Electrical Engineering - Three-Phase Systems in Power Applications - Discussion

Correct answer is D:

as Va and Vb are 90 degree apart so;

Va=120<0; ---> polar form
Vb=120<90 ---> polar form.

Lets convert it into rectangular form
Va=120*(cos(0)+jsin(0)).
Va=120;
Vb=120*(cos(90)+jsin(90)).
Vb=j120;.

Now calculating current;
Ia= 120/90;
Ia=1.33;
Ib= j120/90;
Ib=j1.33;.

Total Current=Ia+Ib.
I=1.33+j1.33;
Converting back to polar form.
magnitude=
I= √((1.33)^2+(j1.33)^2).
I=1.76;.
θ=tan inv(1.33/1.33);.
θ=45.

So;
I=1.76<45;
Thanks; Hope it works

There are two loads, hence the current passes through both the resistors and common Neutral

V= 120V each phase
1/R= 1/90+1/90=2/90= 1/45
R=45Ω

V = IR.
I = V/R.

Hence neutral Current
I=V/R1+V/R2
=(120/90)+(120/90).

I = 1.33+1.33 = 2.66 amp.
Or I = 120/45= 2.66 Amp.

Whereas Generator is two-phase so the Neutral Current is
IN = I * (difference in Phase angle/Total +ve angle)
the difference in phase=120.

Total angle start from zero then high and zero=180
= 2.66 * (120/180).

In two phase system, phase angle difference between two phase is 90 degree.

So, neutral current will be equal magnitude since the both phase voltage are same but there have phase difference between two.

That is why the answer will be,
For one phase (say I1) = 120/90 = 1.33 amp.

And for another phase (say I2) = 120/90 = 1.33 amp.

Now Neutral current will be vector(phasor) sum of I1 & I2

= sqrt(I1^2+I2^2) = sqrt(1.33^2+1.33^2) = 1.88 amp.

The answer should be 1.88 Amp. The current at any instant (also called as instantenous current) flowing through each phase and through each load ckt. is Inst(1) = 1.33 Amp. This same current at any instant will flow through th neutral wire. However the current in a.c circuit is expressed in terms of its r.m.s value and the relationship between these currents is Irms = sqrt. 2 * Inst = 1.141 * 1.33 = 1.88 A. Hence the answer should be option B

Given 3 phase currents; A, B, and C, we say that neutral current is equal to (A^2 + B^2 + C^2)-(AB + BC + CA).

Therefore, given 2 phase currents, A and B, netrual current is equal to (A^2 + B^2)-(AB)
Ip=Vp/Rp ==>> 120/90 = 1.3333A.

Since A=B, then neutral current is equal to (A^4)-(A^2) = A^2.
1.3333^2 = 1.7777.

Consider two coils which are connected to the terminals A and B respectively. The other end of the coils terminated in neutral point N.

Since, Emf generated in each coil is 120 then current Ian (flows from A to N) = 1.33(120/90) and Ibn = 1.33(120/90).

Hence, the current in neutral wire is 1.33+1.33 = 2.66 A.

Phase difference = (360°/number of phases). Therefore we have 180° phase difference in a 2 phase system. Since the load is resistive and balanced. The load currents will also have 180° phase difference.

Therefore the current in neutral which is phasor sum of load currents is also zero.

In three phase circuit when loss is balanced the neutral current is zero due to phase diff of 120 degree in each phase. So this is 2 phase generator the neutral current can not be zero nor it is sum of the individual current. It should be sqrt 2 times the individual circuit current.

I am correcting answer of @Mr Rasood.

I = 120/90*1.414 = 1.88 A since the currents are 90 degrees out of phase.

Could also be written as I = 1.33 + j 1.33 (rectangular form) = 1.88 A at 45 degrees (polar form).

= 1.88*cos 45.

= 1.88*1/square root (2).

= 1.88/1.41 = 1.33.

I hope answer is D, because.

Here I = v*r1 + v*r2.
I =( 120/(sqr(2)* 90))+(same).
I = 0.88 + 0.88.
I = 1.76A.

Here generator is 2phase ,but one common neutral is here, 2 phase can divided into two 1 phases with common neutral,then current passing through neutral is 1.76 A