Physics 9 Fall 2009 Homework 9 - Solutions

Transcription

1 . Chapter 36 - Exercise 5. Physics 9 Fall 2009 Homework 9 - s What are V R and V C if the emf frequency in the figure is 0 khz? The voltage are V R = IR, and V C = IX C, where the capitative reactance is X C = E /ωc = /2πfC. The peak current is I = 0. So, the peak current is I = R2 + X 2 C = R 2 +X 2 C (2π (0 4 ) ) 2 = 0.04 A. So, we can now find V R = IR = = 6.0 V V C = IX C = π(0 4 )( ) = 8.0 V

2 2. Chapter 36 - Exercise 27. A series RLC circuit consists of a 50 Ω resistor, a 3.3 mh inductor, and a 480 nf capacitor. It is connected to an oscillator with a peak voltage of 5.0 V. Determine the impedance, the peak current, and the phase angle at frequencies (a) 3000 Hz, (b) 4000 Hz, and (c) 5000 Hz. The impedance is Z = ( R 2 + (X C X L ) 2 = R 2 + 2πfL ) 2, 2πfC while the peak current is I = /Z, and the phase angle is φ = tan [ X L X C ] [ ] R = tan 2πfL /2πfC. So, all we have to do is to plug in each of values for the different R frequencies. Doing so, we get the following table 3000 Hz 4000 Hz 5000 Hz Z 70 Ω 50 Ω 62 Ω I A 0.00 A A φ

3 3. Chapter 36 - Exercise 3. The motor of an electric drill draws a 3.5 A current at the power-line voltage of 20 V rms. What is the motor s power if the current lags the voltage by 20? The average power supplied by the motor is P source = I rms V rms cos φ, where I rms is the current drawn by the motor (3.5 A, in this case). So, we just plug everything in. P source = I rms V rms cos φ = (3.5) (20) cos 20 = 400 W. 3

4 4. Chapter 36 - Problem 37. (a) Evaluate V C in the figure at emf frequencies, 3, 0, 30, and 00 khz. (b) Graph V C versus frequency. Draw a smooth curve through your five points. (a) The voltage across the capacitor is V C = IX C, where I = =. So, ωc 2πfC, and X C = R 2 +XC 2 V C = IX C = = 2πfC R 2 + (2πfC) 2 + (2πfRC) 2. So, plugging in the values gives f (khz) V C (V) (b) The graph is seen to the right. Notice that the voltage approaches zero as the frequency increases. The circuit acts like a low-pass filter, blocking out higher frequencies. 4

5 5. Chapter 36 - Problem 38. (a) Evaluate V R in the figure at emf frequencies 00, 300, 000, 3000, and 0,000 Hz. (b) Graph V R versus frequency. Draw a smooth curve through your five points. (a) The voltage across the resistor is V R = IR, where I = 2πfC. So, V R = IR = So, plugging in the values gives R 2 + (2πfC) 2. f (Hz) V R (V) , , and X C = = R 2 +XC 2 ωc (b) The graph is seen to the right. Notice that the voltage approaches the full 0 V as the frequency increases, while for low frequencies the voltage drops. The circuit acts like a high-pass filter, blocking out lower frequencies. 5

6 6. Chapter 36 - Problem 42. (a) What is the peak current supplied by the emf in the figure? (b) What is the peak voltage across the 3.0 µf capacitor? (a) The circuit can be redrawn in terms of a single equivalent capacitor. The two parallel capacitors combine together to give an equivalent capacitance of 6 µf, which then combines with the 3 µf capacitor for a net capacitance of 2 µf. Then, the current is I = V X C = = 2πfCE /2πfC 0. Thus, the peak current is I = 2πfC = 2π (200) (2 0 9 ) (0) = 25 ma. (b) All the current passes though the 3 µf capacitor, and so the peak voltage is V C = IX C = I. Notice that since I = 2πfC 2πfC eqv, then V C = I = 2πfCeqv E 2πfC 2πfC 0 =. So, we find C eqv C V C = C eqv C = 2 0 =.67 0 = 6.7 V. 3 6

