261
D
IRECT
C
URRENT
C
IRCUITS
26.1.
IDENTIFY:
The newlyformed wire is a combination of series and parallel resistors.
SET UP:
Each of the three linear segments has resistance
R
/3. The circle is two
R
/6 resistors in parallel.
EXECUTE:
The resistance of the circle is
R
/12 since it consists of two
R
/6 resistors in parallel.
The equivalent
resistance is two
R
/3 resistors in series with an
R
/6 resistor, giving
R
equiv
=
R
/3 +
R
/3 +
R
/12 = 3
R
/4.
EVALUATE:
The equivalent resistance of the original wire has been reduced because the circle’s resistance is less
than it was as a linear wire.
26.2.
IDENTIFY:
It may appear that the meter measures
X
directly. But note that
X
is in parallel with three other
resistors, so the meter measures the equivalent parallel resistance between
ab
.
SET UP:
We use the formula for resistors in parallel.
EXECUTE:
1/(2.00
Ω
) = 1/
X
+ 1/(15.0
Ω
) + 1/(5.0
Ω
) + 1/(10.0
Ω
), so
X
= 7.5
Ω
.
EVALUATE:
X
is
greater
than the equivalent parallel resistance of 2.00
Ω
.
26.3.
(a)
IDENTIFY:
Suppose we have two resistors in parallel, with
12
R
R
<
.
SET UP:
The equivalent resistance is
eq
1
2
111
R
RR
=+
EXECUTE:
It is always true that
121
R
+>
. Therefore
eq
1
11
R
R
>
and
eq
1
R
R
<
.
EVALUATE:
The equivalent resistance is always less than that of the smallest resistor.
(b)
IDENTIFY:
Suppose we have
N
resistors in parallel, with
N
R
<<<
"
.
SET UP:
The equivalent resistance is
eq
1
2
1
N
R
R
=+++
"
EXECUTE:
It is always true that
1
1 1
N
R
R
+++ >
"
. Therefore
eq
1
R
R
>
and
eq
1
R
R
<
.
EVALUATE:
The equivalent resistance is always less than that of the smallest resistor.
26.4.
IDENTIFY:
For resistors in parallel the voltages are the same and equal to the voltage across the equivalent
resistance.
SET UP:
VI
R
=
.
eq
1
2
R
.
EXECUTE:
(a)
1
eq
12.3 .
32
20
R
−
⎛⎞
=
+=
Ω
⎜⎟
ΩΩ
⎝⎠
(b)
eq
240 V
19.5 A.
12.3
V
I
R
==
=
Ω
(c)
32
20
240 V
240 V
7.5 A;
12 A.
32
20
VV
II
=
=
EVALUATE:
More current flows through the resistor that has the smaller
R.
26.5.
IDENTIFY:
The equivalent resistance will vary for the different connections because the seriesparallel
combinations vary, and hence the current will vary.
SET UP:
First calculate the equivalent resistance using the seriesparallel formulas, then use Ohm’s law (
V = RI
)
to find the current.
EXECUTE:
(a)
1/
R
= 1/(15.0
Ω
) + 1/(30.0
Ω
) gives
R
= 10.0
Ω
.
I
=
V
/
R
= (35.0 V)/(10.0
Ω
) = 3.50 A.
(b)
1/
R
= 1/(10.0
Ω
) + 1/(35.0
Ω
) gives
R
= 7.78
Ω
.
I
= (35.0 V)/(7.78
Ω
) = 4.50 A
(c)
1/
R
= 1/(20.0
Ω
) + 1/(25.0
Ω
) gives
R
= 11.11
Ω
, so
I
= (35.0 V)/(11.11
Ω
) = 3.15 A.
26
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document262
Chapter 26
(d)
From part (b), the resistance of the triangle alone is 7.78
Ω
. Adding the 3.00
Ω
internal resistance of the battery
gives an equivalent resistance for the circuit of 10.78
Ω
. Therefore the current is
I
= (35.0 V)/(10.78
Ω
) = 3.25 A
EVALUATE:
It makes a big difference how the triangle is connected to the battery.
26.6.
IDENTIFY:
The potential drop is the same across the resistors in parallel, and the current into the parallel
combination is the same as the current through the 45.0
Ω
resistor.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '08
 Jabbour
 Materials Science And Engineering

Click to edit the document details