HELP need help with these electrical questions about load connection and power factors

Lachlanheath

Joined Oct 20, 2016
2
hi guys i can't seem to get my head around these questions i am new to electrical engineering and can't find the formulas ect any help is much appreciated!

A load connected in star is supplied from a 400V 3 phase system and each arm of the star is a resistance of 10 Ohms.

What is the current in

a) Each arm of the load?

Each line from the source?

A 3 phase 400V supply has a delta connected load with each phase arm having a 10 Ohm resistance. The phase current and line current respectively would be?

A 3 phase 400V supply has a delta connected load with each phase arm having a 10 Ohm resistance.
The total power consumed by the loads will be:

A single phase supply feeding to a load consumes an active power of 40 kW and a reactive power of 30 kVAr. What is the power factor of this load?

tcmtech

Joined Nov 4, 2013
2,867
Sounds suspiciously like homework.

What's your textbooks say on the subject of calculating those values?

Lachlanheath

Joined Oct 20, 2016
2
Sounds suspiciously like homework.

What's your textbooks say on the subject of calculating those values?
Just started doing an online electrical engineering diploma, but am new to all this and I didn't know that they assumed a general knowledge of this stuff
Not a big maths whizz so this is the section im un familiar with
Don't have text books just reading and these are questions I can't find the correct formulas too
And the formulars I have found turned out to be wrong

KeepItSimpleStupid

Joined Mar 4, 2014
5,088
Why can't they use Wye and Delta? They use "stars" and delta. They also use arms for legs, I think.

tcmtech

Joined Nov 4, 2013
2,867
Just started doing an online electrical engineering diploma, but am new to all this and I didn't know that they assumed a general knowledge of this stuff
Not a big maths whizz so this is the section im un familiar with
Don't have text books just reading and these are questions I can't find the correct formulas too
And the formulars I have found turned out to be wrong
I see.

The simplest thing to do is reduce each line down to its basic voltage drop between any two connected and factor things using basic Ohms law for the resistive power loads.

In one application its 10 ohm resistance across a 400 volt source per resistor. In the other is a 10 ohm load across 400 divided by the square root of 3 per resistor.

Power factor is just the true power (wattage) divided by the apparent power (Volt Amperes) KVAr reactive power.

drc_567

Joined Dec 29, 2008
1,156
The so called apparent power is the amount of power that is delivered to or applied to any given electrical load, and has units of volt-amps, or VA. It may be thought of as the hypotenuse of a right triangle. The angle at which this hypotenuse resides, with respect to the real axis, is determined by the resulting height of the reactive components, consisting of positive and negative VARs, or volt-amps-reactive, which form a vector parallel to the imaginary axis. Inductive VARs are positive quantities and capacitive VARs are negative. The algebraic sign of this resulting VAR vector determines the inductive or capacitive nature of the device under consideration. The resistive component of the apparent power triangle lies along the real axis of the complex plane, consumes watts, and is the only component that produces heat, according to Joules Law ... I ^2*R. The algebraic ratio of the VAR vector to the watt vector determines the tangent value of the angle between the hypotenuse and the real power vector. The cosine of this angle is the power factor.
... So, when you see a right angle power triangle, that is what you are looking at.