Physics Solution Manual for 1100 and 2101

# Therefore these equations simplify to the following 2

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Unformatted text preview: uation 23.7. Since the voltage and the current are known, we can obtain the impedance from Equation 23.6 as Z = Vrms/Irms. SOLUTION From Equation 23.7, we can determine the resistance as R = With this expression for the resistance, Equation 20.15b for the power becomes 2 Z2 − XL . 2 2 2 P = I rms R = I rms Z 2 − X L Using Equation 23.6 to express the impedance, we obtain the following value for the dissipated power V F I −X GJ I HK 2 P=I 2 rms Z −X 2 2 L =I rms 2 rms 2 L rms b = 1.75 A 115 g F.75 V I − b .0 Ω g = G A J 52 HK 1 2 2 2 123 W 23. REASONING AND SOLUTION In order to find the voltage and phase angle we need to know the impedance of the circuit. We know that XC = 1 1 = = 133 Ω 3 2 π f C 2 π 4 .80 × 10 Hz 0.250 × 10 –6 F c h c The impedance of the circuit can be found using h 1256 ALTERNATING CURRENT CIRCUITS Z= c R2 + X L − XC 232 0 h = b Ω g+ b− 133 Ω g = 267 Ω 2 2 2 a. The voltage of the generator is V = IZ = (0.0400 A)(267 Ω) = 10.7 V b. The phase angle is φ = tan −1 X F G H L − XC R 0 I = tan F − 133 Ω I = J G232 Ω J KH K −1 −29 .8 ° 24. REASONING The rms current Irms in the circuit in part c of the drawing is equal to the V rms voltage Vrms divided by the impedance Z or I rms = rms (Equation 23.6). The impedance Z Z of a series RC circuit is given by Equation 23.7 with XL = 0 Ω, since there is no 2 inductance in the circuit; Z = R 2 + X C . The resistance R is known and the capacitive reactance XC can be obtained from the relation XC = 1/(2π f C) (Equation 23.2), where C is the capacitance and f is the frequency. The capacitance, however, is related to the time constant τ of the circuit in part a of the drawing. The time constant of an RC circuit is the time for the capacitor to lose 63.2% of its initial charge (see the discussion in Section 20.13), and it is equal to the product of the resistance R and the capacitance C; τ = RC (Equation 20.21). 2 SOLUTION Substituting Z = R 2 + X C into I rms = I rms = Vrms = Z Vrms gives Z Vrms (1) 2 R2 + X C Substituting XC = 1/(2π f C) (Equation 23.2) into Equation (1) allows us to write the rms current in the circuit in part c of the drawing as follows: I rms = Vrms 2 R2 + X C = Vrms 1 R2 + 2π f C 2 Substituting C = τ /R (Equation 20.21) into Equation (1), we arrive at an expression for the rms current: Chapter 23 Problems I rms = = Vrms 1 R2 + 2π f C Vrms = 2 1257 R R2 + 2π f τ 2 24 V 18 Ω (18 Ω )2 + 2π ( 380 Hz )( 3.0 × 10−4 s ) 2 = 0.78 A 25. SSM WWW REASONING We can use the equations for a series RCL circuit to solve this problem, provided that we set X L = 0 since there is no inductance in the circuit. Thus, 2 according to Equations 23.6 and 23.7, the current in the circuit is I rms = V rms / R 2 + X C . When the frequency f is very large, the capacitive reactance is zero, or X C = 0 , in which case the current becomes I rms ( large f ) = V rms / R . When the current Irms in the circuit is one-half the value of I rms ( large f ) that exists when the frequency is very large, we have I rms I rms (large f ) = 1 2 We can use these expressions to write the ratio above in terms of the resistance and the capacitive reactance. Once the capacitive reactance is known, the frequency can be determined. SOLUTION The ratio of the currents is I rms I rms ( large f ) = 2 V rms / R 2 + X C V rms / R R = 2 R2 + X C = 1 2 or Taking the reciprocal of this result gives 2 R2 + XC R2 =4 1+ or 2 XC R2 =4 Therefore, XC R b = g 3 According to Equation 23.2, X C = 1 / 2 π f C , so it follows that R2 1 = 2 2 4 R + XC 1258 ALTERNATING CURRENT CIRCUITS XC = R b 1 / 2π f C R g= 3 Thus, f= 1 2 π RC 3 = bg c 1 2 π 85 Ω 4.0 × 10 –6 F h3 = 270 Hz 26. REASONING To find the frequency at which the current is one-half its value at zero frequency, we first determine the value of the current when f = 0 Hz. We note that at zero frequency the reactive inductance is zero (XL = 0 Ω), since XL = 2π f L (Equation 23.4). The 0 current at zero frequency is I rms = Vrms/R (Equation 20.14), since the inductor does not play a role in determining the current at this frequency. When the frequency is not zero, the current is given by I rms = Vrms/Z (Equation 23.6), where Z is the impedance of the circuit. These last two relations will allow us to find the frequency at which the current is one-half its value at zero frequency. 0 SOLUTION We are given that I rms = 1 I rms , so 2 1 V = rms Z 2 R Vrms or Z = 2R 2 Since the impedance of the circuit is Z = R 2 + X L (Equation 23.7) and XL = 2π f L (Equation 23.4), the relation Z = 2R becomes R 2 + ( 2π f L ) = 2 R 2 Solving for the frequency gives f= 3 (16 Ω ) 3R = = 1.1×103 Hz 2π L 2π 4.0 ×10−3 H ( ) 27. REASONING The instantaneous value of the generator voltage is V(t) = V0 sin 2π ft, where V0 is the peak voltage and f is the frequency. We will see that the inductive reactance is greater than the capacitive reactance, XL > XC, so that the current in the circuit lags the...
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## This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

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