PHY 402G Electronic Instrumentation and Measurement Test #1: September 19, 2007

Transcription

1 Name: PHY 40G Electronic nstrumentation and Measurement Test #: September 9, 007 Graduate students: nswer all questions. Write your work and calculation clearly. Total perfect score is 00 points. Undergraduate students: Option. nswer questions to 3. Your score will be multiply with a factor of 4/3. Total perfect score is 00 points. Option. nswer all questions. Write your work and calculation clearly. Total perfect score is 00 points. Your decision (undergraduate student only, circle one): Option / Option

2 . (Total: 5 points) Consider the following circuit: 00kΩ 0V 400kΩ a. (0 points) ccording to Thevenin s theorem, above circuit between points and is equivalent to the following circuit: Calculate the values of e and r in this equivalent circuit. e r e is given by the voltage across where there is no load across it. 400k e 0 400k + 00k On the other hand, if r R // R V e 8 r r we short and in the voltage divider circuit, the current in the wire f we short and with a wire, the current in the wire will be given by 0 4 will be 0 00k r 8 0 Ω or 80kΩ r lternatively, r is the output impedance of the divider circuit. (400k)(00k) 80kΩ (400k + 00k)

3 b. (5 points) ccording to Norton s theorem, above circuit between points and is equivalent to the following circuit: 0 r' Calculate the values of 0 and r in this equivalent circuit. f we short and with a wire, the current in the wire will or 0.m 00k The voltage across where there is no load is calculated in part (a) as 8V, but this also equals to r r' 8 in this Norton's equivalent circuit. 8 r' Ω or 80kΩ be given by 0. c. (5 points) f a voltmeter of input impedance MΩ is connected across points and, what will be the reading of the voltmeter? 0V 00kΩ voltmeter 400kΩ MΩ

4 400kΩ//MΩ Voltage across the voltmeter 0V 00kΩ + 400kΩ//MΩ (400k)(000k) 400kΩ//MΩ 85.7k (400k + 000k) 85.7k Voltage across the voltmeter 0V 0V 00k k lternatively, we can also use the equivalent circuit of part (a): e8 r80k k 7.4V MΩ voltmeter MΩ Voltage across the voltmeter 8V 80kΩ + MΩ 000 8V V 7.4V 080 You can also use the equivalent circuit of part (b) and get the same answer. Try it! d. (5 points) f a student accidentally connects an ammeter of input impedance 00Ω across points and (not the right way to use an ammeter), what will be the reading of the ammeter? 0V 00kΩ mmeter 400kΩ 00Ω

5 400kΩ//00Ω Voltage across the ammeter 0V 00kΩ + 400kΩ//MΩ (400k)(0.k) 400kΩ//00Ω 0.00k (400k + 0.k) 0.00k Voltage across the ammeter 0V 0.000V 00k k 0.000V -4 Current through the ammeter.0 0, or 0.m 00Ω lternatively, we can also use the equivalent circuit of part (a): e8 r80k 00 ammeter 8V Current through the ammeter 80kΩ + 00Ω 8 80.k , or 0.m You can also use the equivalent circuit of part (b) and get the same answer. Try it!

6 . (Total points: 5 points) Consider the following circuit: R300 V in C0 F V out The input voltage V in is a sinusoidal signal of frequency f00hz and peak to peak amplitude V pp 0V. a. (7 points) What is the amplitude (peak to peak) of current through the circuit? ic X R + X C R + ic C C p-p Vp-p R C R C π f π π 6 400π 0 0 C p-p , or 3.m 6 (400π ) (300) (0 0 ) b. (7 points) What is the amplitude (peak to peak) of output voltage V out across points and? ic From part (a), X R + X C R + ic ic Vout XC Vout ic π f π π V out p-p (400π ) (300) (0 0 6 ) 0 p-p R C 0.56V V in p-p

7 This answer makes sense because the circuit is a The signal frequency of 00Hz is way above f 3d low pass filer with 3d -6 RC rad/s 3d f3d 53. Hz π π 0, and Vp-p should be less than 7 ev. c. (7 points) What is the phase difference between the input voltage V in and output voltage V out? From part (b), Vout Vout V in R C m(v ) - out - φ tan tan R C tan Re(Vout ) RC R C π f π π φ tan - - tan RC 400π tan 6 400π or rad The output voltage lag the input voltage by 4.9 o d. (4 points) For voltage signal, this circuit is a (circle the right answer). Resonance circuit. ow pass filter C. High pass filter D. and pass filter E. Rectifier o -. RC 6

8 3. (5 points) Each part of this question is independent of each other. a. (5 points) n amplifier has an amplification of 4d, how may times it magnify the signal? ns: 3d 4d 8(3d) 4d ( 4d 6 ) The amplifier magnify the signal by 6 times. 8 b. (5 points) Fill in the words small or large in the blanks: n ammeter has a small input impedance. voltmeter has a large input impedance. voltage source has a current source has a small large output impedance. output impedance. c. (5 points) Calculate the impedance of a 0mH coil at frequency f60hz. Please give your answer in complete complex number form. Pay attention to sign and units. X πf π 60 0π X i i(0π )(0 0 3 ) i3.77ω ns:

9 d. (5 points) Write down the two definitions of Q-factor. Provide sketch if necessary. Definition : Energy stored in the system in one cycle Q π Energy dissipated in one cycle Defnintion : With respect to the figure at the right, 0 Q δ e. (5 points) Consider the following circuit: max max 0 R V in C V out Which of the following best describes the dependence of current on frequency (circle the right answer)?.. C. D. E.

10 4. (5 points) + attery R - student measured the voltage across the terminals of a battery (i.e., with R ) and obtained a reading of 9.00V. The student then connect a load resistance of R 00Ω across the terminal of the battery, as shown in the circuit diagram above. The voltage across and drops to V 6.00V. a. (6 points) What is the current from the battery? V Current from the battery R , or 6m b. (9 points) What is the internal resistance (or output impedance) of the battery? f the internal resistance of the battery is R 9 - R out R R R 0.06R out out out Ω 0.06 out ( R V 6V)

11 c. (0 points) The students repeated the measurement of and V for different values of R. Fill in the blank cells in the following table: R V 00Ω nswer of part (b) nswer of part (a) 50 Ω 4.5V Ω 0V Ω 8.00V Ω 3.0V 0.