three similar inductive coil of resistance 20 ohm and reactance 30 ohm are connected in mesh across 3 phade 400 volt supply, what will be the phase current
Answers
I’m surprised nobody has picked this one up yet, aside from a couple of homework objectors. I suspect that this one will not see too many views, but I’ll give it a go anyway.
Let’s work it out relative to the red phase with voltage vred at 0°. The circuit is symmetrical and so the results for this phase will apply to the other phases equally with just a phase difference. I’ll use complex number (phasor) representation of voltage, current and impedance for the calculations.
First we need the current in the red phase. This is just the sum of the currents through the two attached inductors connecting red to the blue and yellow phases respectively, ie: in the delta configuration. If we let v1 and i1 be the voltage and current across the red-blue inductor, and v2 and i2 be the voltage and current across the red-yellow inductor, then we have that the total current being supplied by the red phase is:
i1+i2=v1z+v2z
=v1+v2z(1)
where z is the impedance of each inductor. The voltage for the blue phase is 120º ahead of the red phase, which makes it parallel with the unit phasor:
cos(120)+sin(120)i=−0.5+3√2i.
So the voltage across the inductor connected from red to blue phase is:
v1=vred−vblue=vphase−vphase(−0.5+3√2i)
=vphase(1.5−3√2i)
Similarly, the yellow phase is 120º behind the red phase and hence parallel with unit phasor −0.5−3√2i. So:
v2=vred−vyellow=vphase−vphase(−0.5−3√2i)
=vphase(1.5+3√2i)
Hence using equation (1), the total current being supplied by the red phase is:
i1+i2=vphase(1.5−3√2i)+vphase(1.5+3√2i)z
=3vphasez
In fact this is a standard formula for the delta configuration. Also, given that vline=3–√vphase, we can write the total per phase current as:
3–√vlinez(2)
We’re given that the inductors are 0.05H with 20Ω resistance, so with an AC frequency of 50Hz this has an impedance of:
z=R+2πfLi=20+2∗π∗50∗0.05i
=20+5πi
With vline=415 V, the current using equation (2) works out to be:
=3–√∗41520+5πi=3–√∗415(20−5πi)202+(5π)2
=22.23−17.46i
So the magnitude of the current being drawn from red and therefore from each phase is:
22.232+17.462=28.3 Amps
with power factor :
22.2328.3=0.787
The real power being supplied by each phase is:
22.23vphase=22.234153–√
=5,326 Watts
and so the total real power being supplied/absorbed is:
5326∗3=15,978 Watts=16.0 kW
The total apparent power being supplied is:
16.00.787=20.3 kVA
To double check the result we can calculate the power being dissipated in each inductor from the line voltage across the inductor. This power is:
(vline)2z=415220+5πi
=4152(20−5πi)202+(5π)2
=5326−4183i
The real component of this power, 5,326 Watts, corresponds with the real power taken from each phase, as each inductor consumes the same power by symmetry and the total power dissipated by the three inductors is equal to the total power supplied.