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Unformatted text preview: e case for resistances.
Note that to ensure the independence of the voltage values, it is important that the
inductors not be too close together (the related topic of mutual inductance is treated in
Section 3012). The requirement is that magnetic field lines from one inductor should not
have significant presence in any other.
(b) Just as with resistors, Leq = ∑n =1 Ln .
N 48. (a) Voltage is proportional to inductance (by Eq. 3035) just as, for resistors, it is
proportional to resistance. Now, the (independent) voltages for parallel elements are
equal (V1 = V2), and the currents (which are generally functions of time) add (i1 (t) + i2 (t)
= i(t)). This leads to the Eq. 2721 for resistors. We note that this condition on the
currents implies 1183 bg bg bg di1 t di2 t
di t
+
=
.
dt
dt
dt
Thus, although the inductance equation Eq. 3035 involves the rate of change of current,
as opposed to current itself, the conditions that led to the parallel resistor formula also
apply to inductors. Therefore,
1
1
1
=
+.
Leq L1 L2
Note that to ensure the independence of the voltage values, it is important that the
inductors not be too close together (the related topic of mutual inductance is treated in
Section 3012). The requirement is that the field of one inductor not to have significant
influence (or “coupling’’) in the next.
(b) Just as with resistors, N
1
1
=∑ .
Leq n =1 Ln 49. Using the results from Problems 3047 and 3048, the equivalent resistance is
Leq = L1 + L4 + L23 = L1 + L4 + L2 L3
(50.0 mH)(20.0 mH)
= 30.0 mH + 15.0 mH +
L2 + L3
50.0 mH + 20.0 mH = 59.3 mH. 50. The steady state value of the current is also its maximum value, ε/R, which we denote
as im. We are told that i = im/3 at t0 = 5.00 s. Equation 3041 becomes i = im 1 − e − t0 /τ L , ( ) which leads to τL =− 5.00 s
t0
=−
= 12.3 s.
ln 1 − i / im
ln 1 − 1 / 3 b g b g 51. The current in the circuit is given by i = i0 e −t τ L , where i0 is the current at time t = 0
and τL is the inductive time constant (L/R). We solve for τL. Dividing by i0 and taking the
natural logarithm of both sides, we obtain
ln FG i IJ = − t .
Hi K τ
0 L This yields τL =− .
t
10 s
=−
−3
ln i / i0
ln 10 × 10 A / 10 A
. bg ec Therefore, R = L/τL = 10 H/0.217 s = 46 Ω. = 0.217 s.
h b gj 1184 CHAPTER 30 52. (a) Immediately after the switch is closed, ε – εL = iR. But i = 0 at this instant, so εL =
ε, or εL/ε = 1.00.
(b) ε L (t ) = ε e− t τ L = ε e−2.0τ L τ L = ε e−2.0 = 0.135ε , or εL/ε = 0.135.
(c) From ε L (t ) = ε e −t τ L we obtain
⎛ε
= ln ⎜
τL
⎝ εL
t ⎞
⎟ = ln 2
⎠ ⇒ t = τ L ln 2 = 0.693τ L ⇒ t / τ L = 0.693. 53. (a) If the battery is switched into the circuit at t = 0, then the current at a later time t is
given by
i= ε R c1 − e h ,
− t /τ L where τL = L/R. Our goal is to find the time at which i = 0.800ε/R. This means
0.800 = 1 − e− t /τ L ⇒ e− t /τ L = 0.200 .
Taking the natural logarithm of both sides, we obtain –(t/τL) = ln(0.200) = –1.609. Thus,
t = 1609τ L =
....
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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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