20095ee110_1_HW5_all_solutions

# 20095ee110_1_HW5_all_solutions - HW 5 Solutions for PSPICE...

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HW 5- Solutions for PSPICE Problems Problem 4. (a) The schematic of the circuit in problem 10.62 is shown above. The values of L1 and L2 are chosen (L1 = 16000 H, L2 = 1000 H for example) such that they satisfy the turns ratio. The transformer coupling ratio is set to 1. R_L is swept from 0.1 Ohm to 2Ohm in steps of 0.1 Ohms. The voltage, current and average power curves for R_L of the problem is shown below. From the simulation the maximum average power is 100W and occurs for R_L = 1 Ohm and we use this value for R L for parts (b) & (c) of the problem.

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(b) Measure the values of V across the voltage source and the current “I” going through the voltage source as shown below. The printed values can be seen at the bottom of the netlist as shown below: So, from the above result, the average power delivered by voltage source = V*I* cos(Ѳ v Ѳ i ) = 25 * 9 * cos (≈0) = 225 W.
Note that we the factor 0.5 was not used in the average power calculation, because I directly used 25V (which is the RMS value) for the voltage source. Note that you can also find out the values of V & I by plotting the traces of V and I of the voltage source

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## This note was uploaded on 10/02/2010 for the course EE EE 110 taught by Professor Gupta during the Fall '09 term at UCLA.

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20095ee110_1_HW5_all_solutions - HW 5 Solutions for PSPICE...

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