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?
1.33 A
1.88 A
2.66 A
1.77 A
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
63 comments Page 4 of 7.

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.

Shankar yadav said:   8 years ago
V = IR.
Total resistance of two resistors(is parallel) is:

45ohms,
Voltage is 120v,
Current is =120/45,
Current is 2.66 amps.
(1)

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.

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.

Amjad moawia said:   9 years ago
There is the two phase.

120*120 = 14400 volt.

90*90 = 8100 ohm.

Hence current (i) = v/r.

= 14400/8100 = 1.77 amp.
(2)

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.

Sachin said:   1 decade ago
Please give proper solution.

That justify answer is 1.77.

Because by calculations answer will come 1.88.

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.

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.

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.


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