91. When the switch is open, we have a series LRCcircuit involving just the one capacitor near the upper right corner. Eq. 31-65 leads to o1tan( 20)tan20.ddLCRωφ−==−°=−°Now, when the switch is in position 1, the equivalent capacitance in the circuit is 2C. In this case, we have 112tantan10.0 .ddLCR−==°Finally, with the switch in position 2, the circuit is simply an LCcircuit with current amplitude 2211mmmLCddddIZLCLCεεε===−⎛⎞−⎜⎟⎝⎠where we use the fact that 1()ddCL−>in simplifying the square root (this fact is evident from the description of the first situation, when the switch was open). We solve
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.