Physics 1c practical, 2015 Homework 5 Solutions
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1 Physics c practical, 05 Homework 5 Solutions Serway mh, = 0 7 F, = 0Ω, V max = 00V a The resonant frequency for a series L circuit is f = π L = 3.56 khz At resonance I max = V max = 5 A c Q = ω0l =.4 d V L,max = X L I max = ω 0 LI max =.4 kv Serway V out ) max = N V in ) max = = 97V N 350 V out ) rms = 97 = 687V Serway points) The circuit is simply a voltage divider. V out = Z = V in Z L + ωl /ω) Setting this equal to /, squaring oth sides, and rearranging terms: ωl /ω) = 3
2 Homework 5 Solutions 05 ωl /ω = ± 3 πfl /πf = ± 3 This equation must hold for f = 00 Hz and f = 4 khz. πf L /πf = s 3 πf L /πf = s 3 Where s = + or to replace the ±; the sign will e determined elow. A simple change of variales A = / shows that this is a pair of linear equations in L and A. Solving I used ramer s ule): ) 3 s /f s /f ; A = ) π f /f f /f = s f s f 3π f /f f /f Unfortunately, the signs of s and s are amiguous, ut if we restrict ourselves to positive values of L and negative values are meaningless), then the only possile values are s = and s = +. ) 3 /f + /f ; π f /f f /f = 3π f + f f /f f /f ) Numerically, ) 38 Ω) /00 Hz) + /4 khz) = 580 µh. π 4 khz)/00 Hz) 00 Hz)/4 khz) ) 4 khz)/00 Hz) 00 Hz)/4 khz) = = 54.6 µf. 3π8 Ω) 4 khz) + 00 Hz) 0 points) is given in part a). Points for part ) are included in a). 0.5 point) eferring to the expression at the eginning of part a, V out /V in will e maximum when the denominator is minimum. This will occur when ω 0 L /ω 0 = 0. In that case, V out /V in = 0.5 point) Solving for ω 0 in part c, [ f 0 = π π ω 0 = / L; 3 f /f f /f /f + /f f 0 = ω 0 /π = /π L )] [ 3π f 0 = f f = 00 Hz)4 khz) = 894 Hz. f + f f /f f /f The resonance frequency is the geometric mean of the two half-amplitude frequencies. The circuit actually has the more general property that the resonant frequency is the geometric mean of any two frequencies with equal voltage amplitudes. )]
3 Homework 5 Solutions 05 3 point) The phase for an aritrary frequency f is given y: ) ϕf) = tan XL f) X f) [ ] 3f ϕf) = tan /f + /f 3 f /f f /f f ϕf) = tan 3 f f ) πfl /πf = tan [ [ f f f f f + f f /f f /f ] ) For the values we are interested in: ϕf ) = tan ) 3 = π/3 radians,.05 radians, or 60. ϕf ) = tan + ) 3 = +π/3 radians, +.05 radians, or From part c, f 0 = f f, giving [ 3 f ϕf 0 ) = tan f f ] ) f = tan 0 = 0. f f f f ]) 0.5 point) The average power across the resistor for voltages given as amplitude) is P = V outf)/ By the statement of the prolem, V out f ) = V out f ) = V in /, so P f ) = P f ) = 8 V in = 8 By parts and c, at resonance V out f 0 ) = V in, giving 0.0 V) 8.00 Ω =.56 W. P f 0 ) = V in = 0.0 V) 8.00 Ω = 6.5 W. 0.5 point) From the discussion in the text: Q = ω 0L = πf 0L π894 Hz)580 µh) = = Ω This is a very poor Q ecause the pass and is extremely wide relative to the resonance frequency. QP9 MS quantities elow are denoted y a tilde, for example: Ṽ.
4 Homework 5 Solutions points) Because of Kirchhoff s current rule, the current through this circuit will e the same for each part of the loop. Looking particularly at the capacitor: points) The total impedance of the circuit is Ĩ = ṼY Z X = ṼY Z /ω = πf) ṼY Z Ĩ = π).00 khz).00 µf)5.5 V) = 97.4 ma. Z = + ωl /ω) Ohms Law reads: Solving for L: Ṽ XZ = ĨZ = Ĩ + ωl /ω) ω ± ) ṼXZ ω Ĩ ) π).00 khz).00 µf ± 0. V 35.0 Ω) π).00 khz 97.4 ma 40.8 mh or 9.8 mh It may help to draw a phasor diagram to see where the two possile values of L come from. The magnitude of V XZ and the component along V are fixed. The value of L is chosen so that the phasors V + V + V V XZ. One possile diagram follows elow. The diagram is not necessary for full credit.).5 points) The MS voltage across XY will e For 40.8 mh: For 40.8 mh: Ṽ XY = ĨZ = Ĩ + ω L Ṽ XY = 97.4 ma) [35.0 Ω] + [π).00 khz)] [40.8 mh] = 5. V. Ṽ XY = 97.4 ma) [35.0 Ω] + [π).00 khz)] [9.8 mh] = 6.9 V.
5 Homework 5 Solutions 05 5 QP3 Since urns 00 Watts at 0 V and P = I, to reduce P to 5 Watts, the current I has to e reduced y /. Therefore, total = and adj =. Now that the two resistors have the same resistance, P total = P = 50 Watts. The resistance of is found from 00 = V MS = 44 Ω. In the case when L adj is used, we can look at P = V /. This implies the voltage across the ul V has to e reduced y /. V = V A + ωl adj ) / = ωl adj / ) = 3 L adj 0.66 H + ωl adj ) Since there is no MS power in the adjustale L adj, the total power is 5 Watts. QP8 a A capacitor resists sudden voltage changes. So at t = 0, V = 0. at t there will e no current across the capacitor. So we can effectively remove capacitor from the original circuit and calculate V. We get V = + V B
6 Homework 5 Solutions 05 6 c d V = Q = I dt ) 0 I + I = I ) Kirchoff s law for attery,, loop, and apacitor, loop are: e V B I I = 0 3) 0 I c dt I = 0 4) solving ) and 3) for I we get I = V B I +. Sustituting this value in 4) we get Now using ) we get f time constant = d dt V + I dt V B I c = I dt V B + I = 0 5) ) V = V B 6)
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