Solutions Chapter 11 Problem 1 Part a I = Em /Z, s R2 + ωd L − Z= 1 ωd C 2 As there is no resistor, R = 0. As there is no capacitor, C = ∞ or I= 1 ωd C = 0. So Z = ωd L, and Em = 4.17 × 10−3 A. ωd L Part b i = I sin(ωd t − φ) where φ is given by: tan φ = ωd L − 1 ωd C R = ∞. So, φ = π/2 and i = I sin(ωd t − π/2) = −I cos(ωd t) Hence, i is a maximum when ωd t = π, 3π, 5π, . . . . At these times sin(ωd t) = 0. So, at these times, E = Em sin(ωd t) = 0. Problem 2 Part a I = Em /Z, s 1 2 Z= + ωd L − ωd C As there is no resistor, R = 0. As there is no inductor, L = 0. So Z = 1/(ωd C), and R2 I = Em ωd C = 5.00 × 10−2 A. 1 Part b i = I sin(ωd t − φ) where φ is given by: tan φ = ωd L − 1 ωd C R = −∞. So, φ = −π/2 and i = I sin(ωd t + π/2) = I cos(ωd t) Hence, i is a maximum when ωd t = 0, 2π, 4π, . . . . At these times sin(ωd t) = 0. So, at these times, E = Em sin(ωd t) = 0. Problem 3 Part a The resonant frequency is √ ωd = 1/ LC = 500rad/sec Part b The current amplitude is I= Em Em =p 2 Z R + (XL − XC )2 At resonance, XL = XC , hence, I= Em 10 = = 2.50 A R 4 Also at resonance ωd = √ 1 LC and the voltage amplitude across the inductor is IL VL = IXL = Iωd L = √ =I LC s L = 2500 V C Part c The inductor may have a higher voltage across it than the source because the capacitor voltage can partially (or completely in case of resonance) cancel the inductor voltage. 2 Problem 4 tan φ = XL − XC R Hence, R= 2πfd L − 2πf1d C ωd L − ωd1C XL − XC = = = 140 Ω tan φ tan φ tan φ Problem 5 Part a s Z= R2 + ωd L − 1 ωd C 2 = 67 Ω Part b I= Em = 6.7 A Z Part c The phase constant is given by tan φ = ωd L − 1/(ωd C) = 1.35 R Hence, φ = 53◦ Problem 6 ................................................................................................... .. ... i ... .. ... .> > > r> > > > > ..> . R> > .. ..... ......... > > E ..... ∼ ..... > ............ ...> .. ... .. ... .. .................................................................................................. 3 Power dissipation in R is 2 P = Irms R where Irms = Erms Erms = Z r+R So, R 2 P = Erms (r + R)2 The value of R for which P is maximum, must obey the maxima condition from calculus: This gives: 1 2R 2 Erms − = 0. (r + R)2 (r + R)3 The solution to this equation is: r = R. Problem 7 The voltage of the secondary is Vs = Ns Vp = 4.0 V Np 4 dP dR = 0.