Quarter PHY138 Test Solutions
February 10, 2004
Multiple Choice Questions
1. What is the relationship between the electric field vector at a point and the electric field line going through that point? (A) They are perpendicular to each other. (B) The electric field vector makes an angle of /3 radians with the field line. (C) The electric field vector makes an angle of /2 radians with the field line. (D) The electric field vector is tangent to the field line. (E) The electric field vector has a value 1/r2 times the value of the field line.
2. The potential difference across a 3 F capacitor is 12 V. Its plates are then connected to those of an uncharged 6 F capacitor. When equilibrium is reached, the magnitude of the charge (in C) on the plates of the 3 F capacitor is (A) 12 (B) 18 (C) 24 (D) 27 (E) 36
3. What is the speed (in m/s) of a proton (m = 1.67 10-27 kg, q = 1.60 10-19 C) that has been accelerated from rest through a potential difference of 4.0 kV? (A) 2.8 104 (B) 9.8 105 (C) 8.8 105 (D) 1.2 106 (E) 6.2 105
4. Three balls of very small mass coated with a conducting material hang side-by-side without touching (as shown in the figure opposite). When a positive charge is placed on ball A, B is repelled by A, but C is attracted to B. We can conclude that: (A) B is positively charged and C is negatively charged; (B) B is positively charged and C has no excess charge; (C) B is negatively charged and C is positively charged; (D) B is negatively charged and C has no excess charge; (E) B is positively charged and C is either negatively charged or has no excess charge.
5. A 100 Watt light bulb is connected to a 120 volt source. What is the resistance of the filament in the bulb (in Ohms)? (A) 83.3 (B) 144 (C) 12000 (D) 0.833 (E) 1.20
6. An RC circuit contains a 220 resistor, a 5 F capacitor and a 300 V battery. Initially the switch is open. 11 ms after the switch closes, the current in the resistor (in A) is: (A) 62 (B) 13 (C) 67 (D) 64 (E) 69
7. Given a 6 resistor, a 9 resistor and an 18 V battery, what is the maximum power (in W) that can be dissipated? (A) 54 (B) 71 (C) 90 (D) 80 (E) 22
8. Given the circuit in the diagram on the right, find I2 . (A) 1.8 A (B) 2.4 A (C) 3.3 A (D) 4.2 A (E) 4.8 A
1. The electric field is tangent to the field lines at all points. So the answer here is
D.
2. We know that for a capacitor Q = C V . So the 3 F capacitor will have QTot = 3 F 12 V = 36 C of charge on it. When we connect the two capacitors together they must end up with the same voltage across them. So we are led to the relation: Q1 Q2 = C1 C2 But Q1 + Q2 = QTot as the only source of charge for the final state is the charge that was on C1 initially. Substituting this relationship for the charge into the expression above we get something like: Q1 = C1 (QTot - Q1 ) C2 C1 C1 ) = QTot C2 C2 A.
Q1 (1 +
With C1 /C2 = 0.5 I find Q1 = 0.5 0.67QTot = 12C. So the answer here is 3. The electric potential energy lost by the proton is:
U = 1.60 10-19 C 4000V = 6.4 10-16 J This will result in an increase in the kinetic energy of the proton of: Kin = m v2 = 8.35 10-28 v2 J 2
The increase of kinetic energy comes from the loss of electrical potential energy. So the final speed of the proton is given by: 6.4 10-16 = 8.75 105 m/s 8.35 10-28
v=
So the answer here is
C.
4. When A is positively charged, B is repelled, so it must have a positive charge. The choice is between answers A, B and E. But then we are told that C is attracted to B. So C must either be negatively charged OR be neutral. In the latter case there will be a small split between the negative and positive charges on ball C (we talked about this in the first lecture when we talked about charging objects by induction). This split will bring some of the negative charges to the left side of ball C causing it to be slightly attracted to ball B even if has no excess charge. So the best/most-complete answer is E. 5. We know that P = V 2 /R or R = V 2 /P with a 120 V difference across the bulb and 100 W being dissipated the filament resistance must R be = 1202 /100 = 144 and so the answer is B.
6. The time constant for this circuit is R C = 220 5 10-6 F = 1.1 ms. So 11 ms represents 10 time constants. But we know that the current flowing in the resistor of an RC circuit at any time is given by: I(t = 10) = V0 -t/RC e = 300V/220 e-10 R A.
Plugging this into a calculator I get I(t = 10) = 1.36Ae-10 = 62A. So the answer is
7. Given two resistors there are two possible ways to combine them with a battery in a circuit. But the power dissipated in the circuit will be given by P = V 2 /Req . So to maximise the power dissipation we want to minimise the equivalent resistance. That is done by putting the resistors in parallel. 9 and 6 in parallel give: Req = (1/R1 + 1/R2 )-1 = 3.6 This will give a power of P = 182 /3.6 = 90 W. So the answer is C.
8. This was probably the most involved multiple choice question I could have asked. Applying Kirchoff's node law we find that: I2 = I1 + I3 Although we are interested in I3 I will eliminate it from the Kirchoff loop equations and the use this expression to figure it out at the end. Counter-clockwise around the top loop I get: 18 - 4I1 - 2(I1 + I3 ) = 0 Clockwise around the bottom loop I get: 12 - 2I3 - 2(I1 + I3 ) = 0 Subtracting these two I get: I3 = 2I1 - 3 Plugging this back into the first equation, I get: I1 = 2.4A and I3 = 1.8A Summing these two to get the request current (using the result of Kirchoff's node law above) I get I2 = 4.2 A, so that answer is D.
Worked Problem:
9. A 1.00 m long wire is made by welding two 50.0 cm wire-segments: one section is made of Tungsten ( = 5.50 10-8 m) and the other is made of Constantan ( = 44.0 10-8 m). The resulting wire has a diameter of 3.00 mm along its entire length. The ends of the wire are connected to a 12.0 V battery.
(a) What is the potential drop across each segment of the wire? I chose to compute the resistance of each segment of the composite wire and from there determine the currents and voltage drops. There were many ways you could get started with a problem like this. l Recall that R = A . For the Tungsten part of the wire this gives: RTungsten = For the Constantan part of the wire I get: RConstantan = 44.0 10-8 0.5 = 31.1m (0.0015)2 5.50 10-8 0.5 = 3.89m (0.0015)2
These two pieces of wire just act as resistors in series so the resistance of the full wire is RTot = 3.89 + 31.1 = 35.0m. Thus the current flowing in the circuit (the answer ...