# Generated current in a $3\times450$ VAC $60$ Hz $1150$ kVA generator

Let's say I've a $3\times450$ VAC $60$ Hz $1150$ kVA generator. How much current will there be generated when it runs at maximum capacity?

MY WORK:

The total Active power over a three-fase system is given by:

$$P=\sqrt{3}UI\cos\varphi\tag1$$

Where $U$ is the line voltage and $I$ is the line current.

The total Reactive power is given by:

$$Q=\sqrt{3}UI\sin\varphi\tag2$$

Where $U$ is the line voltage and $I$ is the line current.

The total Complex power is given by:

$$S=\sqrt{3}UI\tag3$$

Where $U$ is the line voltage and $I$ is the line current.

So, I know that:

$$1150\times1000=\sqrt{3}UI\tag3$$

But what is the line voltage? Is it $450$ volts? So, then we get: $1150\times1000=\sqrt{3}\times450I$ which give that the line current equals $I=\frac{1150\times1000}{\sqrt{3}\times450}\approx1475.5$ A.

• Huh, what? "Well, also that is not possible, because I get 3740 A and that current is two times bigger than the fuse that is used!" What fuse? Is this real or homework? – Tyler Dec 24 '17 at 18:10
• @Tyler It is a real schematic for a project I'm working on – asdasdw Dec 24 '17 at 18:10
• The fuse does not need to be sized for the maximum that the system can produce. – Tyler Dec 24 '17 at 18:26
• Anyone messing around with MW power levels shouldn't have to ask here. Conversely, anyone asking here shouldn't be messing with MW power levels. – Olin Lathrop Dec 25 '17 at 14:32

Let $V$ be the RMS power between each of the phases and the neutral, and $V_p$ be its amplitude: $V_p = \sqrt{2}V$.

Let $U$ be the RMS power between two phases, and $U_p$ be its amplitude:

$U = \sqrt{3}V$, $U_p = \sqrt{2}U$.

Finally, let $P = 1150 kVA$ be the apparent total power available from the generator: it is the sum of the powers of each dipole.

It is unclear in the question whose of these voltages is meant by 450V AC. I suppose here that that this is RMS voltage between the phases and the neutral, that is $V$: $V = 450$.

Next, the question of "how much current it runs" is ill-posed. A current flows through two terminals, but here there are 4 terminals. Is it the current flowing through the phase and the neutral? the current flowing through 2 phases (if connected), the total current flowing through the terminals in a balanced "Delta" configuration? or the total current flowing through the terminals in a balanced "Wye" configuration? This need be specified, because the answer is different in each case.

Assuming for example you demand the maximal current between two phases: Then

$I_{max}\ (rms) = P / U$, which is the value you have to use for your "fuse" between 2 phases. The maximum current amplitude is $\sqrt{2}P/U$

If you demand the maximal current between a phase and the neutral, then $I_{max}\ (rms) = P / V$ and the maximum amplitude is $\sqrt{2}P/V$.

The other cases are processed similarly.

• Well, this is not true. Because the voltage and current are always the RMS ones (not the maximum/amplitude). P is in Watt (W) not in VA. Since I only not the things (of the generator) that are listed in my first line of the question I can not use anything about the impedance. Next the current is big and can never be as low as $3.74$ A!!! – asdasdw Dec 24 '17 at 17:26
• Well, also that is not possible, because I get $3740$ A and that current is two times bigger than the fuse that is used! – asdasdw Dec 24 '17 at 17:35
• What is wrong, when I use the formula: $$\text{S}=\sqrt{3}\cdot\text{U}\cdot\text{I}$$ Where $\text{U}$ is the effective line voltage and $\text{I}$ is the effective line current? Or I can write $$\text{S}=\sqrt{3}\cdot\frac{\hat{\text{U}}}{\sqrt{2}}\cdot\frac{\hat{\text{I}}}{\sqrt{2}}=$$ $$\frac{\sqrt{3}}{2}\cdot\hat{\text{U}}\cdot\hat{\text{I}}$$ – asdasdw Dec 24 '17 at 17:44
• My book tells me the formula's I used! – asdasdw Dec 24 '17 at 17:57
• Where did you get that? – asdasdw Dec 24 '17 at 17:59