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SLTNS whw # 3
CH 21
AND MPTWO
Problem #s:
45,51,77,78,79,81,91,93
45.
Picture the Problem
: Four resistors and a switch are connected to a battery as
shown in the diagram at the right.
Strategy:
The symmetric nature of the circuit gives clues to its behavior.
Because the total resistances for the two paths on either side of the switch are
the same, equal currents will flow through each branch.
Because the top two
resistors in each branch are the same, the voltage drops across each will be the
same as well.
That means the electric potential at either side of the switch will
be the same, and therefore the potential difference across the switch is zero and
no current will flow through the switch when it is closed.
Use Ohm’s Law and
the rules for calculating equivalent resistance (equations 217 and 2110) to
verify these observations.
Solution:
1. (a)
Because the resistors are identical, the potential is the same at both sides of the switch regardless of
whether it is open or closed. Therefore the current through the battery will stay the same when the switch is closed.
2.
(b)
Find the current
0
I
through the battery when the switch is
open:
The circuit is two parallel branches of resistors connected
in series.
0
eq
1
2
I
R
R
R
R
3.
Find the current
I
through the battery when the switch is closed.
The circuit is a series connection of two sets of resistors that are
connected in parallel.
0
1
1
1
1
1
1
eq
R
R
R
R
I
I
R
R
Insight:
If the bottom two resistors were each doubled to 2
R
, the current
0
I
would decrease but the current
I
when
the switch is closed would still equal
0
I
.
51.
Picture the Problem
: Four resistors and two batteries are
connected as shown in the circuit diagram at right.
Strategy:
The circuit can be analyzed by applying
Kirchoff’s rules.
First apply the Junction Rule to point A
in the circuit, then apply the Loop Rule to two loops, the
lefthand loop 1 and the outside loop 2 labeled in the
diagram. These three equations can be combined to
algebraically find
1
2
3
,
, and
.
I
I
I
From the currents we
can find the potential difference between the points A and
B.
Solution:
1. (a)
Apply the
Junction Rule to point A:
1
2
3
(i)
I
I
I
K K
2.
Apply the Loop Rule to loop 1,
beginning in lower lefthand
corner, and solve for
3
I
:
1
1
3
3
1
4
1
4
3
1
1
3
3
0
12 V
12 V
13.7
10 A
(ii)
1.2
I R
I R
I R
R
R
I
I
I
R
R
K K
3.
Apply the Loop Rule to loop
2, beginning in lower lefthand
corner, and solve for
2
I
:
1
1
2
2
1
4
1
4
2
1
1
2
2
0
12 V
9.0 V
12.0
9.0 V
3.0 V
13.7
(iii)
6.7
6.7
I R
I R
I R
R
R
I
I
I
R
R
K K
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Substitute equations (ii) and
(iii) into equation (i).
1
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This note was uploaded on 08/12/2009 for the course PHY 122 taught by Professor L.rose during the Fall '07 term at New Mexico Junior College.
 Fall '07
 L.ROSE
 Physics

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