1. (10 points) How many 60 W light bulbs, connected to 120 V in parallel, can be
ID: 3278381 • Letter: 1
Question
1. (10 points) How many 60 W light bulbs, connected to 120 V in parallel, can be used without blowing a 15A fuse? 2, (10 points) Find the direction of the force on a negative charge for each diagram shown in figure where V is the velocity of the charge and B is the direction of the magnetic field means the vector points inward O rneans it points outward, toward you.) f 3. (10 points) What is the direction of the induced current in the circular loop due to the current shown in each part of figure I increasing I constant increasing 1 decreasing 4. (15 points) A proton moves through a region of space where there is a magnetic field B (0.20i 0.10) T. At a given instant, the proton's velocity is V= (2.0i + 1.0-3.01) x 10s m/s. Determine the components of the total force on the proton. Charge of the proton is equal to 1.60 x 101" C 5. (10 points) A coaxial cable consists of a solid inner conductor of radius R1, surrounded by a concentric cylindrical tube of inner radius R2 and outer radius R3 (see the figure). The conductors carry equal and opposite currents lo distributed uniformly across their cross sections. Let lo 1.00 A, R1 = 2.00 mm, R2 = 5.00 mm, and R3 = 6.00 mm. Find B for (a) (5 points) = 2.50 mm; (b) (5 points) r = 7.50 mm 6. (15 points) a) (10 points) Determine the equivalent resistance of the "ladder" of equal 15 resistors shown in figure In other words, what resistance would an ohmmeter read if connected between points A and B? b) (5 points) What is the current through each of the two resistors on the left (the one connected to point A and the other connected to point B) if a 24.0 V battery is connected between points A and B? B OVER PLEASEExplanation / Answer
6) R , R and R are in series
R1 = 15 + 15 + 15 = 45 ohm
R1 and R are in parallel
R2 = 15*45 / 15+45 = 10.25 ohm
10.25 .15 , 15 are inseries
so R3 = 10.25 + 15+15 = 40.25 ohm
15 and 40.25 are in parallel
R3 = 15*40.25 / 15 + 40.25 = 10.93 ohm
10.93 , 15,15 ohm are in series
So net resistance = 10.93 + 15 + 15 = 40.93 ohm
b) current through the circuit = V/R = 24 / 40.93 = 0.586 A
4) F = q(V X B)
= 1.6*10^-19 [ i j k]
[ 2 1 -3]
[0.2 0.1 0]
1.6*10^-19 [ i(0-(-0.3) -j (0 - (-0.6)) + k(0.2 - 0.2)]
= 1.6*10^-18 [3i - 6j]
3) a) anticlockwise b) clockwise
c) anticlockwise d) clockwise
2) F = q(vXB)
so the force will act perpendicular to V and B
a) to the right
b) to the left
c) in the upward direction
d) in the page
e) no force will act
f) in the upward direction
1) power = VI = 120*15 = 1800 W
no. of bulbs = 1800/60 = 30 bulbs
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