hw6 - homework 06 LIN KEVIN Due 4:00 am c Question 1 chap...

homework 06 – LIN, KEVIN – Due: Feb 24 2008, 4:00 am 1
Question 1, chap 28, sect 2.
part 1 of 2 10 points
A battery with an emf of 5 . 3 V and internal
resistance of 0 . 36 Ω is connected across a load
resistor R .
If the current in the circuit is 1 . 99 A, what
is the value of R ?
Correct answer: 2 . 30332 Ω (tolerance ± 1
%).
Explanation:
Let : E = 5 . 3 V ,
I = 1 . 99 A , and
R
i
= 0 . 36 Ω .
The electromotive force E is given by
E = I ( R + R
i
)
R =
E
I
− R
i
=
5 . 3 V
1 . 99 A
− 0 . 36 Ω
= 2 . 30332 Ω .
Question 2, chap 28, sect 2.
part 2 of 2 10 points
What power is dissipated in the internal
resistance of the battery?
Correct answer: 1 . 42564 W (tolerance ± 1
%).
Explanation:
The power dissipation due to the internal
resistance is
P = I
2
R
i
= (1 . 99 A)
2
(0 . 36 Ω)
= 1 . 42564 W .
Question 3, chap 28, sect 4.
part 1 of 2 10 points
Four resistors are connected as shown in
the figure.
92 V
S
1
c
d
a
b
22 Ω
32 Ω
57 Ω
84 Ω
Find the resistance between points a and b .
Correct answer: 38 . 1928 Ω (tolerance ± 1
%).
Explanation:
E
B
S
1
c
d
a
b
R
1
R
2
R
3
R
4
Let : R
1
= 22 Ω ,
R
2
= 32 Ω ,
R
3
= 57 Ω ,
R
4
= 84 Ω , and
E
B
= 92 V .
Ohm’s law is V = I R .
A good rule of thumb is to eliminate junc-
tions connected by zero resistance.
E
B
c
a
b
R
1
R
2
R
3
R
4
The parallel connection of R
1
and R
2
gives
the equivalent resistance
1
R
12
=
1
R
1
+
1
R
2
=
R
2
+ R
1
R
1
R
2
R
12
=
R
1
R
2
R
1
+ R
2

homework 06 – LIN, KEVIN – Due: Feb 24 2008, 4:00 am 2
=
(22 Ω) (32 Ω)
22 Ω + 32 Ω
= 13 . 037 Ω .
E
c
a
b
R
12
R
3
R
4
The series connection of R
12
and R
3
gives
the equivalent resistance
R
123
= R
12
+ R
3
= 13 . 037 Ω + 57 Ω
= 70 . 037 Ω .
E
B
a
b
R
123
R
4
The parallel connection of R
123
and R
4
gives the equivalent resistance
1
R
ab
=
1
R
123
+
1
R
4
=
R
4
+ R
123
R
123
R
4
R
ab
=
R
123
R
4
R
123
+ R
4
=
(70 . 037 Ω) (84 Ω)
70 . 037 Ω + 84 Ω
= 38 . 1928 Ω .
or combining the above steps, the equivalent
resistance is
R
ab
=
parenleftbigg
R
1
R
2
R
1
+ R
2
+ R
3
parenrightbigg
R
4
R
1
R
2
R
1
+ R
2
+ R
3
+ R
4
=
bracketleftbigg
(22 Ω) (32 Ω)
22 Ω + 32 Ω
+ 57 Ω
bracketrightbigg
(84 Ω)
(22 Ω) (32 Ω)
22 Ω + 32 Ω
+ 57 Ω + 84 Ω
= 38 . 1928 Ω .
Question 4, chap 28, sect 4.
part 2 of 2 10 points
What is the current through the 57 Ω resis-
tor?
Correct answer: 1 . 31359 A (tolerance ± 1
%).
Explanation:
The voltages across R
12
and R
3
, respec-
tively, (the voltage between a and b ) is
V
ab
= V
12
+ V
3
= I ( R
12
+ R
3
)
= I R
123
.
where I = I
12
= I
3
is the current through
either resistor R
12
or R
3
. Hence, the current
through R
3
is
I
3
=
V
ab
R
12
+ R
3
=
V
ab
R
123
=
92 V
70 . 037 Ω
= 1 . 31359 A .
Question 5, chap 28, sect 4.
part 1 of 1 10 points
The following diagram shows a closed elec-
trical circuit. The ammeter in the center of
the resistive network reads zero amperes.
E
S
1
A
4 Ω
6 Ω
2 Ω
R
x
Find the electric resistance R
X
.
1. R
X
= 5 Ω .
2. R
X
= 20 Ω .

homework 06 – LIN, KEVIN – Due: Feb 24 2008, 4:00 am 3
3. R
X
= 4 Ω .
4. R
X
= 8 Ω .
5. R
X
= 6 Ω .
6. R
X
= 17 Ω .
7. R
X
= 19 Ω .
8. R
X
= 10 Ω .
9. R
X
= 3 Ω . correct
10. R
X
= 11 Ω .
Explanation:
E
S
1
u
ℓ
x
y
I
u
R
1
I
u
R
2
I
ℓ
R
3
I
ℓ
R
x
I
A
0 A
Let : R
1
= 4 Ω ,
R
2
= 6 Ω , and
R
3
= 2 Ω .