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Phy 2049
Summer 2009
Exam II Solutions
1.
A circuit is drawn with four resistors of 4 Ohms and one resistor of 2 Ohm. An ideal 12V
battery is connected to the circuit as shown.
Two of the resistors are labeled A and B. What is
the current through the resistor labeled B?
The equivalent resistance is 6Ώ.
Thus the current drawn from the battery is 12V/6Ώ =
2A.
That is the
current flowing through the resistor B.
2.
In the figure in the problem above, what is the power dissipated in the resistor marked A?
It is best done in plain words.
There are 2A coming out of the battery.
When they get to the junction,
there is the same resistance on both branches.
The current must split equally and therefore the current
through the 2Ώ resistor is 1A.
Finally the next junction also consists of equal resistances and the
current splits equally.
The current through the resistor A is ½A.
The power absorbed I
2
R =
1W.
The formal way to do this problem is to note:
A) the current out of the battery must be the same as the current going into the battery which is
same as the current through the resistor A.
The potential difference across A is IR = 2x4 = 8V.
B) Therefore the potential difference across the complex must be 4V.
The resistance down the right
arm is 4Ώ therefore the current through that arm is 1A.
C) It goes through the 2Ώ resistor and
is then split is two equal parts each fo ½A.
..
The total power is IV = 2
x
12 = 24W.
The power through resistor A is 16W.
That through the 2Ώ
resistor is 2W.
Power in the 4Ώ on the left of the complex is 4W and in the two 4Ώ (parallel including
resistor A) is 1W each.
They all add up correctly.
3.
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 Spring '08
 Any
 Physics, Current

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