Impedance/Reactance Problems

Transcription

1 Impedance/Reactance Problems. Consider the circuit below. An AC sinusoidal voltage of amplitude V and frequency ω is applied to the three capacitors, each of the same capacitance C. What is the total reactance of the circuit? b) X tot = 2 3 ω C c) X tot = 2 ω C d) X tot = ω C e) X tot = 3 2 ω C f) X tot =3 ω C = + = 2 X X eq X C X C X tot =X C + C 2 X C= 3 2 The impedance is the same. The phase shift is -π/2. 2. Consider the circuit from # made of all inductors of inductance L instead of capacitors. What will be the impedance of the circuit? a) Z tot = 2 3 ω L b) Z tot = 2 ω L c) Z tot =ω L d) Z tot = R 2 +(ω L) 2 e) Z tot = 3 2 ω L ωc = + = 2 X X eq X L X L X tot =X L + L 2 X = 3 L 2 ω L The impedance is the same. The phase shift is +π/2. Page

2 3. Consider an inductor and a capacitor in parallel. What is the phase shift of the circuit? a) 0 o b) 45 o c) 90 o d) -90 o e) Not enough information is provided. = + = X + X L C X X eq X C X L X L X tot = X X L C C X L + X C = L/C ω L ωc The impedance of the circuit is the same. If X L > X C, then the quantity is negative and so the phase shift will be -π/2. If X C > X L then the quantity is positive and so the phase shift will be +π/2. 4. We place a kω resistor into an AC driven circuit (frequency 80 Hz). We use an oscilloscope to measure the current and find that it is out of phase with the voltage input by 45 o (with the current lagging the voltage). What is the reactance of this circuit? What component does this imply exists in the circuit? a) kω Capacitors b) kω Inductor c) kω Resistor d) 45 kω Capacitor e) 45 kω Inductor There must be a net inductive reactance in the circuit to have a lagging current. tan ϕ= X R X L=R tan ϕ= k Ω tan (45 o )=k Ω Alternating Current Questions. We measure that the average power dissipated in an AC circuit is 200 W. If the amplitude of the applied voltage is 20 V, what is the amplitude of the current? a).67 A b) 2.00 A c) 2.36 A d) 3.00 A e) 3.33 A P=V rms I rms I rms = I 2 = P = P 2 I V rms V max =2 P =2 ( 200W ) max V max (20V ) = A I rms = A 2 Page

3 2. We send a square wave of amplitude V and frequency f = Hz into a circuit. The width of the square wave pulses are measured to be t = π/2. What is V rms? a).44 V b) V c) 2 V d) 0.5 V e).57 V Note that the pulse width is one quarter (not ½ like a sine wave) of the period. Thus the average of the squares is half of what it would otherwise be : = V ( )2 +(0 ) 2 +( 0) 2 +( 0) 2 rms = 4 2 V AC Circuit Component Problems. We have an AC driven LRC circuit. We use an oscilloscope to find that the frequency with the largest transmission amplitude is ω = 200π s -. If we know that the circuit contains an inductor of L = μh and a resistor of R = kω, what is the value of the capacitance of this circuit? a) F b) 3.95 μf c) μf d) 2.53 F e) μf ω 2 0 = LC C = L ω = 2 0 ( 0 6 H ) (200π s ) F 2 2. If we build a low pass filter by measuring a voltage across a capacitor which of the following describes the whole circuit? a) L, R, and C are in series with the AC voltage source as well as each other. b) There is an inductor in parallel with the capacitor, both in series with the source. c) A resistor is in series with the source, the capacitor in parallel. d) There is an inductor in series with the capacitor, both in series with the source. e) There is a resistor in parallel with the source, the capacitor in series. If the capacitor is in parallel with the source, then it will stop any low frequencies from going through it forcing them to go through the circuit desired. To address the other options : a resistor in parallel with the source, which would be in series with the capacitor would allow the capacitor to block the low frequencies from the source (it would be a high-pass filter). An inductor in series with the capacitor and in series with the source would be a resonance circuit (allowing only the resonance frequency to pass). An inductor in parallel with the capacitor would simply allow high frequencies through one branch and low frequencies through the other. And the LRC would simply oscillate and allow a band of frequencies through. 3 Page

