Physics Solution Manual for 1100 and 2101

Reasoning and solution according to equation 168 we

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Unformatted text preview: gle φ are given by Z = R2 + ( X L − X C ) 2 (23.7) tan φ = XL − XC R (23.8) where R is the resistance, XL is the inductive reactance, and XC is the capacitive reactance. In the present case, there is no inductance, so X L = 0 Ω . Therefore, these equations simplify to the following: 2 Z = R2 + X C (1) tan φ = − XC Values are given for Z and φ, so we can solve for R and XC. R (2) 1274 ALTERNATING CURRENT CIRCUITS SOLUTION Using the value given for the phase angle in Equation (2), we find that XC tan φ = tan ( −9.80° ) = −0.173 = − R X C = 0.173R or (3) Substituting this result for XC into Equation (1) gives 2 2 2 Z = R 2 + X C = R 2 + ( 0.173 R ) = 1 + ( 0.173) R Solving for R reveals that R= Z 1 + ( 0.173) 2 = 4.50 ×102 Ω 1 + ( 0.173) 2 = 443 Ω Using this value for R in Equation (3), we find that X C = 0.173R = 0.173 ( 443 Ω ) = 76.6 Ω 51. SSM REASONING At the resonant frequency f0, we have C = 1 / ( 4 π 2 f 02 L ) . We want to determine some series combination of capacitors whose equivalent capacitance C s′ is such that f 0′ = 3 f 0 . Thus, C s′ = 1 1 1 = = 2 2 2 9 4 π f 0′ L 4π ( 3 f 0 ) L 2 F 1 I= Gπ f L J 4 HK 2 2 0 1 C 9 The equivalent capacitance of a series combination of capacitors is 1 / C s′ = 1 / C1 + 1 / C 2 + ... If we require that all the capacitors have the same capacitance C, the equivalent capacitance is 1 11 n = + +L = C s′ CC C where n is the total number of identical capacitors. Using the result above, we find that 1 = C s′ 1 9 1 n = C C or n=9 Therefore, the number of additional capacitors that must be inserted in series in the circuit so that the frequency triples is n ′ = n − 1 = 8 . Chapter 23 Problems 1275 52. REASONING A resistance R and an inductance L are connected in series to the generator, but there is no capacitance in the circuit. Therefore, the impedance Z of the circuit is given by Z = R 2 + ( X L − X C ) 2 (Equation 23.7), where XL is the inductive reactance of the inductor, and the capacitive reactance XC is zero: R2 + ( X L − 0) = Z 2 2 R2 + X L = Z or (1) The impedance Z is related to the rms current Irms and the rms generator voltage Vrms by Vrms = I rms Z (Equation 23.6). The rms voltage VL, rms across the inductor is given by I rms = VL, rms XL (Equation 23.3). We will determine the inductive reactance from the generator frequency f and the inductance by means of X L = 2π f L (Equation 23.4). SOLUTION Solving Vrms = I rms Z (Equation 23.6) for Z, we obtain Z= Vrms (2) I rms Substituting Equation (2) into Equation (1) yields 2 R2 + X L = Substituting I rms = VL, rms XL Vrms (3) I rms (Equation 23.3) into Equation (3), we obtain 2 R2 + X L = Vrms VL, rms X L 2 R2 + X L = or Vrms X L VL, rms (4) Squaring both sides of Equation (4) and solving for R2, we find that V X 2 R 2 + X L = rms L VL, rms 2 or V X R 2 = rms L V L, rms 2 2 − XL Taking the square root of both sides of Equation (5) yields or 2 2 Vrms R2 = X L 2 − 1 VL, rms (5) 1276 ALTERNATING CURRENT CIRCUITS R = XL 2 Vrms 2 VL, rms −1 (6) Substituting X L = 2π f L (Equation 23.4) into Equation (6), we obtain the resistance R: R = 2π f L 2 Vrms 2 VL, rms − 1 = 2π (130 Hz )( 0.032 H ) ( 8.0 V )2 − 1 = ( 2.6 V )2 76 Ω CHAPTER 24 ELECTROMAGNETIC WAVES ANSWERS TO FOCUS ON CONCEPTS QUESTIONS 1. (b) The loop can only detect the wave if the wave’s magnetic field has a component perpendicular to the plane of the loop, that is, along the y axis. Only then will there be a changing magnetic flux through the loop. The changing flux is needed, so that an induced emf will arise in the loop according to Faraday’s law of electromagnetic induction. The electric and magnetic fields of an electromagnetic wave are mutually perpendicular and are both perpendicular to the direction in which the wave travels. Thus, when the wave travels along the z axis with its electric field along the x axis, the magnetic field will be along the y axis as needed. 2. (c) The wavelength λ, frequency f, and speed c of an electromagnetic wave are related according to c = λ f , where c is the same for any electromagnetic wave traveling in a vacuum and is independent of λ and f. Since c is constant, λ and f are inversely proportional. When f is reduced by a factor of three, λ increases by a factor of three. 3. (b) The magnitudes of the electric and magnetic fields of the wave are proportional to each other, according to E = cB (Equation 24.3). As Section 24.4 discusses, the wave carries equal amounts of electric and magnetic energy. 4. (a) The magnitudes of the electric and magnetic fields of the wave are proportional, according to E = cB (Equation 24.3). Thus, when E doubles, so does B. The total energy density and the intensity are each proportional to the square of the electric field magnitude, according to u = ε 0 E 2 (Equation 24.2b) and S = cε 0 E 2 (Equation 24.5b), Therefore, when E doubles, u and S both increase by a factor of 22 = 4. 5. 698 J/(s·m2) v fo 6. (d) The observed frequency is= fs 1...
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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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