HW6answers - HW 6 answers Using the parallel Z formula for...

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HW 6 answers Using the parallel Z formula for two elements in parallel: 1 / ( 1/Z 1 + 1/Z 2 ) = Z 1 Z 2 /(Z 1 +Z 2 ) and Z 1 = j ϖ L and Z 2 = 1/ j ϖ C = – j (1/ ϖ C) then: ( j ϖ L)( j (–1/ ϖ C)) ( ϖ L/ ϖ C) j ϖ L ( j ϖ L)+ j (–1/ ϖ C) j ( ϖ L–1/ ϖ C) (1– ϖ 2 LC) for ϖ =500 Z = j (500)(0.1)/(1 – 500 2 (0.1)(10e–6)) = j 66.67 ohms for ϖ =1000 Z = j (1000)(0.1)/(1 – 1000 2 (0.1)(10e–6)) = j ohms (resonant) For parallel L and C that are “resonant” the overall impedance becomes infinite if the L and C have no losses. In practice, the L and C do have some losses, so the impedance is very high, but not infinite. The very high impedance of the resonant parallel L-C appears to the rest of the network as an open circuit. This is a useful property that is used to block current flow for a specific frequency, i.e. it is a filter that can stop a signal at that frequency from getting from one part of a circuit to another part of the circuit. This is sometimes referred to as a “trap” (like an animal trap contains the animal). The electronic “trap” keeps the signal from getting outside of the trap area. for ϖ =2000 Z = j (2000)(0.1)/(1 – 2000 2 (0.1)(10e–6)) = j 66.67 ohms Note that Z for ϖ =500 and Z for ϖ =2000 differ only in that they are 180 ° apart from each other. 1 = =
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5+ j 15 - j 15 I 1 20 = - j 15 j 15- j 10 I 2 -(-10) Putting this into my calculator gives: I 1 = 1.644 80.54° I 2 = 2.977 74.20° The next page shows a Fortran program solving this problem using Cramer’s rule as well as solving the problem a different way, solving for V1 using KCL and then solving for the I1 and I2 currents using the V1 voltage. Writing a program like this avoids making
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This note was uploaded on 09/24/2011 for the course E E 331 taught by Professor Staff during the Fall '08 term at University of Texas at Austin.

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HW6answers - HW 6 answers Using the parallel Z formula for...

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