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.
Durga prasad said:
1 decade ago
Wrong answer...correct answer is (1.88) I in neutral= (120/90)+((120/90)<90)=1.88 magnitude
K.n.lakshmi said:
1 decade ago
I=V/R
2 RESISTORS ARE THERE.
NEUTRAL CURRENT=120/9+120/9==2.66
I THINK THE ANSWER IS C=2.66A
2 RESISTORS ARE THERE.
NEUTRAL CURRENT=120/9+120/9==2.66
I THINK THE ANSWER IS C=2.66A
Nitin said:
3 years ago
The 2 resistors are connected in parallel or series, the result (I) may vary accordingly.
Sukhvinder singh said:
1 decade ago
This is wrong the answer is c. The neutral will have sum of both the line currents.
Aditya bakshi said:
9 years ago
I think the answer is c, if the answer is D please give me the correct explanation.
Kalai said:
10 years ago
Parallel connection, voltage same v = 120 v, R = 45 v, I = 120/45 = 2.66 A.
Neha said:
1 decade ago
The answer will be 2.66 because neutral is the sum of line currents.
Awesome said:
9 years ago
@Amjad.
How have you done this. Is there any formula like that?
How have you done this. Is there any formula like that?
Phantom said:
3 years ago
The right answer is option *C*
120/90 = 4/3.
4/3+1.3 = 2.63.
120/90 = 4/3.
4/3+1.3 = 2.63.
Suryabai said:
9 years ago
Which is the correct answer? Please tell the correct option.
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