Physics 4 Homework lutions - Walker hapter 4 onceptual Exercises. The inductive reactance is given by ω π f At very high frequencies (i.e. as f frequencies well above onance) ( gets very large. ). This means that at high frequencies the inductor looks like an open circuit! The effective istance of the inductor becomes so large that it is as though the inductor were replaced by a break in the circuit. The reactance of the capacitor is given by ω π f At very high frequencies. This means that the capacitor looks like a short! The effective istance of the cap becomes so small that it is as though the cap were just replaced by a wire. n light of the above imagine replacing with an open circuit and with a short in the figure. Then no current at all would flow in the branch containing and the top. All the current would flow through the branch containing and the bottom. Furthermore with replaced by a short the entire circuit would look just like the single bottom connected across the voltage source. Therefore by Ohm s law the current that would flow would be ( ) high f in which repents the voltage supplied by the source. At very low frequencies (as f i.e. frequencies well below onance) and. This means that the inductor looks like a short and the cap looks like an open circuit. f you imagine replacing with a wire and with an open circuit (a break) there would be no current flowing in the branch containing. All the current would flow through the branch containing. And with replaced by a short the entire circuit would look like a single (the one in the top branch) connected across the A source. Therefore by Ohm s law the current that would flow would be ( ) low f omparing the two expsions for ( ) and ( ) high f low f we see that the same current would flow. 8. (a.) The onant frequency is given by f π which does not depend on. Therefore if the istance in the circuit were doubled nothing would happen to the onant frequency. The maximum current would decrease assuming that the voltage supplied by the source stays the same. The reason is as follows. First of all the amplitude of the current is given by the A form of Ohm s law: max + ( ) in which ve used the fact that the impedance of the is ( ) +.
Physics 4 Homework lutions - Walker hapter 4 When the circuit is driven at its onant frequency. This means that the denominator in the above expsion is smaller than it is at any other frequency. Therefore the amplitude of the current max is larger than it is at any other frequency. n particular at the onant frequency + and the amplitude of the current is ( max ) The ult of all of this is that the current is at its maximum amplitude at onance. Now if were doubled ( ) which means ( ) would become max ( ) max ( ) max would drop by a factor of two. Therefore the maximum current in the circuit would decrease. Notice that used the same max max. This means m assuming that the voltage supplied by the source stays the same as gets replaced by. But this doesn t always happen in reality. No real voltage source is capable of supplying literally the same voltage between its two terminals no matter what istance you stick across it. eal voltage sources always have some internal istance and this limits the current they can supply. When you connect very small load istances across the source you re asking the source to supply a very large current to the load to keep the voltage across the load fixed. But at some point because of the internal istance of the source the source simply isn t capable of supplying enough current to the load to keep the voltage across it the same as it was before. Thus when a very small load istance is connected across the voltage source we say that the voltage supplied by the source sags meaning that it falls below what it was for very large load istances. You will see this effect called loading effect in the lab. in both expsions for ( ) Problems. We re told that and max.8 A. But this max is the amplitude of the current not the current. From Ohm s law max But so max max.8 A 6 Ω 6. (a.) The average power dissipated by the istor is P so but lving for get av P av
