1
4
9.
I
2
I
1
=
1
16
10.
I
2
I
1
=
1
12
Explanation:
Platt, David – Quiz 2 – Due: Oct 18 2005, 10:00 pm – Inst: Ken Shih
14
I
2
I
1
=
V
2
R
2
V
1
R
1
=
R
1
R
2
=
ρ
1
‘
1
A
1
¶
ρ
2
‘
2
A
2
¶
=
‘
1
A
1
2
‘
1
2
A
1
=
1
.
Platt, David – Quiz 3 – Due: Nov 15 2005, 10:00 pm – Inst: Ken Shih
1
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printout
should
have
27
questions.