Finding current and power consumption

1. Jul 20, 2013

MissP.25_5

I need help in solving this, I am stuck half way. Please check if what I am doing is right and tell me what to do next.
By the way, this work should be done without using the calculator, so let values such as √2 as they are. And values like sin(2t + θ) should be just left like that, too.

Current i1(t)= √2sin2t [A]
1) Find i2(t).
2) Calculate the effective power of the entire circuit (power consumption).

In the attachments are the diagram and my partial solution.

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2. Jul 20, 2013

Staff: Mentor

Welcome to Physics Forums, MissP.23_5.

Your work to this point is good except for the very last line (!); check the signs of the terms.

Since you now know the current entering the second parallel section you should be able to find the voltage across that section and hence the current through the inductor.

After that, consider that you have the individual voltages across the two parallel sections and the total current... How might you determine the power?

3. Jul 20, 2013

MissP.25_5

Thanks for replying! Ok, here I have finished my solution, please check if it's correct.

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4. Jul 20, 2013

Staff: Mentor

In your determination of I2, I don't see where the extra √2 is coming from. If you didn't apply a conversion from peak to RMS for i1 when you determined its phasor to be √2, then you've tacitly assumed that the values are already RMS. If this is not the case, then you'll have to adjust your i1 phasor to RMS and redo the calculations for i2 and power. The "extra" √2 in converting I2 to i2 (RMS phasor to peak time domain) would then make sense.

5. Jul 20, 2013

MissP.25_5

The √2 for I2 comes from the value of V2, since V2= -3√2.

6. Jul 20, 2013

Staff: Mentor

So you are assuming that in the given expression i1(t) = √2 sin(2t)A that the √2 represents the RMS value, not the peak value. Is that correct?

7. Jul 20, 2013

MissP.25_5

Nope, √2 is the peak value.

8. Jul 20, 2013

Staff: Mentor

Aha! So then, what's the RMS magnitude of your I1 phasor?

EDIT: I should mention that it's okay to work with peak phasors if you're looking for peak values for currents and voltages. After all, it's just a scalar constant factor. But when it comes time to determine power, you need to know whether you've got RMS or peak values for voltages and currents; you want to use RMS values for power.

Last edited: Jul 20, 2013
9. Jul 20, 2013

MissP.25_5

Oh, is it 1???

10. Jul 20, 2013

Staff: Mentor

Yes indeed. That should simplify some of the subsequent calculations...

11. Jul 20, 2013

MissP.25_5

Ok, so I have corrected all my mistakes and here's what I got:

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12. Jul 20, 2013

Staff: Mentor

What happened to arg(I2) in the i2(t) calculation? You found I2 = -3/j = 3j, right? That shouldn't have a zero angle. Otherwise it looks good.

The power calculation looks okay.

13. Jul 20, 2013

MissP.25_5

Yes, I2=3j and the magnitude of I2 would be 3, right? How can I find arg I2?
I know that tan^-1 (y/x) is the formula but since x=0, wouldn't that be undefined?

14. Jul 20, 2013

Staff: Mentor

Yup. There are a couple of angles where tan is undefined, but the angles are perfectly good. Try plotting the phasor on the complex plane. What angle does it make with the + real axis?

15. Jul 20, 2013

MissP.25_5

On a complex plane, the coordinate would be (0,3), so that makes a straight line along the Imaginary axis and thus, the angle is ∏/2. Am I right? So, i(t) = 3√2sin(2t+∏/2), correct?

16. Jul 20, 2013

Staff: Mentor

You are right

It's often handy to quickly sketch the phasor on the complex plane to get an idea of what the angle should be.

Also for future reference, remember that the arctan function has a restricted domain so it can't distinguish between arguments that should be in the 2nd or 4th quadrants, or those in the 1st and 3rd quadrants. The atan2() function does not have this limitation, and neither does the built-in rectangular to polar conversion function, if your calculator has either of those.

17. Jul 20, 2013

MissP.25_5

Finally, I got it right. Thank you so much for your patience. I am actually studying for my exams next week, so I'm having a lot of questions. I hope you don't mind helping me with my other thread.
Thanks!

18. Jul 20, 2013

Staff: Mentor

No worries, that's why we're here