Electrical Engineering - Three-Phase Systems in Power Applications - Discussion

I=V/R
2 RESISTORS ARE THERE.
NEUTRAL CURRENT=120/9+120/9==2.66
I THINK THE ANSWER IS C=2.66A

The answer should be 1.88 Amp. The current at any instant (also called as instantenous current) flowing through each phase and through each load ckt. is Inst(1) = 1.33 Amp. This same current at any instant will flow through th neutral wire. However the current in a.c circuit is expressed in terms of its r.m.s value and the relationship between these currents is Irms = sqrt. 2 * Inst = 1.141 * 1.33 = 1.88 A. Hence the answer should be option B

Give me the formula please.

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.

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.

When the load is balance Zero current flows through the neutral line.since neutral is common to both load.

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.

The answer will be 2.66 because neutral is the sum of line currents.

Can you give the circuit diagrm?

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).