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1. You hook up a light bulb (unknown resistance) to a battery (unknown voltage), and
measure the current using an ammeter, as shown in Circuit 1. The ammeter has no
internal resistance (i.e. it has no effect on the circuit other than to measure the current).
You find that the current in
Circuit 1 is 0.4 A
. You then add an identical light bulb as
shown in Circuit 2.
a) Using just the information given above,
what current will you measure in Circuit 2?
Use the energy density equation to justify your answer.
Going around the Circuit 1 in a complete loop, we get
± ² ³ ´ µ
¶
· ² ¸
, which leads to
I
1
²¹³/R
. Since the current in
circuit 1 is .6 A, we now know that
³
/R = .4 A.
Going around Circuit 2 in a complete loop, remembering that the battery is the same, and that we have added another
identical resistor, we get
± ² ³ ´ µ
º
· ´ µ
º
· ² ³ ´ »µ
º
· ² ¸
.
So now we have
I
2
=
³
/2R = (.4 A)/2 = .2 A.
b) You now use a voltmeter to measure the voltage difference between points 1 and 2 and find that
± ² ´¼½¹±
. Using your
answer to part a), and the appropriate energy density equation, find the
resistance of the light bulb in Ohms (Ω).
Between points 1 and 2 in Circuit 2, there is one resistor, so ΔV = V
2
 V
1
= I
2
R. Using the current from part a), this leads
to
.6 V = (.2A)*R
R = 3 Ω
Note that no points were taken off for having the incorrect current from part a), as long as the current used to find the resistance
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 Spring '07
 JOHNCONWAY
 Energy, Work, Volt, Quadratic equation, Electrical resistance, Energy density, Energy Density Equation

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