# Calculate Currents & Power in 3-Phase Systems w/ Star-Connected Load

• Mathn00b!
In summary, the homework statement says that a balanced three phase 4 wire supply has r.m.s. line voltage of 415V and supplies a star-connected load made of 3 impedances. The problem asks for the current phase and neutral currents, as well as the total power dissipation. The student has found an error in the angle calculated for Zc.
Mathn00b!

## Homework Statement

Hello, I am given the following problem.
A balanced three phase 4 wire supply has r.m.s. line voltage of 415V and supplies a star-connected load made of 3 impedances:
Za = 10 - j10; Zb = 10 + j10; Zc = 0 + j8;

Calculate the current phase and neutral currents, also the total power dissipation. [NOTE: Assume that Va is drawn vertically up, i.e. its phase angle is 90]

## Homework Equations

I've used Vline = √3 x Vphase
Also, P = Ea x I a x cos(Φa) + Eb x I b x cos(Φb) + Ec x I c x cos(Φc)

## The Attempt at a Solution

Please find the attached file, as my solution to the problem. As it could be seen the answer is quite different. In fact their answer is just the sum of the 1st two bits of the power, ignoring the last one. I assume they used cos(90) to get 0 and hence lower power. If this is the case, then the Zc should look like 8∠90° .
But isn't it found from the equation : Zc = √02 + 82 ∠tan-1 (8/0) ?

#### Attachments

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Mathn00b! said:
Please find the attached file, as my solution to the problem.
Error: The requested attachment could not be found.

You must find the center voltage, En, at the center of the star-load.

Then:

Ia = ( Ea - En ) / Za , etc.

( I think? I cannot see if the center of the star-load is connected to neutral )

Last edited:
Strange, just sent it to one of my friends, he was able to open it. Will post a photobucket link just in a sec.
I am asked to find the neutral current, which could be found from the sum of all currents, no?

I've been able to open it now (error disappeared).

You may argue: If neutral is not connected to center of the star-load, then In = 0.
But I've found an error: Zc = ( 0 + j8 ) = 8 / 90°

Could you please, explain to me how you found it to be 8 ∠90°? What formula did you use? Isnt't it the same I used at the 1st post of the thread Zc = √02 + 82 ∠tan-1 (8/0) ? and tan-1 (0) = 0.
I presume it is connected since I've been asked to find it.

I'm not using any formula: Draw the vector ( 0 + j8 ) in the complex plane and measure the angle ( or you can see it by intuition ).

( Contrary: ( 8 + j0 ) = 8 / 0° )

The angle you've associated with Zc is incorrect. As a result the angle calculated for ##I_c## is incorrect.
The angle calculated for ##I_a## is incorrect.

gneill said:
The angle you've associated with Zc is incorrect. As a result the angle calculated for ##I_c## is incorrect.
The angle calculated for ##I_a## is incorrect.
Yes you are right, it should be 135, but really don't understand the Zc's angle. Should I do it always like that? Drawing a vector, when the real part is zero?

Mathn00b! said:
Yes you are right, it should be 135, but really don't understand the Zc's angle. Should I do it always like that? Drawing a vector, when the real part is zero?
It certainly can be helpful to sketch a vector for a complex value in order to get an idea of what the angle should be (such as the quadrant it lies in). But you should know that any purely imaginary number must have an angle of either + or - 90 degrees.

Thank you both for the replies!

## 1. How do you calculate the total current in a 3-phase system with a star-connected load?

To calculate the total current in a 3-phase system with a star-connected load, you can use the formula I = √3 * V / Z, where I is the total current, V is the line voltage, and Z is the total impedance of the load. This formula takes into account the fact that in a 3-phase system, the three phases are 120 degrees out of phase with each other, resulting in a higher total current compared to a single-phase system.

## 2. What is the power factor in a 3-phase system with a star-connected load?

The power factor in a 3-phase system with a star-connected load is the ratio of the real power (measured in watts) to the apparent power (measured in volt-amperes). It is expressed as a decimal or a percentage and indicates the efficiency of the system in converting electrical energy into useful work.

## 3. How do you calculate the total power in a 3-phase system with a star-connected load?

The total power in a 3-phase system with a star-connected load can be calculated by multiplying the line voltage by the total current by the power factor. This gives you the real power in watts. To calculate the apparent power, you can use the formula S = √3 * V * I, where S is the apparent power. The total power can also be calculated by multiplying the line voltage by the line current.

## 4. Can you explain the difference between single-phase and 3-phase systems?

In a single-phase system, there is only one alternating current (AC) waveform, whereas in a 3-phase system, there are three AC waveforms that are 120 degrees out of phase with each other. This allows for a more efficient distribution of power and allows for higher power loads to be carried. 3-phase systems are typically used in industrial and commercial settings, while single-phase systems are used in residential settings.

## 5. How do you convert a star-connected load to a delta-connected load in a 3-phase system?

To convert a star-connected load to a delta-connected load in a 3-phase system, you can use the formula Zd = Zs / √3, where Zd is the delta impedance and Zs is the star impedance. This conversion is commonly used in electrical systems to balance the load across all three phases and improve the efficiency of the system.

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