advertisement

UNIVERSITY OF MASSACHUSETTS DARTMOUTH COLLEGE OF ENGINEERING EGR 101 INTRODUCTION TO ENGINEERING I This assignment is due Thursday, October 11, 2012. Use engineering papers. Show all of your steps. Present your work neatly and clearly. Box your final answers. Problem 1 Resistors are rated not only in terms of resistance, but also in terms of their ability to dissipate power without burning. For example, if you use a resistor rated to dissipate 2 W in a circuit where the voltage and current conditions require that the wire dissipates 3 W then the wire will burn, and the circuit may be damaged. The ratings of resistors are given in section 2.3.5. on p. 56. Chapter 3: solve problem 15 on p. 93, and refer to section 2.3.5 on p. 56 in order to select the required power rating. I = V/R = 10V/10kΩ = 1 mA P = V*I = 10V*1mA = 10 mW = 0.010 W so 1/8W OK Problem 2 Chapter 3: problem 27 on p. 94. The range of possible current values is found by calculating the minimum and maximum V values of the current. If you consider Ohm’s law in the form of I , then you can R infer that for fixed voltage, as R increases, I decreases, and vice versa. Use the 4-band color code of the resistor to find its nominal value and tolerance. From this information find the minimum and maximum values of resistance and the corresponding currents. Red Red Red Gold = 2200 Ω 5% tolerance resistance range = 2090-2310 Ω Imax = 8V/Rmin = 8V/2090Ω = 3.83 mA Imin = 8V/Rmax = 8V/2310Ω = 3.46 mA Problem 4 Refer to Figure 3.2 (a) on p. 77. What is the % setting of the potentiometer if its resistance rating is 2.5k ? R = V/I = 10V / 100 mA = 100 Ω 100/1500x100 = 6.67% Problem 5 Chapter 4: problem 5 on p. 122. R3 = 53k -12k -33k = 8kΩ 8k/10k x 100 = 80% Problem 6 Chapter 4: problem 9 on p. 122. Rt = 51 + 39 + 10 = 100 Ω I = V/Rt = 1V / 100 Ω = 10 mA Problem 7 Chapter 4: problem 13 on p. 122. Rt = V/I = 12V / 560 µA = 21,429 Ω. 3629/5000 x 100 = 72.58% 21429 - 6800 -11000 = Rpot = 3629 Ω Problem 8 Chapter 4: problem 17 on p. 123. Rt = 51 + 39 + 20 = 110 Ω Vs = I*R = 50 mA * 110 Ω = 5.5 V