PY106 TEST 1
Page 1 of 4
Feb 23, 2000
Name_________________
ID___________________
PROBLEM 4. [20 points] – Light bulbs
The bulbs in the circuit are identical.
V
V
V
V
(1) [6 points] In the circuit on the left, rank bulbs 1-6 in order of decreasing brightness.
Briefly explain your answers.
Brightness of a bulb is determined the power dissipated through the bulb and is
given by I2R. Since R is the same for all the bulbs, their brightness will be rated by
the current passing through each one of them. The more the current passing
through the bulb, the brighter it will be.
If the current drawn from the battery is I, then bulb 6 sees the full current passing
through it. Bulbs 4 and 5 are in parallel and have the same resistance, so they share
this current and half of the current passes through each one of them. Bulbs 1 and 2
are series with each other and parallel to bulb 3, thus the path through 2 and 3 has
twice the resistance compared to the path with bulb 3. Therefore the current splits
itself and two-thirds current passes through bulb 3, while only one-third of the
current flows through bulbs 1 and 2.
Ordering the bulbs according to icreasing current (or brighness) we get:
6>3>5=4>1=2
(2) [6 points] If each bulb has a resistance of 12 ohms, what is the equivalent
resistance of the circuit on the left?
Req = R6 + (R5 // R4) + (R3 // (R1+R2))
= R6 + (R5 * R4)/(R5+R4) + R3 * (R1+R2)/(R1+R2+R3)
= R + R/2 + 2 R/3 (use the fact that all resistance are the same)
= 13 R /6 = 13*12/6
= 26 Ohms.
PY106 TEST 1
Page 2 of 4
Feb 23, 2000
Name_________________
ID___________________
(3) [8 points] When bulb 1 is removed, leaving a break in the circuit….
a) The brightness of bulb 3 __________
[ X ] increases
[ ] decreases
[ ] stays the same
Briefly explain your answer:
When bulb 1 is removed all of the current now flows through 3, increasing its
brightness
b) The brightness of bulb 6 __________
[ ] increases
[ X] decreases
[ ] stays the same
Briefly explain your answer:
The overall resistance of the circuit increases, therefore the current decreases, leading to a
decreased brightness of bulb 6
PY106 TEST 1
Page 3 of 4
Feb 23, 2000
Name_________________
ID___________________
PROBLEM 6. [20 points] – A bubble chamber
In bubble chamber experiments, charged particles leave small hydrogen gas bubbles
along their paths. The paths of three such charged particles are shown below; the paths
are in the plane of the page. There is a uniform magnetic field directed into the page, and
the three particles have charges of equal magnitude.
a) [6 points] Are the charged particles negative or positive?
Use right hand rule
Particle 1 is [ X] positive
Particle 2 is [ ] positive
Particle 3 is [ X] positive
[ ] negative
[ X ] negative
[ ] negative
b) [3 points] If the masses of the particles are equal, which one has the highest speed?
[ ] Particle 1
[ ] Particle 2
[ X] Particle 3
Speed of a particle moving in a magnetic field v = qrB/m.
Since B, m and q are same for all the 3 particles, the one with the largest radius of
curvature has the largest speed.
c) [5 points] If the radius of the path followed by particle 3 is 4 times larger than the
radius of particle 1’s path, determine the ratio of their kinetic energies. (Assume they
have the same mass.)
Since the masses are equal, the ratio of the Kinetic energies:
KE3/KE1 = ( v3/v1)2 = (r3/r1)2. (use the expression from part b)
= (4r/r) 2. = 16
PY106 TEST 1
Page 4 of 4
Feb 23, 2000
Name_________________
ID___________________
d) [6 points] Particle 3 enters the chamber at point A with a speed of 2.4 x 105 m/s. If
the magnetic field in the chamber is 0.13 T, what is the magnitude of the electric field
that can be introduced so particle 3 moves as if it does not experience any force?
Sketch the electric field on the diagram above, clearly showing its direction.
The particle will feeel two forces: Force due to the magnetic filed = qvB,
and the force due to the electric field = qE. Since we want the net force to be zero,
qE=qvB
or E = vB = 2.4 105 m/s * 0.13 T = 3.1 x 104 N/C