University of Florida Department of Physics
PHY2049 Fall 2007 Physics 2 with Calculus
TA: Ivan Furi´c, [email protected]fl.edu
Quiz #7, October 29th, 2007
Candidate Name: (please print large and legibly. No UFID is necessary.)
Total: 20 points
This is a closed-book quiz; no reference materials should be used. Use of calculators is allowed. To be awarded credit, numerical results need justification, either by means of a formula/derivation or reasoning of applicable symmetry. For full credit on a result, sign conventions and units need to be correct, in addition to the numerical value.
University of Florida Honor Code: ”We, the members of the University of Florida community, pledge to hold ourselves and our peers to the highest standards of honesty and integrity.” On all work submitted for credit by students at the University of Florida, the following pledge is either required or implied: ”On my honor, I have neither given nor received unauthorized aid in doing this assignment.”
The quiz and solutions will be posted at 5 pm today to the following URL: http://www.phys.ufl.edu/ ikfuric/teaching/2049 fall 2007.html ∼
DO NOT OPEN QUIZ UNTIL INSTRUCTED Problem 1: Units, Definitions and Oscillatiors
(a) What is the unit of capacitance (C)? (1 points )
Answer: The unit of capacitance in the MKSA/SI system is 1 Farad [F ]
(b) What is the unit of magnetic inductance? (L)? (1 points )
Answer: The unit of inductance in the MKSA/SI system is 1 Henry [H]
(c) What is the unit of electric resistance? (R)? (1 points )
Answer: The unit of inductance in the MKSA/SI system is 1 Ohm [Ω]
(d) What is the unit of time? (t)? (1 points )
Answer: The unit of frequency in the MKSA/SI system is 1 second, 1 s.
(e) What is the unit of frequency? (f)? (1 points )
Answer: The unit of frequency in the MKSA/SI system is 1 Hertz [Hz] = 1/s.
(f) An LC oscillating circuit consists of a capacitor C = 16 mF and an inductance L = 25 mH connected in series. What is the resonant frequency ω of this oscillator? (3 points )
Answer: The resonant frequency of the LC circuit is 1 1 ω = = = 50 Hz (1) √LC (√16 10−3 25 10−3) s · × ·
(g) We add a resistor of resistance R = 80Ω to the LC circuit in series. What is the new resonant frequency? (2 points )
Answer: Question was ill-posed as the system will go into resonance when driven by the frequency from the previuos question. The idea was to calculate the new natural frequency of the system, however the system is now over-damped. Everyone was awarded 2 points. For reference, the new natural frequency of the LRC circuit is
1 R 2 1 2 80 2 2 ω = ( 2 ) = 16·10−3×25·10−3 Hz ( 2×25·10−3 ) Hz (2) q LC − L q − = (2500 (1600)2)Hz2 = imaginary number (3) q −
2 i
Emf X
Figure 1: Circuit with a changeable component X.
Problem 2: Consider the circuit depicted in Figure 1. Regardless of what component X is, the source always delivers a voltage: mf = V0 sin(ωt) (4) E where V0 is 110 Volt, and ω = 300 Hz.
(a) Let the component X in the circuit be a resistor R = 100 Ω. What is the current through the resistor, as a function of time? (2 points )
Answer: The current for resistor R is:
i = V/R = 1.1 A sin(ωt) (5)
(b) Let the component X in the circuit be a capacitor, C = 50 µF. What is the current through the capacitor, as a function of time? (3 points )
Answer: The charge on the capacitor is:
Qc = C Vc = V0 C sin(ωt) (6) · · · we find the current by differentiating with respect to t:
∂Qc −6 ic = = V0 C ω cos(ωt) = 110 V 50 10 F 300 Hz cos(ωt) = 1.65 A cos(ωt) (7) ∂t · · · · × · ·
3 (c) Let the component X in the circuit be an inductor, L = 20 mH. What is the current through the inductor, as a function of time? (3 points )
Answer: The potential drop across the inductor is: ∂i V = L (8) L ∂t from which ∂i V V0 = L = sin(ωt) (9) ∂t L L · to solve for i, we integrate with respect to time:
∂i V0 V0 i = dt = sin(ωt)dt = cos(ωt) = (10) Z ∂t Z L · −Lω · 110 V cos(ωt) = 18.33 A cos(ωt) (11) −300.0 Hz 20 10−3 H − · · ·
(d) Let the component X in the circuit be an ideal transformer with Ns = 3000 coils on the secondary and Np = 100 coils on the primary. What is the amplitude of the potential on the secondary coil? (2 points )
Answer: For an ideal transformer, the amplitude of the potential on the secondary coil is: Ns 3000 Vs = Vp = 110 V = 3300 V (12) Np 100
4