7 7. Chapter 36 - Problem 43. You have a resistor and a capacitor of unknown values. First, you charge the capacitor and discharge it through the resistor. By monitoring the capacitor voltage on an oscilloscope, you see that the voltage decays to half its initial value in 2.5 ms. You then use the resistor and capacitor to make a low-pass filter. What is the crossover frequency f c? The crossover frequency is f c = ωc =. The voltage across a capacitor falls off 2π 2πRC exponentially fast as V = V 0 e t/rc. If the voltage falls to half its initial value in a time t 0, then V (t 0 ) = V 0 2 = V 0e t 0/RC 2 = e t 0/RC. Taking natural logs of both sides and solving gives RC = t 0. Then, since f ln 2 c = we have f c = ln 2. 2πt 0 With numbers, f c = ln 2 2πt 0 = π ( ) 44 Hz. 2πRC, 7

8 8. Chapter 36 - Problem 5. For the circuit in the figure, (a) What is the resonance frequency, in both rad/s and Hz? (b) Find V R and V C at resonance. (c) How can V C be larger than? Explain. (a) The resonance frequency is ω R = LC, and so f R = ωc 2π = 2π LC. Thus, ω c = LC = = rad/s. f c = 2π = LC 2π = Hz. (b) At resonance, X C = X L, and so Z = R, giving I = /R, or amp. This gives V R = IR = 0 = 0 V. Furthermore, V C = IX C = I/ω R C. We found the resonance frequency in part (a), and so substituting in gives V C = = V. (c) We know that, for the instantaneous values, v R + v C + v L = E, but this isn t true for the peak values. At resonance, v L and v C cancel each other out, but can each be big, if they are compensated by an equally big v R. This lets the peak values be larger than the peak current, due to the cancellation. 8

9 9. Chapter 36 - Problem 57. Show that the impedance of a series RLC circuit can be written Z = R 2 + ω 2 L 2 ( ω0/ω 2 2 ) 2. The impedance Z = R 2 + (X L X C ) 2, where X L ωl, and X C =. Now, ωc (X L X C ) 2 = ( ) ωl 2 ωc = ω 2 L ( ) 2. 2 ω 2 LC But, LC =, where ω ω0 2 0 is the natural resonant frequency. Thus, (X L X C ) 2 = ω 2 L ( 2 ω2 0 ). So, we find that ω 2 Z = ( ) R 2 + ω 2 L 2 ω2 0. ω 2 9

10 0. Chapter 36 - Problem 68. (a) Show that the average power loss in a series RLC circuit is P avg = ω 2 E 2 rmsr ω 2 R 2 + L 2 (ω 2 ω 2 0) 2. (b) Prove that the energy dissipation is a maximum at ω = ω 0. (a) The average power dissipated is P avg = I rms E rms cos φ, where I rms = E rms /Z, and cos φ = R/Z. Thus, P avg = E 2 rmsr/z 2. Now, since /LC = ω 2 0. So, Z 2 = R 2 + (X L X C ) 2 ) 2 = R 2 + ( ωl ( ωc ) = R 2 + L2 ω ω LC = R 2 + L2 (ω 2 ω ω 0) 2 2, 2 P avg = ErmsR 2 rmsrω 2 2 R 2 + L2 (ω ω 2 ω0) 2 2 = E R 2 ω 2 + L 2 (ω 2 ω0) (b) The power dissipated depends on the frequency, ω. It is a maximum when d P dω avg = 0. So, taking the derivative gives ( 2ErmsRω 2 R 2 ω 2 + L 2 (ω 2 ω0) 2 2) ErmsRω 2 2 (2ωR 2 + 2L 2 (ω 2 ω0) 2 2ω) (R 2 ω 2 + L 2 (ω 2 ω 20) 2) 2 = 0. This reduces to (ω 2 ω 2 0) 2 2ω 2 (ω 2 ω 2 0) = 0, which is solved by ω = ω 0. So, the power is maximized when ω = ω 0, i.e., when the system is at resonance! 0