Capacitor c to have power factor equal to .707

Discussion in 'Homework Help' started by led23, Apr 28, 2009.

1. led23 Thread Starter New Member

Mar 10, 2009
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for the attachement i am trying to determine
1. the value of c to have a power factor equal to .707
2. for this value of capacitance, determine the real power transferred to the 10 K ohm load
3. what is the apparent power demanded from the source

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2. t_n_k AAC Fanatic!

Mar 6, 2009
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Not much interest in this problem it seems.

I have ...

1. C = 39.79 nF

2. Power = 10 mW

3. Apparent Power = 14.14 mVar

3. t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
Oops - Maybe I should have

3. Apparent Power = 14.14 mVA

4. led23 Thread Starter New Member

Mar 10, 2009
6
0
what is the formula you used to come to c= 39.79 nF? if I may ask

5. t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
To achieve a power factor of 0.707 (cos 45°) would require the capacitive reactance to equal the load resistance. If you are not sure why this is the case, draw a load (or power) triangle representing the resistive and reactive terms.

The required capacitive reactance magnitude is then known from the problem statement as