Need help with question 2 please. Struggling to get to college with working shifts and this is referred work from last year. I'm guessing I will need to find the current in each loop and go from there? Simultaneous equations perhaps? Could do with a rough step by step of what to do s I don't know really where to start. Thanks jn advance jonny

 

Most people here are not going to bother downloading and opening a 4.8MB file when, in all likelihood, you could rescale it to something like 400 to 600 pixels wide and have something that is perfectly readable and less than 1% the size. For instance, in just 47 kB you can have:



By "referred work from last year" I take you mean that this is review from a prerequisite course you took last year.

If that's the case, and if you truly have no idea how to start, I strongly recommend that you drop the current course you are taking and that you retake the course that covered this material. If you don't, you will just get progressively more lost because you lack the foundational circuit analysis skills.

Otherwise, you need to show YOUR best attempt to work YOUR homework as far as you can. The further the better, keeping in mind that you don't have to be correct, but we need to see how you are going about things in order to help you fill in the gaps in your understanding.

 

Step 1.
Convert all values into complex form.
R1 is 3+0j=3 Ohm
R2 is 4+0j=4 Ohm
R3 is 5+0j=5 Ohm
C1 is given to you, 0+(-4j)=-4j Ohm
L1 is given to you 0+4j=4j Ohm
You need to convert V1 and V2 into complex form, they are in phasor form right now (I think).

Step 2.
You have 3 loops, you will have 3 simultaneous equations.
Set them up and solve them to find the 3 mesh currents.

Step 3.
Use mesh currents that you found to find \(I_{C}, I_{L},\) p.d. across 4 Ohm resistor.

By the way. What is p.d.? Power drop? I don't think I ever encountered this abbreviation.

You might want to invest into graphing calculator. My Casio can do math using complex numbers, solve simultaneous equations and Casio generally cheaper than TI.

 

By the way. What is p.d.? Power drop? I don't think I ever encountered this abbreviation.

It is almost certainly means "potential drop" or more commonly, voltage drop.

 

The first step in the actual analysis is to clearly define your mesh currents, which means not only what the name of the current is, but what direction is it circulating in.

 

The first step in the actual analysis is to clearly define your mesh currents, which means not only what the name of the current is, but what direction is it circulating in.

Thank you. Potential drop does make sense. I think my instructors mostly used voltage drop.