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Unformatted text preview: Physics 315: Oscillations and Waves Homework 4: Solutions 1. Consider the i th loop from the left in the circuit, which consists of the i th capacitor, the i th inductor, and the i + 1th capacitor. The current flowing through the i th inductor is I i . Thus, the potential drop across the inductor is L I i . From Kirchoffs first circuital law, the current flowing through the i th capacitor (with the same sense of circulation about the i th loop as the current flowing through the i th inductor) is I i- I i 1 . Hence, the potential drop across the capacitor is integraltext t [ I i ( t )- I i 1 ( t )] dt /C . Likewise, the current flowing through the i + 1th capacitor (with the same sense of circulation) is I i- I i +1 . Thus, the potential drop across the capacitor is integraltext t [ I i ( t )- I i +1 ( t )] dt /C . According to Kirchoffs second circuital law, the sum of the potential drops around any closed loop in a circuit is zero. Hence, applying this law to the i th loop, we get L dI i ( t ) dt + integraldisplay t [ I i ( t )- I i 1 ( t )] dt /C + integraldisplay t [ I i ( t )- I i +1 ( t )] dt /C = 0 . (1) Differentiating with respect to t , and rearranging, we obtain d 2 I i dt 2 = 2 ( I i 1- 2 I i + I i +1 ) , (2) where = 1 / L C . Thus, we have a set of N coupled ordinary differential equations in which the loop index i ranges from 1 to N . However, the first and last equations are special cases, since they govern the behavior of the leftmost and rightmost circuit loops, respectively, and these loops are different from the interior loops. Consider the leftmost circuit loop. In this loop, the current flowing through the 1st capacitor is I 1 , rather than I 1- I . We can incorporate this additional information into the 1st differential equation by setting I to zero. Likewise, in the rightmost circuit loop, the current flowing through the N + 1th capacitor is I N , rather than I N- I N +1 . Hence, it is necessary to set I N +1 to zero in the N th differential equation. In other words, we need to solve Eqs. (2), for i = 1 , N , subject to the constraints I = I N +1 = 0....
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