# 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 wL and Z 2 = 1/ j wC = – j (1/wC) then: ( j wL)( j (–1/wC)) (wL/wC) j wL ( j wL)+ j (–1/wC) j (wL–1/wC) (1–w 2 LC) for w=500 Z = j (500)(0.1)/(1 – 500 2 (0.1)(10e–6)) = j 66.67 ohms for w=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 w=2000 Z = j (2000)(0.1)/(1 – 2000 2 (0.1)(10e–6)) = j 66.67 ohms Note that Z for w=500 and Z for w=2000 differ only in that they are 180° apart from each other.

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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:

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