# 2.21 (2).pdf

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33-228 Electronics Carnegie Mellon University Spring 2014 Department of Physics Homework 3 SOLUTIONS Problems 2.17, 2.21, 2.27, 2.28, 2.29 and 2.D42. 2.17 For a capacitor and an inductor in series, the equivalent impedance is just Zeq = ZC + ZL . Expanding this , we have that 1 Zeq = + jωL jωC 1 − ω 2 LC Zeq = jωC 1 − ω 2 LC Zeq = (−j) ωC 1 . The magnitude of this impedance is zero when ω = √LC 2.21 For three components in parallel, the impedances add as 1 1 1 1 = + + . Zeq R ZC ZL 1 1 1 +
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33-228 Electronics Spring 2014 Carnegie Mellon University Department of Physics Homework 3 SOLUTIONS Problems 2.17, 2.21, 2.27, 2.28, 2.29 and 2.D42. 2.17  For a capacitor and an inductor in series, the equivalent impedance is just  Z  eq  = Z  C   +  Z  L . Expanding this , we have that Z  eq  = 1  jωC   +  jωLZ  eq  = 1 − ω 2 LC  jωC Z  eq  = ( −  j )1 − ω 2 LC ωC  The magnitude of this impedance is zero when  ω  =  1 √  LC  . 2.21  For three components in parallel, the impedances add as1 Z  eq = 1 R  + 1 Z  C  + 1 Z  L . 1 Z  eq = 1 R  +  jωC   + 1  jωL 1 Z  eq = 1 R  + 1 − ω 2 LC  jωL 1 Z  eq =  jωL  +  R (1 − ω 2 LC  )  jωLR Inverting, we get that Z  eq  =  jωLRR (1 − ω 2 LC  ) +  jωLZ  eq  = (  jωLR )( R (1 − ω 2 LC  ) −  jωL ) R 2 (1 − ω 2 LC  ) 2 +  ω 2 L 2 Z  eq  =  ω 2 L 2 RR 2 (1 − ω 2 LC  ) 2 +  ω 2 L 2 +  j ωLR 2 (1 − ω 2 LC  ) R 2 (1 − ω 2 LC  ) 2 +  ω 2 L 2 As we care looking for the case when  Z  eq  is real, we only need to ﬁnd the case whenthe imaginary part of   1 Z  eq is zero. Thus, we have1 Z  eq = 1 R  +  jωC   + 1  jωL 1 Z  eq = 1 R  + 1 − ω 2 LC  jωL 1  The imaginary part is zero when  ω 2 − LC   = 0, or  ω  =  1 √  LC  . At this frequency, themagnitude of the impedance is just  Z   =  R . 2.27  Assuming a linear fall-oﬀ of attenuation versus frequency on a log-log plot, we wantto show that a 20 dB  change over a factor of 10 in frequency corresponds to a 6 dB change over a factor of 2 in frequency. This corrsponds to20 dB log 10 (10) =  xdB log 10 (2)20 dB 1 =  xdB (0 . 301) x  = 6 dB 2.28  If we have an attenuation of 60 dB  per decade, then the power is given as:  p  =  − 60 dB 20 dB p  =  − 3which means that we have | G | ∼  f  − 3 2.29  If we have a gain that goes as  G ( ω ) =  A/ √  ω , the constant  A  must have units of   s − 12 .We also see that | G | ∼  f  − 12 so  p  = − 12  and the attenuation per deacde is then − 12 =  − A 20 dBA  = 10 dB/decade 2.D42  The basic idea would be to gave the capacitors rapidly discharge through the hotdogs,thus cooking them. The energy stored in a fully charged capacitor would be E   = 12 CV   2 E   = 12(0 . 5 F  )(35 V   ) 2 E   = 306 J  In order to cook the hot dog, we need to raise its temperature to about 65 C from arefrigerated temperature of about 10 C, or ∆ T   = 55 C. The energy needed to do this2  is E   =  m · c · ∆ T E   = (50 g ) · (4 . 2 J/ ( gC  )) · (55 C  ) E   = 11 . 5 kJ . Thus, we need the energy stored in 38 capacitors to be able to cook the hot dog. Todo this, we could put the 38 capacitors in parallel, to yield a total capacitance of 17 F  .We now need to discharge this (quickly) through the hotdog. The discharge time isgiven by tha characteristic time of the circuit, or  τ  RC   =  RC  . This time is 1700 s orabout half and hour. This would seem far too long to cook the hot dog.In order to cook the hot dog, we need the charateristic time to be on the order of 50s, which for a ﬁxed resistance of 100Ω, yields a capacitance of about 0 . 5 F. To store11 . 5 kJ in this capacitor, we would need the voltage across the capacitor to be V   =   2 · E/C V   = √  46000 V V   = 214 V . To hold 214 V with 35 V capacitors, we need to place about seven of them in series.Unfortunately, the equivalent capacitance of these in series is  17  of the individual ca-pacitances. We can recover this by building the circuit shown in Figure 1. Thus, inorder to be able to do this, your friend’s uncle will need 49 of the capacitors and a DCvoltage source that can provide 200 V as well as switching capability for 200 V DC.                        0 . 5 F        0 . 5 F        0 . 5 F        0 . 5 F        0 . 5 F        0 . 5 F        0 . 5 F                             0 . 5 F        0 . 5 F        0 . 5 F        0 . 5 F        0 . 5 F        0 . 5 F        0 . 5 F        0 . 5 F        0 . 5 F        0 . 5 F        0 . 5 F        0 . 5 F        0 . 5 F        0 . 5 F        0 . 5 F        0 . 5 F        0 . 5 F        0 . 5 F        0 . 5 F        0 . 5 F        0 . 5 F                         0 . 5 F        0 . 5 F        0 . 5 F        0 . 5 F        0 . 5 F        0 . 5 F        0 . 5 F                         0 . 5 F        0 . 5 F        0 . 5 F        0 . 5 F        0 . 5 F        0 . 5 F        0 . 5 F        0 . 5 F        0 . 5 F        0 . 5 F        0 . 5 F        0 . 5 F        0 . 5 F        0 . 5 F    Figure 1:  The capacitor array for problem D2.31 3

Jul 22, 2017

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