# chap06 - PROBLEM 6.1 KNOWN A current loop having N = 20 A =...

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PROBLEM 6.1 KNOWN: A current loop having N = 20 A = 1 in 2 = 0.000645 m 2 I = 0.02 A B = 0.4 Wb/m 2 FIND: Torque on the current loop, T . ASSUMPTIONS: The magnetic field is oriented at an angle of 90 o to the current flow direction. SOLUTION: From (6.3) T NIAB sin The maximum torque occurs when sin = 1, which yields     max 22 20 0.02 A 0.000645 m 0.4 Wb/m T NIAB max 4 1.03 10 N-m T 

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PROBLEM 6.2 KNOWN: A voltage dividing circuit with R T = 5000 (as shown in Figure 6.16) FIND: R m such that the loading error is less than 12% of the full scale value. SOLUTION: Since the full-scale output is E i (with e as the error)   e E R R R R R R R R R R R R i T T m T T T m 1 1 1 2 1 1 1 1 A plot of the error as a function of meter resistance and 1 R is shown below Error, e/E i 0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0 0.2 0.4 0.6 0.8 1 R 1 /R T Error R m /R T = 1 R m /R T = 2 R m /R T = 4
PROBLEM 6.3 KNOWN: 12 500 T R 10,000 m  1 0.5 10 V Ti kR k E FIND: a) Loading error as a percentage of the output b) Loading error as a percentage of the full scale output SOLUTION: The loading error may be defined as e E E o o   and in terms of a percentage of the output   E E E E E o o o o o 1 which yields, in terms of k   k k k k R R T m 1 1 1 1 The loading error for this condition is 0.0125 or 1.25%. As a percentage of full-scale output, i ,     E E E k k k k kR R o o i T m 1 1 which produces a value for loading error of 0.62%. Since E o is one-half of E i , the loading errors are each 0.062 Volts.

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PROBLEM 6.4 KNOWN: R 3 = R 4 = 200 R 2 = variable resistor 1 40 100 Rx FIND: a) R 2 to balance the bridge with x = 0 b) Find a general expression for 2 () x f R ASSUMPTIONS: Zero Galvanometer error SOLUTION: At balanced conditions R R R R 2 1 4 3 Thus    22 1 100 and 100 RR and in general 2 40 100 .
PROBLEM 6.5 KNOWN: A Wheatstone bridge with 2 1 20 Rx (x is a measured variable) 34 100 RR 2 46 R  at balanced conditions FIND: x SOLUTION: Since at balanced conditions R R R R 2 1 4 3 1 Thus 46 20 1 46 20 152 2 x x .

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PROBLEM 6.6 KNOWN: A sensor has a resistance of 500 under conditions of no load, and a static sensitivity of 0.5 /N. The bridge circuit has 1 2 3 4 500 R R R R (initially) FIND: a ) E o for applied loads of 100, 200, and 350 N. b) I 1 c) Repeat parts (a) and (b) with 10 k 500 m s R R  SOLUTION: For a bridge in which all resistances are initally equal   E E R R R R o i 4 2 and with a load of magnitude F L R F L 05 . E E F F o i L L 500 4 2 500 . / .
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chap06 - PROBLEM 6.1 KNOWN A current loop having N = 20 A =...

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