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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
selfinductance is L = N Φ/i = 5 × 10−3 H. From the derivative of Eq. 3041 (or a
combination of that equation and Eq. 3039) 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 batteryfuseinductor 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. 3015.
58. Applying the loop theorem, ε−L FG di IJ = iR ,
H dt K we solve for the (timedependent) 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. 3049 and Eq. 3041, 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. 2622 and Eq. 3041, 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. 2814 (with Eq. 3041).
63. From Eq. 3049 and Eq. 3041, 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. 2622 and Eq. 3041, 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.
 Fall '08
 schuller
 Magnetism, Work

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