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 -load power -load reactive power -power drawn from supply -reactive power drawn from supply -power factor of supply -power lost in connecting cable
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.
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 ... Vload will be (I_load * Z_load) where Z_load is 5.0 (angle: 36.87°) ohms and I_load = 68.66 amps. I think you forgot to include the load inductance in the total load calculation.
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?
I get ..... Load power = I^2*R_load = (68.66)^2*4 = 18.85kW Source power = I^2*R_total = (68.66)^2*4.5 = 21.21kW Load Reactive Power = I^2*X_load = (68.66)^2*3.0 = 14.14kVar 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