Homework 10 Solutions

5108 500 103 s05108 979 103 s and the inductance

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Unformatted text preview: 1609 L 1609(6.30 × 10−6 H) . . = = 8.45 × 10−9 s . R 120 × 103 Ω . (b) At t = 1.0τL the current in the circuit is i= ε 14.0 V (1 − e ) = ⎛ 1.20 ×10 Ω ⎞ (1 − e ⎜ ⎟ R ⎝ ⎠ −1.0 3 The current as a function of t / τ L is plotted below. −1.0 ) = 7.37 × 10−3 A . 1186 CHAPTER 30 i= ε c1 − e h, R − t /τ L where τL = L/R is the inductive time constant and ε is the battery emf. To calculate the time at which i = 0.9990ε/R, we solve for t: 0.990 ε R = ε (1− e R − t /τ L ) ⇒ ln ( 0.0010 ) = − ( t / τ ) ⇒ t / τ L = 6.91. The current (in terms of i / i0 ) as a function of t / τ L is plotted below. 56. From the graph we get Φ/i = 2 ×10−4 in SI units. Therefore, with N = 25, we find the self-inductance is L = N Φ/i = 5 × 10−3 H. From the derivative of Eq. 30-41 (or a combination of that equation and Eq. 30-39) we find (using the symbol V to stand for the battery emf) di VR t Vt = L e− /τL = e− /τL = 7.1 × 102 A/s . R dt L 57. (a) Before the fuse blows, the current through the resistor remains zero. We apply the loop theorem to the battery-fuse-inductor loop: ε – L di/dt = 0. So i = εt/L. As the fuse blows at t = t0, i = i0 = 3.0 A. Thus, t0 = i0 L ε = ( 3.0 A ) ( 5.0 H ) = 1.5 s. 10 V (b) We do not show the graph here; qualitatively, it would be similar to Fig. 30-15. 58. Applying the loop theorem, ε−L FG di IJ = iR , H dt K we solve for the (time-dependent) emf, with SI units understood: 1189 τL = t/0.5108 = (5.00 × 10–3 s)/0.5108 = 9.79 × 10–3 s and the inductance is c hc h L = τ L R = 9.79 × 10−3 s 10.0 × 103 Ω = 97.9 H . (b) The energy stored in the coil is UB = b gc 121 Li = 97.9 H 2.00 × 10−3 A 2 2 h 2 = 196 × 10−4 J . . 62. (a) From Eq. 30-49 and Eq. 30-41, the rate at which the energy is being stored in the inductor is 2 1 2 dU B d ( 2 Li ) di ⎛ε ⎞ ⎛ ε 1 −t τ L ⎞ ε = = Li = L ⎜ (1 − e − t τ L ) ⎟ ⎜ = (1 − e −t τ L ) e− t τ L . e ⎟ dt dt dt ⎝R ⎠⎝ R τL ⎠R Now, τL = L/R = 2.0 H/10 Ω = 0.20 s and ε = 100 V, so the above expression yields dUB/dt = 2.4 × 102 W when t = 0.10 s. (b) From Eq. 26-22 and Eq. 30-41, the rate at which the resistor is generating thermal energy is Pthermal = i 2 R = ε2 R 2 c1 − e h R = εR c1 − e h . −t τ L 2 2 −t τ L 2 At t = 0.10 s, this yields Pthermal = 1.5 × 102 W. (c) By energy conservation, the rate of energy being supplied to the circuit by the battery is dU B Pbattery = Pthermal + = 3.9 × 102 W. dt We note that this result could alternatively have been found from Eq. 28-14 (with Eq. 3041). 63. From Eq. 30-49 and Eq. 30-41, the rate at which the energy is being stored in the inductor is dU B d ( Li / 2 ) ε2 di ⎛ε −t τ L ⎞ ⎛ ε 1 −t τ L ⎞ = = Li = L ⎜ (1 − e ) ⎟ ⎜ R τ e ⎟ = R (1− e−t τ L ) e−t τ L dt dt dt ⎝R ⎠⎝ L ⎠ 2 1190 CHAPTER 30 where τL = L/R has been used. From Eq. 26-22 and Eq. 30-41, the rate at which the resistor is g...
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This document was uploaded on 02/26/2014 for the course PHYS 2b at UCSD.

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