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.
Anusha said:
1 decade ago
Give me the formula please.
Mohammed Javed said:
1 decade ago
I think two loading rheostats are there. the current passes through two rheostats.given voltage is 120v(each coil).
v=ir
i=v/r
i=(120/90)+(120/90)=1.33+1.33=2.66 amp.
v=ir
i=v/r
i=(120/90)+(120/90)=1.33+1.33=2.66 amp.
V.aruna said:
1 decade ago
There are two loads, hence the current passes through both the rheostats.
The given voltage is 120v.
For finding current,
V = IR.
I = V/R.
Hence I = (120/90) + (120/90) = 1.33+1.33 = 2.66 amp.
The given voltage is 120v.
For finding current,
V = IR.
I = V/R.
Hence I = (120/90) + (120/90) = 1.33+1.33 = 2.66 amp.
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.
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.
Neha said:
1 decade ago
The answer will be 2.66 because neutral is the sum of line currents.
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.
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).
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.
Ganapathi said:
1 decade ago
In = (120/90)+(120<-90/90).
= 240/90.
= 2.66.
= 240/90.
= 2.66.
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