Recent content by Aristotle

Woops didnt mean to say the answer was wrong. ?:) But yeah the ε in the arctan in that step is odd. Think they forgot a 't' in the numerator.

Thank you so much for your help! I also got the same. Possibly their answer is incorrect...

Wouldn't that get you arctan(t) and not arctan(ε)?

The only way I see taking ε2 out of the denominator is dividing that number by itself for numerator and denominator. But you would get ∫(1/ε)⋅( (dt) / ( (t2/ε2)+1) )

If you don't mind me asking, where did you get t/ε from?

Hi, I was just looking at an example for a certain problem and noticed that in the second step they went to arctan(epsilon). I know there's a form that is equal to arctan but am a little unsure. I've come across formulas on the web such as arctan(x) = ∫(dt)/(a2+t2) but nothing else that would...

There is many ways in finding power dissipation--you can also do Px=(Iacross-4Ω)2 x R4Ω where Iacross-4Ω= 36 volts / 4Ω = 8 Amps (since they are in parallel connection, voltage is same across each resistor!) Then square this and multiply it by the 4Ω.. and still get the correct answer--just...

Current will want to flow across the path of less resistance. So a resistor with high resistance will have less current going through it, while there is more current flow across a resistor with less resistance. The only time current will not flow through a resistor is if there is a shorted wire...

By using Ohm's Law of course! Voltage drop across resistor (V) = Current flowing across a resistor (I) x Resistor's Value in Ohms (R)

The nice thing about resistors being in parallel nicely like that is that if you know the voltage drop across one of them, you'll know the voltage drop across the other resistors and the supply source--they share the same voltage :smile: Just keep in mind current flow will differ across each...

This formula you speak of is called Current Division rule, useful when you want to find the current of a branch in a circuit... especially in this scenario--since current splits off in the resistors that are in parallel. The 10.8 A current is the total current that flows out of R2 and that flows...

Right..so once the voltage source is transformed into a current source with R1 in parallel, the current sources can be added together :)

Cnh is correct! :wink: If my memory serves me right, when you want to do source transformation, you always want to consider Ohm's Law Vst=Ist*R (where st can be source transformation). You are correct about R3 being in series with R2, but the combination isn't parallel with R1 because you still...

Yes I've solved it, turned out part b's solution of the book is incorrect according to my professor. Thanks!

Homework Statement Homework Equations Wye: V_Line = sqrt(3) x V_Phase Delta: V_Line = V_Phase The Attempt at a Solution My answer doesn't quite match up with the solutions in the book, and was wondering what I went wrong here? Any help is greatly appreciated!