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# chapt31 - Intro to Biological Physics mid-term exam 2...

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Chapter 31: Electromagnetic Oscillations and Alternating Current Problem 1 In a certain oscillating LC circuit, the total energy is converted from electrical energy in the capacitor to magnetic energy in the inductor in 1.50 μ s. What are (a) the period of oscillation and (b) the frequency of oscillation? (c) How long after the magnetic energy is a maximum will it be a maximum again? Answer (a) 6.00 μ s. (b) 167 kHz. (c) 3.00 μ s. Solution (a) The time it takes for the energy in the capacitor to drop from a maximum to zero is T/4 (see Figure 31-4). Therefore the period for this circuit is 4*(1.5 μ s) = 6.0 μ s. (b) The frequency is f = 1/T = 1/(6.0 μ s) = 167 kHz. (c) the energy in the inductor goes from a maximum to zero and back to a maximum in half a period or 3.0 μ s (again, see Figure 31-4). Problem 2 What is the capacitance of an oscillating LC circuit if the maximum charge on the capacitor is 1.60 μ C and the total energy is 140 μ J? Answer 9.14 nF. Solution When the capacitor has maximum charge it also has maximum energy and the inductor has zero stored energy therefore the energy in the capacitor is the total energy. U tot = 1 2 q max 2 C or the capacitance is C = 1 2 q max 2 U tot = 1 2 1.6 μ C ( ) 2 140 μ J = 9.14 nF

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31-2 Chapter 30: Electromagnetic Oscillations and Alternating Current Problem 3 In an oscillating LC circuit, L = 1.10 mH and C = 4.00 μ F. The maximum charge on the capacitor is 3.00 μ C. Find the maximum current. Answer 45.2 mA. Solution The maximum energy stored in the inductor is equal to the maximum energy stored in the capacitor 1 2 Li max 2 = 1 2 q max 2 C Therefore the maximum current is i max = q max LC = 3.0 μ C 1.1 mH ( ) * 4.0 μ F ( ) = 45.2 mA Problem 8 LC oscillators have been used in circuits connected to loudspeakers to create some of the sounds of electronic music. What inductance must be used with a 6.7 μ F capacitor to produce a frequency of 10 kHz, which is near the middle of the audible range of frequencies? Answer 37.8 μ H Solution The frequency of an LC oscillator is f = 1 2 π 1 LC Therefore the required inductance is L = 1 4 π 2 f C = 1 4 π 2 10 kHz ( ) 2 * 6.7 μ F ( ) = 37.8 μ H
Chapter 31: Electromagnetic Oscillations and Alternating Current 31-3 Problem 10 A single loop consists of inductors (L1, L2, ...), capacitors (C1, C2,...), and resistors (R1, R2,...) connected in series as shown, for example, in Fig. 31-27a. Show that, regardless of the sequence of these circuit elements in the loop, the behavior of this circuit is identical to that of a simple damped LC circuit shown in Fig. 31-27b. (Hint: Consider the loop rule and see Problem 47 in Chapter 30.) L 1 C 1 L 2 R 1 C 2 R 2 L C R (a) (b) Figure 31-27 Solution If we write the loop equation for the circuit of Fig. 31-27a we get L 1 di dt q C 1 L 2 di dt iR 1 q C 2 iR 2 = 0 Note that since all the components are in series the charge on the two capacitors is the same, the currents in the two resistors are the same and the rates of change of currents in the two inductors are the same. Regrouping the equation we get L 1 + L 2 ( ) di dt q 1 C 1 + 1 C 2 i R 1 + R 2 ( ) = 0 or L eq di dt q C eq iR eq = 0

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