0

I had a doubt, in a balanced star three phase arrangement:

Using two wattmeter method, we obtain power readings as W1 and W2. (which may be unequal).

Let us take W1=5 watt and W2= 6 watt for instance and phase current to be 10 ampere.

However from the definition of real power, we know that power will be dissipated in resistor only.

So if I use power= i^2 *r,

In the first case, when power is 5 watt,

r=5/100 ohm while in second case it is 6/100 ohm

However, we said that it is a balanced arrangement, then how come the value of resistances is different.

Any help is appreciated.

programmer
  • 11
  • 2
  • if there's neutral current you will need three wattmeters. – Jasen Слава Україні Feb 07 '21 at 08:01
  • there isnt neutral current @Jasen – programmer Feb 07 '21 at 08:10
  • How the current is got for your calculation? Is the three phase outlet symmetric (=equal phase voltages and 120 degrees shifts)? What happens if you swap the meters? Do they still keep their readings? Is the load said to be symmetric or is it a measured fact? If you have a 3 phase transformer it can quite far from ideal. Is there one? The specs of it? –  Feb 07 '21 at 09:25

0 Answers0