PHY2049
Spring 2005
1
Prof. Darin Acosta
Prof. Paul Avery
March 9, 2005
PHY2049, Spring 2005
Exam 2 Solutions
1. Two identical light bulbs A and B are connected in series to a constant voltage source.
A wire is then connected across B. What is the brightness of A relative to its former
brightness?
A light bulb will have a certain resistance R.
The light it emits will be proportional to the
energy dissipated by this resistance:
2
PiR
=
.
So when two identical light bulbs are
connected in series to a constant voltage V, the current passing through the circuit is
/2
iV R
=
and the power dissipated in light bulb A is:
22
1
2
44
VV
PR
R
R
==
.
Now when a wire is connected across light bulb B, that means that the resistance is
bypassed by a wire in parallel to the bulb (i.e. the bulb is shorted out by a jumper). The
resistance of bulb B becomes zero, and the current in the circuit is
/
iVR
=
. This current
is twice as high as before, but the power dissipated by light bulb A will be four times
more:
2
21
4
V
PP
R
Answer: 4
2. In the circuit shown, what is the current (in amps) flowing through the 18V battery?
Let
1
i be the current through the 18V battery,
2
i the current through the middle branch
with the 6V battery, and
3
i the current through the rightmost branch.
By Kirchoff’s junction rule:
123
iii
=+
.
By Kirchoff’s voltage sum rule applied to the rightmost loop:
23
32
6V 12
24
0.5
0.25
ii
+Ω= Ω
⇒=
+
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Spring 2005
2
By Kirchoff’s voltage sum rule applied to the leftmost loop:
()
12
23
2
22
2
2
32
123
18V 6
6V 12
18 6
6 12
18 6
6 0.25 0.5
6 12
18 1.5 9
6 12
10.5
0.5
21
0.25 0.5
0.5
1.0
ii
i
i
i
iii
−Ω=
+ Ω
−+
=
+
−−
+
=
+
−−=
+
⇒=
=
+
=
⇒=+=
Answer: 1.0 Amps through 18V battery
3. An electron is accelerated from rest by a potential difference of 8.0 kV.
It then enters a
uniform magnetic field of magnitude 1.0 T with its velocity perpendicular to the direction
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 Spring '08
 Avery
 Physics, Light, Magnetic Field, uniform magnetic field, Prof. Darin Acosta Prof. Paul Avery

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