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 µ . ASSUMPTIONS: The magnetic field is oriented at an angle of 90 o to the current flow direction. SOLUTION: From (6.3) T NIAB µ α = sin The maximum torque occurs when sin α = 1, which yields () ( ) max 22 20 0.02 A 0.000645 m 0.4 Wb/m TN I A B = = 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 RR R R R R R R R R R R i TT m T T T m = −+ + +− + 11 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
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PROBLEM 6.3 KNOWN: 12 500 T RR R =+= 10,000 m R = 1 0.5 10 V Ti Rk R 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 oo = ′ − and in terms of a percentage of the output ′ − = E E E E E o o o 1 which yields, in terms of k () k k k k R R T m 1 1 11 + + The loading error for this condition is 0.0125 or 1.25%. As a percentage of full-scale output, i E , ′ − =− +− + E E E k k kk kR R i T m 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 =+Ω FIND: a) R 2 to balance the bridge with x = 0 b) Find a general expression for 2 () x fR = 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 .
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PROBLEM 6.5 KNOWN: A Wheatstone bridge with 2 1 20 R x = (x is a measured variable) 34 100 RR == 2 46 R =Ω 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 1234 500 RRRR ==== (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 = + δ 42 and with a load of magnitude F L R F L = 05 . E E F F o i L L = + 500 500 . / . This yields F L [N] δ R [ ] E o [V] 100 50 0.238 200 100 0.455 350 150 0.652
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The current flow through the sensor is I 1 and for an infinite meter resistance is I E RR i 1 12 = + which yields F L [N] R 1 + R 2 [ ] I 1 [Amps] 100 1050 0.00952 200 1100 0.00909 350 1150 0.00870 c) Consider the circuit shown below, with values of δ R equal to those for part (a) A circuit analysis of this bridge yields the following simultaneous equations governing the currents and potentials: () 11 2 2 11 33 44 22 0 0 is s m mm EI RI I R IR IR IR =+ + +− = + = 34 13 0 ms I II III +−= s om m R E I R =++ = Solving these equations simultaneously yields R 1 [
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This note was uploaded on 03/17/2008 for the course EML 3301C taught by Professor Chesney during the Spring '08 term at University of Florida.

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

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