2153:
Picture the Problem
: Six resistors are connected in the manner
indicated in the diagram, and a 9.0V battery is connected to
terminals A and B.
Strategy:
Find the equivalent resistance
of the righthand
portion of this circuit by combining resistors 4, 5, and 6 in parallel,
and then combining that group in series with resistor 3.
The
voltage drops across resistors 1, 2, and
are each 9.0 V.
Ohm’s
Law can then be used to find
eq
R
eq
R
3
d .
12
, , an
I
II
Use Ohm’s Law
together with
3
I
to find the voltage drop across resistor 3, then
subtract that result from 9.0 V to find the voltage drop across
resistors 4, 5, and 6.
Ohm’s Law can then be applied to each of
those resistors to find
45
, ,
6
and .
I
Solution:
1. (a)
Use Ohm’s Law to find
1
:
I
1
9.0 V
6.0 A
1.5
I
2.
Repeat step 1 to find
2
:
I
2
9.0 V
3.6 A
2.5
I
3.
Find
for the righthand portion of the circuit:
eq
R
1
eq
111
6.3
7.9
4.8
3.3
8.1
R
4.
Apply Ohm’s Law to the righthand branch:
3
9.0 V
1.14 A
1.1 A
7.9
I
5.
Find
across resistors 4, 5, and 6:
V
33
9.0 V
1.14 A
6.3
1.8 V
VI
R
6.
Apply Ohm’s Law to find
4
:
I
4
1.8 V
0.38 A
4.8
I
7.
Repeat step 6 to find
5
:
I
5
1.8 V
0.55 A
3.3
I
8.
Repeat step 6 to find
6
:
I
6
1.8 V
0.22 A
8.1
I
9. (b)
There is a potential drop across resistors 4, 5, and 6 (1.8 V in this case), resulting in a potential
difference across the 6.3
Ω
resistor that is less than the 9.0 V that is across the 1.5
Ω
resistor.
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 Fall '06
 Alexandrakis
 Physics, Resistance, Resistor, Inductor, Electrical resistance, Electrical impedance

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