Pages from W 06 ex1 soln

# Pages from W 06 ex1 soln - EECS 314 Winter 2006 Exam 1 EECS...

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Instructor: Alexander Ganago Problem 6 (Suggested by Sing-Rong Li; modified) The circuit on this diagram includes a voltage source and a potentiometer; it is used as a sensor for the angular position (angle α ) of a robotic arm. At α = 0, the potentiometer’s tap is in its middle position so that R X = R 0 = R P /2 At an arbitrary, non-zero angle α (in degrees) the resistance R X equals R X = R P /2 + K α where K is constant measured in k Ω /degrees. Assume R P = 100 k Ω and determine the maximal value of K such that the angles measured with this sensor cover the range from – 40 (negative 40) to +40 degrees. The maximal value of K equals … A. 0.625 k Ω /degree B. 1.25 k Ω /degree C. 2. 5 k Ω /degree D. 5 k Ω /degree E. Depends on the source voltage, which is not given. Solution As the angle becomes negative and changes from 0 to -40 (negative 40) degrees, the resistance decreases from R X = R P /2 = 50 k Ω to R X = R P /2 - K 40. For the sensor to work properly, R X = R P /2 + K α should remain non-negative. Thus, at the maximal value of K, at the limit position of – 40 (negative 40) degrees the resistance should equal zero; thus 50 k Ω +(- 40 degrees) K (k Ω /degrees) = 0, or K = 1.25 k Ω /degree. At positive angle α that increases from 0 to +40 degrees, the resistance R X grows from 50 k Ω to R X = R P /2 + K 40; at the maximal value of K, at the limit position of +40 degrees R X should be 100 k Ω . Thus 50 k Ω +(- 40 degrees) K (k Ω /degrees) = 100 k Ω , or K = 1.25 k Ω /degree. Answer: B EECS 314 Old exam problems with solutions, for practice

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Problem 7 In the circuit shown on this diagram, the current source absorbs … A. – 36 W B. – 24 W C. – 12 W D. +12 W E. +36 W Solution To solve the problem, we need reference marks, as shown on this diagram: Note that the polarity of voltage V X across the current source is chosen in agreement with the passive sign convention, so that the current enters the positive terminal of the source. The voltage V R across the resistor equals, according to Ohm’s law, (3 Ω ) (2 A) = 6 V The V R polarity is such that the current enters the more positive terminal of the resistor. Also, note that the same 2 A current enters the positive terminal of the voltage source. KVL equation around the loop is: -12 V – V R V X = 0 or -12 V – 6 V – V X = 0 Thus V X = – 18 V According to the passive sign convention, the current source absorbs (2 A) V X = – 36 W To double-check the answer, verify the power balance: the voltage source absorbs (12 V) (2 A) = +24 W the resistor absorbs (3 Ω ) (2 A) (2 A) = +12 W Thus the total power in the circuit equals (– 36 W) +24 W +12 W = 0, as expected.
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Pages from W 06 ex1 soln - EECS 314 Winter 2006 Exam 1 EECS...

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