Two A.C Sources in one circuit. Magnitude and Frequency the same but one is Sin function + one Cos.

Thread Starter

Neil Hayes

Joined May 23, 2015
14
Hi there folks!

I'm just doing a question at the moment and the question asks to find the current through a load which is being fed from two voltage sources via their respective reactances ( inductive ). I need to find this current through a center branch using three different techniques which isn't an issue. My am just slightly confused by the fact that the voltages are quoted with respect to time and are also different functions, for eg: 415sin(100pit) and 415cos(100pit)

There is no mention of time so would I be correct to assume that I just take the magnitude in each case and assume that it is giving an out of that constantly?

Any help would be greatly appreciated : )

Neil
 
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Thread Starter

Neil Hayes

Joined May 23, 2015
14
Thanks a million for your reply WBahn!

I sure did do phasors. The sine function will be 0 degrees in polar with magnitude 415 and the cosine will lead by 90 degrees, is that the right approach? I had a brain freeze on that question!
 

Thread Starter

Neil Hayes

Joined May 23, 2015
14
Hi WBahn,

Sorry for the delayed response here but one thing I've noticed on this question is that it quotes that the system is acting with a (.7 p.f. Lagging).

The question asks to find the current through a particular load. I'm not being asked anything about the power dissipated or anything like that so I'm not sure of the relevance of the p.f.

Maybe it's there to cause confusion? If so it is working lol

Neil
 

MrAl

Joined Jun 17, 2014
7,656
Hello,

It would be better if you showed the circuit.

There are different ways to handle this depending on the circuit.
Sine and cosine functions can add to form one function sometimes. Other times we can just handle it as a two source circuit using superposition.
 

Thread Starter

Neil Hayes

Joined May 23, 2015
14
Hi there MrAI,

Thanks for your reply. I've actually just figured it out there this morning. Being told the p.f. enabled me to break the 50ohm load into its real and imaginary parts. I took the arccos of .70 which gave me the phase angle and this let me find the resistance and reactance of the load.

Thanks for getting back to me!

Neil
 

MrAl

Joined Jun 17, 2014
7,656
Hi,

Congrats :)

Sometimes it is nice to see the circuit anyway even if already solved, for the benefit of other readers who might be working on the same problem.
 
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