Physics 234 Homework Chapter 31 Q8, Q11, P11, P32, P36, P50 Q8) The two equations that characterize the voltage across the generator and the current in the circuit are, m sin d t and i I sin d t . A possible phasor diagram for the RLC circuit is shown below for the case when VL > VC. VL VL VC m I VR VC For this particular case, the phase constant, , is positive. For cases where VC > VL, the phase constant would be negative. a) For cases when the load is primarily capacitive implies that VC > VL, the case where the phase constant is negative. Cases (1) and (4) would apply in this case. b) The phasor diagram shown above is consistent with a situation where the current lags the voltage across the generator (the alternating emf, m). The phasor diagram is also the case corresponding to a positive phase constant, and would therefore apply to cases (2) and (3). Q11) The situation in Figure 31-25, shows the current slightly leading a voltage across the generator. It therefore represents a negative phase constant, , and can be represented by the phasor diagram, below. VL I VC VL VR m VC In this case, the voltage across the capacitor is slightly larger than the voltage across the inductor. a) If one slightly increases the inductance of the circuit, the difference, VC – VL, will become smaller (since VL = dL) and the phase constant will also become smaller, which is apparent from examining the phasor diagram. Another way to look at this is, VC VL I ( X C X L ) I ( 1 d L ). d C As dL approaches 1/dC, VC – VL eventually becomes zero at resonance. Therefore, a slight increase in inductance will shift the current and voltage, , curves closer to one another. In other words, it will shift the current curve rightward. The amplitude of the current in the circuit can be determined by the equation, I m Z m R (XC X L) 2 2 m 1 R2 ( d L) 2 d C . Therefore, a slight increase in L will decrease the difference, XC – XL, decreasing the denominator in the equation above, and increasing the current amplitude, I. b) Increasing C slightly decreases XC, since XC = 1/dC. Similar to the situation in part a), this will decrease the difference, VC – VL. The same reasoning as in part a) will still apply so that the current curve will shift rightward and the current amplitude will increase. c) Increasing d increases XL = d L and decreases XC = 1/dC. This brings XL and XC even closer together than in parts a) and b) and decreases the difference, VC – VL, even more so than in parts a and b. The effects are the same qualitatively so that the current curve will shift rightward and the current amplitude will increase. P11) (a) Since the frequency of oscillation f is related to the inductance L and capacitance C by f 1 / 2 LC , the smaller value of C gives the larger value of f. Consequently, f max 1/ 2 LCmin , f min 1/ 2 LCmax , and Cmax 365 pF f max 6.0. f min Cmin 10 pF (b) An additional capacitance C is chosen so the ratio of the frequencies is r 160 . MHz 2.96. 0.54 MHz Since the additional capacitor is in parallel with the tuning capacitor, its capacitance adds to that of the tuning capacitor. If C is in picofarads (pF), then C 365 pF C 10 pF 2.96. The solution for C is, C 365 pF 2.96 2 C 10 pF C (1 2.96 2 ) 2.96 2 10 pF 365 pF 87.616 pF 365 pF 35.7 pF ~ 36 pF . (1 2.96 2 ) C (c) We solve f 1 / 2 LC for L. For the minimum frequency, C = 365 pF + 36 pF = 401 pF and f = 0.54 MHz. Thus L 1 2 2 Cf 2 2 2 40110 1 12 F 0.54 106 Hz 2 2.2 104 H. P32) (a) The circuit consists of one generator across one inductor; therefore, m = VL. The current amplitude is I m XL m 25.0 V 5.22 103 A . d L (377 rad/s)(12.7 H) (b) When the current is at a maximum, its derivative is zero. Thus, Eq. 30-35 gives L = 0 at that instant. Stated another way, since (t) and i(t) have a 90° phase difference, then (t) must be zero when i(t) = I. The fact that = 90° = /2 rad is used in part (c). (c) Consider Eq. 31-28 with m / 2 . In order to satisfy this equation, we require sin(dt) = –1/2. This occurs when the value of dt is either 210º or at 330 º. Now we note that the problem states that is increasing in magnitude, which (since it is already negative) means that it is becoming more negative. The sin function is becoming more negative at 210º (with increasing angle) and this is, therefore, the correct value of dt. Solving for i, we have, i I sin d t 5.22 mAsin 210 90 5.22mAsin 120 4.52 mA. P36) (a) The circuit has a resistor and a capacitor (but no inductor). Since the capacitive reactance decreases with frequency, then the asymptotic value of Z must be the resistance: R = 500 . (b) We describe three methods here (each using information from different points on the graph): method 1: At d = 50 rad/s, we have Z 700 , which gives C = (d Z2 - R2 )1 = 41 F. method 2: At d = 50 rad/s, we have XC 500 , which gives C = (d XC)1 = 40 F. method 3: At d = 250 rad/s, we have XC 100 , which gives C = (d XC)1 = 40 F. P50) (a) The phasor diagram is shown below. VL VR (b) We have VR = VL, which implies I R = I XL → R = d L which yields f = d/2 = R/2L = 318 Hz. (c) = tan1(VL /VR) = +45. (d) d = R/L = 2.00×103 rad/s. (e) I = (6 V)/ R2 + XL2 = 3/(40 2) 53.0 mA.