Hi I have the following question:

The following are connected in delta to a 600 Volt three-phase supply.

• A-B has a 20 Amp load with a power factor of .9 lag.
• B-C has a 30 Amp load with a power factor of 1
• C-A has a 40 Amp load with a power factor of .9 lag.

Calculate the current in line A:
Calculate the current in line B:
Calculate the current in line C:

I have drawn out the circuit in delta form to try and understand where the current is going but without the neutral of a star system I am confused when using kirchoffs law.

I know that two of the currents are not in phase so I have:

A-B 20A /_ 25.85 deg
B-C 30A /_ 0 deg
C-A 40A /_ 25.85 deg

I can't use the formula I line = I phase X √3 as the loads are not balanced. So I am unsure how the current is distributed between the phases.

 

Any of the line currents would be the vector summation of the two associated delta currents. So Ia=Iab-Ica and so forth.

 

So to find the current in line B I would do the following?

Ib= Iba-Ibc

20/_ 35.85 deg - 30/_ 0 deg
18+j8.717 - 30+j0 = -12-j8.717 = 14.83/_ -144 which isnt the correct answer.

Think im missing something here.

 

But

Ib=Ibc-Iab

etc....

 

Also how could a lagging pf current give a +ve imaginary part.

 

I don't understand, the correct answers I have are B = 37.2A and C = 67A.

Would you be able to show me the vector sums so I can see clearly how i've gone wrong?

 

Am I missing the fact that all of the phases are 120 deg out of phase with eachother? Do I need to take this into account?

 

Am I missing the fact that all of the phases are 120 deg out of phase with eachother? Do I need to take this into account?

Yes - absolutely you must remember the individual line voltage phase displacements are an important consideration.

 

But how do I add currents which are normally 120 deg out of phase with eachother when two of them are out of phase with themselves?

 

I'm giving my solution since we've hashed it out enough. Don't you think?

Here's the way I see it. Given Iab, Ibc, & Ica, find Ia, Ib, & Ic.

Given:

Iab = 20A @ pf 0.9 lag
Ibc = 30A @ pf 0.0
Ica = 40A @ pf 0.9 lag

Since all currents will be in Amperes, I will drop the "A."

Iab = 20.0 ang(-25.85 deg)
Ibc = 30.0 ang(0.00 deg)
Ica = 40.0 ang(-25.85 deg)

I shall define Ia as the current entering node A, etc. As t_n_k said (at least for Ib):

Ia = Iab+Iac = Iab-Ica
Ib = Ibc+Iba = Ibc-Iab
Ic = Ica+Icb = Ica-Ibc

Ia = Iab-Ica = -18.00 + j8.72 = 20.0 ang(154.2 deg)
Ib = Ibc-Iab = +12.00 + j8.72 = 14.83 ang(36.0 deg)
Ic = Ica-Ibc = +6.00 - j14.44 = 18.00 ang(-71.0 deg)

I think you'll find if you sum Ia, Ib, & Ic, you will get zero.

If you draw a phasor diagram, you'll see the 3 currents, Ia, Ib, & Ic are almost 120 degrees apart. I say "almost," because A & B are ~118 deg apart, B & C are ~107 deg apart, and C & A are ~135 deg apart.

 

Thanks for the reply this is what I understand according to hat tnk has said. I do not understand why the textbook answers are line currents B=37.2A and C=67A?

 

Thanks for the reply this is what I understand according to hat tnk has said. I do not understand why the textbook answers are line currents B=37.2A and C=67A?

Is that the entire problem? I noticed the supply voltage was given as 600 volts. Is there something else given? What does it say for the A current?

 

That is the entire problem in the first post the question only asked to work out lines A and B so I've been spending alot of time trying to figure out the answers given.

 

I'm giving my solution since we've hashed it out enough. Don't you think?

Here's the way I see it. Given Iab, Ibc, & Ica, find Ia, Ib, & Ic.

Given:

Iab = 20A @ pf 0.9 lag
Ibc = 30A @ pf 0.0
Ica = 40A @ pf 0.9 lag

Since all currents will be in Amperes, I will drop the "A."

Iab = 20.0 ang(-25.85 deg)
Ibc = 30.0 ang(0.00 deg)
Ica = 40.0 ang(-25.85 deg)

I shall define Ia as the current entering node A, etc. As t_n_k said (at least for Ib):

Ia = Iab+Iac = Iab-Ica
Ib = Ibc+Iba = Ibc-Iab
Ic = Ica+Icb = Ica-Ibc

Ia = Iab-Ica = -18.00 + j8.72 = 20.0 ang(154.2 deg)
Ib = Ibc-Iab = +12.00 + j8.72 = 14.83 ang(36.0 deg)
Ic = Ica-Ibc = +6.00 - j14.44 = 18.00 ang(-71.0 deg)

I think you'll find if you sum Ia, Ib, & Ic, you will get zero.

If you draw a phasor diagram, you'll see the 3 currents, Ia, Ib, & Ic are almost 120 degrees apart. I say "almost," because A & B are ~118 deg apart, B & C are ~107 deg apart, and C & A are ~135 deg apart.

Unfortunately this solution is incorrect - the Op's answers are good.

 

What did I do wrong?

 

I have been on this same question for 2 days now and still cannot come up with the answers. This is as close as I can get:

B = 20/_ -25.84 - 30/_ 120 = 18+j8.717 - -15+j25.98 = 33+j34.697 = 47.88 /_46.435

 

OK so i've finally realised what I did wrong.

B = 20/_ -25.84 - 30/_ 120 = 18+j8.717 - -15+j25.98 = 33-j17.26 = 37.24/_-27.61A

FINALLY.

 

C= 30/_ 0 - 40/_ 145.84 = 30+j0 - -33.09+j22.46 = 63.09-j22.46 = 66.96/_19.59A

 

Correct answers are

\( I_a=52.92 \angle{-66.74^o} \ A \)

\( I_b=37.2 \angle{ -152.5^o} \ A\)

\( I_c=66.9 \angle{ 79.58^o} \ A \)

 

how did you get those different angles?