PHY303L Solutions%206 - homework 06 FAUSAK TAYLOR Due 4:00...

homework 06 – FAUSAK, TAYLOR – Due: Feb 25 2008, 4:00 am 1
Question 1, chap 28, sect 4.
part 1 of 2 10 points
A 28 V battery has an internal resistance r .
1 A
28 V
r
42 Ω
56 Ω
internal
resistance
What is the value of r ?
Correct answer: 4 Ω (tolerance ± 1 %).
Explanation:
I
1
I
2
I
3
E
r
R
2
R
3
internal
resistance
Let : E = 28 V ,
R
2
= 42 Ω ,
R
3
= 56 Ω , and
I
1
= 1 A .
Since R
2
and R
3
are connected parallel,
their equivalent resistance R
23
is
1
R
23
=
1
R
2
+
1
R
3
=
R
3
+ R
2
R
2
R
3
R
23
=
R
2
R
3
R
2
+ R
3
=
(42 Ω) (56 Ω)
42 Ω + 56 Ω
= 24 Ω .
Using Ohm’s law, we have
E = I
1
r + I
1
R
23
r =
E − I
1
R
23
I
1
=
28 V − (1 A) (24 Ω)
1 A
= 28 Ω − 24 Ω
= 4 Ω .
Question 2, chap 28, sect 4.
part 2 of 2 10 points
Determine the magnitude of the current
through the 56 Ω resistor on the right-hand
side of the circuit.
Correct answer: 0 . 428571 A (tolerance ± 1
%).
Explanation:
The potential drop across the 56 Ω resistor
on the right-hand side of the circuit is
E
3
= E − I
1
r
= 28 V − (1 A) (4 Ω)
= 28 V − 4 V
= 24 V ,
so the current through the resistor is
I
3
=
E
3
r
3
=
24 V
56 Ω
=
3
7
A
= 0 . 428571 A ,
and
I
2
= I
1
− I
3
= 1 A −
3
7
A
=
4
7
A
= 0 . 571429 A .
Question 3, chap 28, sect 4.
part 1 of 3 10 points
7 . 2 V 1 V
4 . 7 V
I
1
1 . 3 Ω 2 Ω
I
2
7 . 7 Ω
I
3
9 . 7 Ω

homework 06 – FAUSAK, TAYLOR – Due: Feb 25 2008, 4:00 am 2
Find the current I
1
in the 1 . 3 Ω resistor
at the bottom of the circuit between the two
power supplies.
Correct answer: 0 . 734918 A (tolerance ± 1
%).
Explanation:
E
1
E
2
E
3
I
1
R
A
R
B
I
2
R
C
I
3
R
D
At a junction (Conservation of Charge)
I
1
+ I
2
− I
3
= 0 . (1)
Kirchhoff’s law on the large outside loop gives
( R
A
+ R
B
) I
1
+ R
D
I
3
= E
1
+ E
2
. (2)
Kirchhoff’s law on the right-hand small loop
gives
R
C
I
2
+ R
D
I
3
= E
3
. (3)
Let : R
A
= 1 . 3 Ω ,
R
B
= 2 Ω ,
R
C
= 7 . 7 Ω ,
R
D
= 9 . 7 Ω ,
E
1
= 7 . 2 V ,
E
2
= 1 V , and
E
3
= 4 . 7 V .
Using determinants,
I
1
=
vextendsingle
vextendsingle
vextendsingle
vextendsingle
vextendsingle
vextendsingle
0 1 − 1
E
1
+ E
2
0 R
D
E
3
R
C
R
D
vextendsingle
vextendsingle
vextendsingle
vextendsingle
vextendsingle
vextendsingle
vextendsingle
vextendsingle
vextendsingle
vextendsingle
vextendsingle
vextendsingle
1 1 − 1
R
A
+ R
B
0 R
D
0 R
C
R
D
vextendsingle
vextendsingle
vextendsingle
vextendsingle
vextendsingle
vextendsingle
Expanding along the first row, the numera-
tor is
D
1
=
vextendsingle
vextendsingle
vextendsingle
vextendsingle
vextendsingle
vextendsingle
0 1 − 1
E
1
+ E
2
0 R
D
E
3
R
C
R
D
vextendsingle
vextendsingle
vextendsingle
vextendsingle
vextendsingle
vextendsingle
= 0 − 1
vextendsingle
vextendsingle
vextendsingle
vextendsingle
E
1
+ E
2
R
D
E
3
R
D
vextendsingle
vextendsingle
vextendsingle
vextendsingle
+ ( − 1)
vextendsingle
vextendsingle
vextendsingle
vextendsingle
E
1
+ E
2
0
E
3
R
C
vextendsingle
vextendsingle
vextendsingle
vextendsingle
= − [( E
1
+ E
2
) R
D
− E
3
R
D
]
− [ R
C
( E
1
+ E
2
) − 0]
= R
D
( E
3
− E
1
− E
2
) − R
C
( E
1
+ E
2
)
= (9 . 7 Ω) (4 . 7 V − 7 . 2 V − 1 V)
− (7 . 7 Ω) (7 . 2 V + 1 V)
= − 97 . 09 V Ω .