- #1

- 36

- 0

## Homework Statement

The frequency is 50Hz (we've been told to always assume in our class that the frequency is 50Hz) and thus omega is 100pi.

## The Attempt at a Solution

I don't have time to scan an image of my attempt (complete attempt) however what my query is: the way i went about to answer this question, i firstly zeroed the current source, then i used the voltage divider to find the voltage across the 5ohm resistor, after this i used ohm's law (I=V/R) to find the current, which is (according to my calcuations): 0.021e^(-j1.56) Amps.

I then zeroed the voltage source, and used the current divider to find the current across the 5ohm resister, which was found to be: 0.0318e^(-j1.53) Amps. Adding these two (well i added their rectangular forms then converted back to exponential form), since that's what the superposition rule requires (right?), i get: 0.05e^(-j1.54) Amps.

I'm not too bothered whether anyone checks if this is right, but does the procedure seem right? As in, is it ok that i used the voltage divider, I=V/R then current divider? It seems rather simple to just use those three concepts/rules..it was my first attempt and that's what stood out. However someone i know in the same class went about it some other way (i don't know how, we exchanged text messages), and they ended up getting some 97Amps, which is kinda high i thought for such a small resistor value and even voltage value..thanks.

edit: i think i used the current divider incorrectly, the way i did it was: I = (Z(R 5ohm) / Z(R) + Z(L) + Z(C) + Z(R 10ohm)) * Isource. With Z being the impedances of the respective components..? I think i should have summed the inductor, capacitor and resistor(10ohm) and put that sum on the numerator? Then divide by the 5ohm resistor and summed impedances..

Last edited: