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Chapter 28 Problem 51 † Given C = 0.15 µF L = 20 mH R = 1.6 Ω Solution Find the number of cycles it takes to have the peak voltage drop to half its original value. The formula describing charge on a capacitor in a damped oscillator circuit is q = qo e−Rt/2L cos ωt Charge on a capacitor is related to the voltage on the capacitor by V = CQ = Cqo e−Rt/2L cos ωt = Vo e−Rt/2L cos ωt Ignoring the cosine function, which does not cause the voltage in the circuit to decay, we can determine the time for the voltage to decay to half its value. V = e−Rt/2L Vo −2L V t= ln R Vo Substituting in the appropriate values gives us a time of 1 −2(0.020 H) ln t= = 0.0173 s 1.6 Ω 2 The angular frequency of the circuit is ω=√ 1 1 =p = 1.83 × 104 rad/s LC (0.020 H)(1.5 × 10−7 F ) This gives a time period for each oscillation of T = 2π 2π = = 3.44 × 10−4 s ω (1.83 × 104 rad/s) The number of cycles before reaching 1/2 voltage is then 1cycle t = 0.0173 s = 50.3 cycles 3.44 × 10−4 s † Problem from Essential University Physics, Wolfson