# AC power problems.

Discussion in 'Homework Help' started by eleceng, Aug 3, 2009.

1. ### eleceng Thread Starter New Member

Aug 3, 2009
4
0
Studying for a big final exam and have no solutions for this questions and unable to work out the last few questions. The circuit diagram is attached and i need to calculate the following:
-total impeadance of the circuit (i calculate 5.8 ohm)
-the load current (i calculate 68.65A)
-rms load voltage (Vl) (i caculate 242.7v)
-the phase angle of load voltage relative to the supply voltage
-power drawn from supply
-reactive power drawn from supply
-power factor of supply
-power lost in connecting cable

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2. ### yourownfree Active Member

Jul 16, 2008
89
0
When I calculate separate I get .741401 for the 2.23 mh and resistor and 5 for the 9.55 and 4 ohm total of 5.741401 Ohms. So I would have to say 5.74 Ohms. or 5.7 Ohms.

Last edited: Aug 3, 2009
3. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
The total complex impedance is 5.83 (angle:39.43°) ohms - so you are correct eleceng.

You have the RMS current correct but not the load RMS voltage.

I have V_load = 343.3 volts

Since ...

4. ### eleceng Thread Starter New Member

Aug 3, 2009
4
0
I intially had a the voltage of 343.3 Volts but assumed that to get the RMS i had to multiply the value by 1/√2 - clearly i was wrong. Therefore the load power is 23.57KW ? (P = I^2 * R) and the power drawn from supply is 27.342KW. The load reactive power is (P = I(68.66)^2 * (2.93)XL) therefore 13.817Kvar and reactive power drawn from supply is (P = I(68.66)^2 * (3.7)XL) therefore 17.446KVAr?

5. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783

I get .....

Source power = I^2*R_total = (68.66)^2*4.5 = 21.21kW
Source Reactive Power = I^2*X_total = (68.66)^2*3.7 = 17.44kVar
Source apparent power = V_source*I=400*68.66=27.46kVA
Source pf = Real Power / Apparent Power = 21.21/27.46 = 0.772

Verify pf using total impedance angle = 39.43°

pf = cos(39.43°)= 0.772

6. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
pf is lagging BTW

7. ### eleceng Thread Starter New Member

Aug 3, 2009
4
0
Thank you very much for your help, can now see where I have been going wrong! Thanks.