Is this correct to find the supply current Is?

 

You have got the right answer for Zt. I did mention to you in post #33 that the convention is: "The angle should be expressed as one falling in the range -180° < Θ ≤ 180°"

You might also carry a few more digits in your calculations. Using my HP50g which has built-in the capability to do these sorts of calculations with 12 digit precision and accuracy, I get the final result Zt = 8.7517 - j8.2207 or 12.0072 ∠-43.208°

or exactly (1269 - j1192)/145

 

Is this correct to find the supply current Is?

View attachment 142694

At the very end of this you made a couple of sign errrors. The angle of the current should be -317°, not +317°. Also, the rectangular form of 8.31 ∠317° is not the correct answer because it's just the rectangular form of the wrong answer.

 

why should it be -317degrees? I dont understand. Thanks

 

Is this correct to find the supply current Is?

So your answer for the supply current has units of ohms?

 

Look at your work where you have Is = 100 (0-317). The thing in parentheses, (0-317) evaluates to -317. Do you see why?

 

So your answer for the supply current has units of ohms?

Boy howdy! I sure missed that, didn't I?

 

still working on tracking my units. is the working correct now?

It takes effort -- and having your mistakes pointed out -- to get in the habit.



Remember, treat units just like you would a variable. If you had a variable 'y' and had the expression y * y then that would become y², right?

Well, that means that Ω·Ω = Ω².

So your second line should have Ω², and not Ω, in the numerator and Ω in the denominator, so the end result would be just Ω.

In your third line you have Ω in both numerator and denominator, which is consistent with the error you made in the prior line, and in the fourth line you cancel them out, which is consistent with the prior line. But at this point, you have an expression for Zt that is dimensionless. THIS is where red flags should be going up. You KNOW that an impedance MUST have units of ohms -- as evidenced by the fact that the correct units just happened to magically appear on the final two lines -- so you KNOW that you have made some kind of a mistake. In this case, it was a simple units goof which is easily fixed. So fix it an move on. But in many cases it WON'T be a simple units goof -- it will be something that messes up the answer. For instance, if you had added the two impedances in the numerator instead of multiplying them.

 

yes I see, apart from the values the calculations are correct ?

for supply current it should have been measured in Amps.

 

Hi, yes I should have tracked my ohms units all the way through instead of just magically appearing at the last two lines. Thanks.

is the formula I'm using correct to find supply current?

What formula should I use to find I1 as I don't know where to begin.

 

Hi, yes I should have tracked my ohms units all the way through instead of just magically appearing at the last two lines. Thanks.

is the formula I'm using correct to find supply current?

What formula should I use to find I1 as I don't know where to begin.

At the very beginning of post #41, you have V = I Z. That's a very famous formula called Ohm's law. When impedances with both real and imaginary parts are involved it's sometimes called Ohm's law for AC circuits. That's the formula to use.

To find the current I1, you convert the two impedances in the branch carrying I1 into their equivalent and divide the applied voltage by that equivalent. Since you've already calculated the equivalent of those two series connected impedances there's not much left for you to do but carry out the division and show the details of your work.

It's the same thing you already did to find Is by converting all four impedances into a single equivalent and dividing Vs by the value of the equivalent. To find I1, you only need to convert the bottom two impedances into their equivalent. That equivalent will have the same source voltage applied to it.