Electrical Engineering - Three-Phase Systems in Power Applications - Discussion
Discussion Forum : Three-Phase Systems in Power Applications - General Questions (Q.No. 9)
9.
A two-phase generator is connected to two 90 
load resistors. Each coil generates 120 V ac. A common neutral line exists. How much current flows through the common neutral line?
load resistors. Each coil generates 120 V ac. A common neutral line exists. How much current flows through the common neutral line?Discussion:
63 comments Page 5 of 7.
Suvankar said:
1 decade ago
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.
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.
Ashok said:
1 decade ago
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
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
Sunil Shingnapurkar said:
1 decade ago
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.
Ganapathi said:
1 decade ago
In = (120/90)+(120<-90/90).
= 240/90.
= 2.66.
= 240/90.
= 2.66.
Minakshi Khushoo said:
1 decade ago
I agree with @Dawid. The answer should be B. Because the two currents will be in quadrature owing to the given 2 phase supply.
Dawid Roos said:
1 decade ago
The answer should be B;
I = 120/90*1.414 = 1.88A since the currents are 90 degrees out of phase
Could also be written as I = 1.33+j1.33(rectangular form) = 1.88A @ 45 degrees(polar form).
I = 120/90*1.414 = 1.88A since the currents are 90 degrees out of phase
Could also be written as I = 1.33+j1.33(rectangular form) = 1.88A @ 45 degrees(polar form).
Subhankar said:
1 decade ago
The guys saying that for balance load no current follows through the neutral should think that neutral is common here. So answer 2.66.
Neha said:
1 decade ago
The answer will be 2.66 because neutral is the sum of line currents.
Alankar said:
1 decade ago
Here the answer will be 2. 66. The guys saying that for balance load no current follows through the neutral should think that neutral is common here.
Mac. F. Crasto said:
1 decade ago
When the load is balance Zero current flows through the neutral line.since neutral is common to both load.
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