Homework 10 Solutions

# The requirement is that the field of one inductor not

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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 30-12). 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. 30-35) 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. 27-21 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. 30-35 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 30-12). 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 30-47 and 30-48, 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 30-41 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.

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