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 3 of 7.
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.
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
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.
Ram T Shelke said:
1 decade ago
In two phase system assuming phase diff equal to 90, the resultant voltage across a given resistor,
SQ ROOT(120*120+120*120) =170V.
Therefore current through neutral = current through given resistor = 170/90 = 1.888A.
SQ ROOT(120*120+120*120) =170V.
Therefore current through neutral = current through given resistor = 170/90 = 1.888A.
Khizar hayat said:
1 decade ago
It is double phase you have simply to do is,
v = 120 and R = 90.
(120.120) = 14400.
(90.90) = 1800.
We have i = v/R = 14400/1800 = 1.77.
D is the correct answer.
v = 120 and R = 90.
(120.120) = 14400.
(90.90) = 1800.
We have i = v/R = 14400/1800 = 1.77.
D is the correct answer.
Sachin said:
1 decade ago
Please give proper solution.
That justify answer is 1.77.
Because by calculations answer will come 1.88.
That justify answer is 1.77.
Because by calculations answer will come 1.88.
Parveen kumar said:
1 decade ago
There is 2 phase so voltage 120x120 = 14400.
2R = 90X90 = 8100.
I = V/R.
S0 = 14400/8100 = 1.77.
2R = 90X90 = 8100.
I = V/R.
S0 = 14400/8100 = 1.77.
Sandeep wasnik said:
1 decade ago
There are the two phase and hence,
120*120 = 14400 volt.
90*90 = 8100 ohm.
Hence current (i) = v/r.
= 14400/8100 = 1.77 amp.
120*120 = 14400 volt.
90*90 = 8100 ohm.
Hence current (i) = v/r.
= 14400/8100 = 1.77 amp.
RUPESH S BORADE said:
1 decade ago
If 2 resistance are in parallel then.
(90*90)/180 = 45.
I = V/R.
I = 120/45.
I = 2.66.
That's answer is C.
(90*90)/180 = 45.
I = V/R.
I = 120/45.
I = 2.66.
That's answer is C.
Saheb said:
1 decade ago
Suppose, open delta connection neutral i = i/3^1/2.
Here, i = (120/90) + (120/90) = 2.66 A.
Neutral i = 2.66/3^1/2 = 1.54~1.77 A.
Here, i = (120/90) + (120/90) = 2.66 A.
Neutral i = 2.66/3^1/2 = 1.54~1.77 A.
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