4 Superconductivity questions. Which of the following are reasons why a superconductor is not a perfect conductor? a) A superconductor has no electrical resistance. b) A superconductor does not have an internal magnetic field. c) A superconductor can have a non-zero electric field inside of it. d) A perfect conductor does not have an internal magnetic field. e) A perfect conductor always has a non-zero voltage on its surface. Electromagnetic Induction / Faraday's Law / Lenz's Law. Consider two loops of wire, both with the same area, same number of loops, and the same orientation in space situated so that their axes line up. If we were to suddenly create a current in one loop, which direction would the induced current flow in the second? a) In the same direction b) In the opposite direction c) It wouldn't create an induced current d) The answer depends on something not given 2. Suppose we have a circular loop of wire of radius cm made with 00 loops of copper wire (M=56 g/mol). This loop is placed inside of a magnetic field B= (0 T / s) t ŷ. If the angle 2 between the plane of the loop and the x axis is 30 o, what is the induced emf in the loop at time t=0 seconds. a).36 mv b).57 mv c) 0.79 mv d) 500 mv e) 433 mv 3. We have a current of A (AC) go into a transformer and a current of 0A coming out. What is the ratio of turns a) /0 b) 0/ c) 5/ d) /5 e) / N in N out The voltage is directly correlated with the number of turns : V V 2 = N N 2 further, given that energy is conserved : V I =V 2 I 2 number of turns is /0. V V 2 = I 2 I = N N 2 thus the ratio of the 4 Page

5 EM Wave problems. Suppose professor frishman has a window m tall and 25 cm wide in his office. If the average intensity of the visible light measured was kw/m 2 outside the building, what is the amount of energy coming through the window each second? You can assume the sunlight is perpendicular to the window. a) 00 J/s b) 250 J/s c) 500 J/s d),000 J/s e) 25,000 J/s S =I, Intensity is the power per unit area I = P A and so the energy per unit second (power) is P=I A=(,000W /m 2 ) (m 0.25 m)=250 J / s 2. Assume we want to build a solar sail (that is propelled by the momentum from sunlight). If we build the sail from reflective material, how large of a sail would we need to accelerate 00 kg of mass at a rate of m/s 2? (The intensity of sunlight in our region of space is.5 kw/m 2 ) Choose the answer that seems the most correct. a) 0 m 2 b) 0 4 m 2 c) 0 7 m 2 d) 0 m 2 e) No amount of area would make this work. dp dt =F =ma, the pressure due from the radiation is P =2 S rad and pressure is c P= F A thus A= F P = m ac = (00kg ) ( ) m/s2 (3 0 8 m/ s ) m 2 (For 2 I 2( W /m 2 ) those who are curious this is 0 km 2 or ~ 2,500 acres or 3.86 mi A beam of light of intensity kw/m 2 is incident on a stationary electron. What is the largest force on the electron due to the electric component of the beam? a) 0 b).02 x 0-6 N c).39 x 0-6 N d) 3.62 x 0-6 N e) 5.3 x 0-6 N I = E 2μ 0 B 0 = 0 2μ 0 c E 2 0, Force on a charge is F =q E =e 2μ 0 c I =( C ) (,000 W /m 2 )2(4π 0 7 )(3 0 8 m/ s ) N 5 Page

6 4. A beam of light of intensity kw/m 2 is incident on a stationary electron. What is the timeaveraged force on the electron due to the electric component of the beam? a) 0 b).02 x 0-6 N c).39 x 0-6 N d) 3.62 x 0-6 N e) 5.3 x 0-6 N The electric field is negative half of the time and positive the other, the net force is zero over time. Optics Problems. If light enters the the atmosphere directly from the sun at a time of year that it would have an angle of zero incidence with the equator, what angle does the sunlight appear to come from (with respect to the horizon) in Iowa? (assume a latitude o ). The index of refraction of air is n air =.00 and the thickness of the atmosphere is negligibly small. a) o b) o c) o d) o e) o Use snell's law : n sin θ =n 2 sin θ 2 sin (40 o )=(.00 ) sin θ θ=39.95 o with respect to the normal ~ o with respect to the horizontal (Note that the height of the sun is apparently higher than it truly is) 2. Consider two lenses, one converging and one diverging. Our goal is to take parallel rays of light from a nearby laser and make the beam narrower by first sending the beam into a converging lens and subsequently through a diverging lens. If the focal length of the converging lens is 0 cm and the focal length of the diverging lens is 5 cm, how far apart should the lenses be placed? a) 5 cm b) 0 cm c) 5 cm d) 20 cm e) We cannot know The light rays must move from the converging lens to the diverging lens appearing as though they will pass through the focus of the diverging lens. (IE The focal points must lie at the same position) 6 Page