Physics 4 Homework lutions - Walker hapter 4 ( ).9 Ω P 75 W av ( ).88 A.9 Ω max max (c.) The maximum power used by the bulb at any given time occurs when () t and () t are at their peak values max and max. At this time the power dissipated by the bulb is Pmax max ( )( ) but Pav 75 W so: P max 75 ( W) 5 W 5. (a.) From the A form of Ohm s law for the cap.5.6 Ω.6 kω -.4 A (c.) At f. khz π f π f 7.7 F.7 μf π. Hz.6 ( )( Ω) 58 Ω π. Hz.7 F 7 ( )( ).5 8.6 4 A.86 ma 58 Ω By a similar procedure for f. khz find 4. ma. 9. + + π f ( 5.5 Ω ) + 6 π ( 6. Hz)( 95. F) 7.8 Ω
Physics 4 Homework lutions - Walker hapter 4. (a.) 5 + π f (.5 ) Ω + 6 π ( 65. Hz)(.5 F).9 A.9 ma φ tan.6 k π f Ω.6 kω φ tan 6..5 kω This is the angle by which the voltage lags the current. o 6. (a.) The power factor is given by cosφ cosφ + + π f 4. cosφ.8 ( 4. ) + 6 π ( 5)(.5 ) f f is increased will get smaller which means will get smaller. Therefore the power factor cosφ will increase. Physically what this means is that the voltage and current get closer to being in phase which means more power is delivered to the load. 4. (a.) The impedance of an circuit is given by + ( ) ( ) 7. Ω + 5 Ω 7 Ω 4 4 A 7 Ω (c.) Pav cosφ in which the power factor cosφ is given by cosφ 4
Physics 4 Homework lutions - Walker hapter 4 7. Ω Pav ( 4 A)( 4 ).5 W.5 kw 7 Ω 4. The impedance of the circuit is given by + ( ) + π f π f ( ) ( )(.5 + π 6. 55 ) 6 π ( 6.)(.5 ).5 Ω.5 kω 48. (a.) f you connect the voltmeter between terminals A and B the voltmeter will measure the voltage across just the inductor. This is given by But the current is given by in which is the voltage supplied to the whole circuit by the voltage source and is the impedance of the entire circuit. Now and Therefore And finally π f π Hz. H 57 Ω 5 Ω π f π Hz. F 6 ( )( ) ( ) ( ) ( ) +.5 Ω + 57 Ω 5 Ω 4. Ω 6..4 A 4. Ω.4 A 57 Ω 79 f you put the voltmeter between Terminals B and you ll measure the voltage across just the cap. This is given by (.4 A)( 5 Ω ) 74 (c.) f you put the voltmeter between Terminals A and you ll measure the voltage across and. This voltage is given by in which means the impedance of the series combination. To get this impedance of and in series think about the impedance of the entire circuit: 5
Physics 4 Homework lutions - Walker hapter 4 ( ) + Now to remove the effect of the istor should set equal to zero. Doing so gives ( ) since >. ( ) (.4 A)( 57 Ω 5 Ω ) 4.9 (d.) f you put the voltmeter between Terminals A and D you ll measure the voltage across the entire circuit. This voltage is the voltage supplied by the source 6.. 54. At onance so This gives 67 Ω.8 A 55. (a.) The current is a maximum at onance. The frequency at which this happens is therefore the onant frequency: f 49.6 Hz π 6 π 57.6 H 79 F The impedance is a minimum at onance. Therefore the answer to this question is also 49.6 Hz. 57. (a.) They have the same reactance (i.e. ) at the onant frequency f. For a -μ F cap and a -m H inductor this occurs at f.5 Hz.5 khz π 6 π H F The onant frequency is the frequency at which the inductor and capacitor have the same reactance. it s the frequency just found.5 khz. 78. (a.) f π 6 And want this to equal 85 MHz which is 85 Hz. solving for get 4π f π ( ) ( ) 6 6 4 85 Hz.8 H 7.9 F 7.9 pf The impedance of the is ( ) + 6
Physics 4 Homework lutions - Walker hapter 4 emember that is a minimum at onance. Therefore anything you do to change the onant frequency makes larger. f is made larger this makes smaller because is inversely proportional to : π f When gets smaller and won t cancel at 85 MHz anymore. Therefore in the impedance ( ) + the term in parentheses won t be zero anymore. Because this term appears squared in the expsion for the impedance any deviation of away from zero makes larger. (c.) At onance the impedance is just equal to the istance: Therefore 5. Ω (d.) The capacitance found in Part (a.) makes at a frequency of 85 MHz. et us call this value of capacitance and the corponding value of the capacitive reactance ( ). Then for this value of capacitance ( ) π f and this equals at onance: ( ) ( ) Now if is made % higher than this means.. Because is inversely proportional to this means that the new value of will be: ( ) π f ( ) π f.. π f. Plugging this into the formula for the impedance get: + +. ( ) ( ) ( ) in which ve recognized that ( ) because the inductance and the frequency didn t change (only the capacitance). Simplifying this expsion a little get. ( ) ( ) + +.. But the capacitive reactance at onance is equal to ( ) π. [ π f] +. f so. 6 6 ( 5. Ω ) + π ( 85 Hz)(.8 H). 6 Ω 7