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 2 of 7.
NASR said:
10 years ago
Right answer is D because:
1 = 120/90a with angle 0+120/90 with angle 120.
= 1.33+j0-0.665+j1.152.
1 = 120/90a with angle 0+120/90 with angle 120.
= 1.33+j0-0.665+j1.152.
Chandrima said:
10 years ago
Answer should be 1.88 amp.
V/i = 1.33 amp, current flowing through neutral 2*1.33 = 2.66, but RMS value of the current will be 2.66/sqrt2 = 1.88 amp.
V/i = 1.33 amp, current flowing through neutral 2*1.33 = 2.66, but RMS value of the current will be 2.66/sqrt2 = 1.88 amp.
Vjagannadharao said:
10 years ago
Resistance will be 45.
And I = V/R.
V = 40, i = 120.
I = 120/45 = 2. 66 A a write method.
Current neutral wire is 1.33+1.33 = 2.66 A.
Then I = V/R = 2.66 A.
And I = V/R.
V = 40, i = 120.
I = 120/45 = 2. 66 A a write method.
Current neutral wire is 1.33+1.33 = 2.66 A.
Then I = V/R = 2.66 A.
Aditya bakshi said:
9 years ago
I think the answer is c, if the answer is D please give me the correct explanation.
Somnath said:
1 decade ago
I think it's 2.66.
2 phase generator. So connect load of 2*90 ohm resistance in parallel.
Resultant resistance will be 45.
Then I = V/R.
= 120/45 = 2.66 A.
2 phase generator. So connect load of 2*90 ohm resistance in parallel.
Resultant resistance will be 45.
Then I = V/R.
= 120/45 = 2.66 A.
Manivel said:
1 decade ago
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.
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.
KeerthiRaj said:
1 decade ago
@Mr Parveen Kumar and Mr Sandeep Wasnik:
Although your ideas were wrong, but I think you people even don't know how to calculate phase to phase voltage, if you don't know please keep quiet, don't post wrong things.
Although your ideas were wrong, but I think you people even don't know how to calculate phase to phase voltage, if you don't know please keep quiet, don't post wrong things.
Sayed Tousif Pasha said:
1 decade ago
Whether the load is in series or parallel.
Hami said:
1 decade ago
I think 1.88 A is right answer. Because there is common neutral wire which means 1.33 A. After multiplying with 1.141, we get 1.88 A.
Rachappa said:
9 years ago
Is there any formula to find this answer?
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