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This is a homework question where i am stuck. Please help me out in this:

Three similar impedances are connected in star across 2000 V, 50 Hz three phase supply. Power absorbed is 300 kW. Current taken is 100 A lagging. Find the values of circuit parameters in each phase.

I have found out the power factor to be 0.866.

I found out the impedance Z to be |Vph| / |Iph| = 11.243 ohms.

But now how do we find the inductance of the circuit?

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    \$\begingroup\$ Same question can be seen here. \$\endgroup\$ – Rohat Kılıç May 16 at 10:45
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    \$\begingroup\$ Show us what you have done so far. What is the apparent power of the load (kva)? \$\endgroup\$ – relayman357 May 16 at 10:46
  • \$\begingroup\$ But the other question's answers dont talk about inductance. \$\endgroup\$ – Chandradhar Rao May 16 at 10:48
  • \$\begingroup\$ Apparent power i found it out to be 346410.1615 va. \$\endgroup\$ – Chandradhar Rao May 16 at 10:50
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    \$\begingroup\$ But now how do we find the inductance of the circuit? - you can't because you don't know if the inductance is parallel to the resistor or in series with the resistor. The load is not defined as a series-elements or parallel-elements. \$\endgroup\$ – Andy aka May 16 at 11:24
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You should determine the impedance of your load using complex numbers. Once you know the impedance, you can find out the reactive component \$X\$ directly:

$$Z = R + jX$$

The sign of the reactive part of your impedance indicates whether you have an more capacitive (negative reactance) or a more inductive (positive reactance) load. You can then convert from reactance to capacitance or inductance if you know frequency:

$$X_C = (-j)\frac{1}{\omega C}$$ $$X_L = (j)\omega L$$

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  • \$\begingroup\$ Thanks but how do i find out the angle of phase current? I found out the angle of phase voltage is -30 degrees. \$\endgroup\$ – Chandradhar Rao May 16 at 10:53
  • \$\begingroup\$ You already have it - the power factor is cosine of the the phase angle between voltage and current. \$\endgroup\$ – Tom Carpenter May 16 at 10:59
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    \$\begingroup\$ Ok thanks.I figured out resistance and then inductance.Thanks \$\endgroup\$ – Chandradhar Rao May 16 at 11:12

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