Homework Solution
Assignment (1-77)
Read (20.9), Do PROBLEMS # (61, 67, 68) Ch. 20 + AP 1997 #4 (handout)
61.
REASONING
‰ When two or more resistors are in series, the equivalent
resistance is given by the equation
Rs = R1 + R2 + R3 + . . . .
‰ Likewise, when resistors are in parallel, the expression
to be solved to find the equivalent resistance is given by
the equation
1
1
1
1
=
+
+
+ ... .
Rp R1 R2 R3
We will successively apply these relationships to the individual resistors in the schematic diagram
beginning with the resistors on the right side of the figure.
SOLUTION
R3
A
Since the 4.0-Ω and the 6.0-Ω resistors are in series,
the equivalent resistance of the combination of those
two resistors is 10.0 Ω.
R4 + R6
R20
B
R9 // R8
The 9.0-Ω and 8.0-Ω resistors are in parallel;
their equivalent resistance is 4.24 Ω.
R3
A
R20
R4 + R6
B
R9 // R8
The equivalent resistances of the parallel combination
(9.0 Ω and 8.0 Ω) and the series combination
(4.0 Ω and the 6.0 Ω) are in parallel;
therefore, their equivalent resistance is 2.98 Ω.
[R4 + R6] // [R9 // R8]
R3
A
The 2.98-Ω combination is in series with the
3.0-Ω resistor, so that equivalent resistance
is 5.98 Ω.
Finally, the 5.98-Ω combination and the
20.0-Ω resistor are in parallel, so the
equivalent resistance between the
points A and B is 4 .6 Ω .
R20
[R4 + R6] // [R9 // R8]
B
A
[[R4 + R6] // [R9 // R8]] + [R4 + R6]
B
________________________________________________________________________________________________________________________________________________________________________________________________________________________
Homework Solution (cont.)
Assignment (1-77)
Read (20.9), Do PROBLEMS # (61, 67, 68) Ch. 20 + AP 1997 #4 (handout)
________________________________________________________________________________________________________________________________________________________________________________________________________________________
67.
REASONING AND SOLUTION
rint
The terminal voltage of the battery is:
VT = Emf – I rint
Since IT = V?Term
R?T
I
ε=9 v
L
O
A
D
VT = 8.3 v
RL = 1.4 Ω
V = Emf – VTerm rint
VTT = ???
RT
8.3 v = 9 v – 8.3 v rint
A little “algorithm”
1.4 Ω
razzmatazz
8.3 v = 9 v – 5.93 rint
= 9…v – 8.3 v = 0.70
Revealsrintthat
rint v =
5.93 A
5.93 A
0.12 Ω
________________________________________________________________________________________________________________________________________________________________________________________________________________________
68.
REASONING AND SOLUTION
Simply stated, Ohm’s law gives:
Since EMF = 1.5 V and I = 28 A
? ? Α) = 0.054 Ω
r = V/I = (1.5 ?V)/(28
D
Size