# Supply current - PART 2

Discussion in 'Homework Help' started by Hello, Mar 25, 2009.

1. ### Hello Thread Starter Active Member

Dec 18, 2008
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Can you please have a look at Q3 part (iv) in the attached file.
Not sure if i have the right answer!

Is = I1 + I2 + I3
= 2.50 + 2.65 + 0.0811-j0.359
= 5.23 - j0.359
= 5.24A

Any help is appreciated

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2. ### thatoneguy AAC Fanatic!

Feb 19, 2009
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Capacitive Impedance:
$I_C=\frac{1}{2 \cdot \pi \cdot f \cdot C}$

Inductive impedance is also imaginary.

3. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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The question asks for the peak currents and it is most likely the source voltage is an RMS value. You have attempted to find only the RMS current values.

Your value of I2 is wrong. It is purely inductive current so it should have only an imaginary value ...

The same is true for I3 but the error is only in the matter of the sign for the imaginary (j operator) part - it should be +ve.

Try again - keeping in mind you are required to give peak values.

4. ### Hello Thread Starter Active Member

Dec 18, 2008
82
0
Thanks for the corrections, here are my new values:

I = I1 + I2 + I3
= 2.50 - j2.65 + 0.0811 + j0.359
= 2.58 - j2.29
= 3.45A @ -41.6°

Peak currents:

I1 = 2.50 A
I2 = 2.65 A
I3 = 3.45 A

Im assuming those are the peak values, but not sure which are the rms values.

5. ### thatoneguy AAC Fanatic!

Feb 19, 2009
6,357
727
Please show the middle step of impedance. It isn't looking right intuitively.