Power factor

Take 10 minutes to prepare this exercise.

Then, if you lack ideas to begin, look at the given clue and start searching for the solution.

A detailed solution is then proposed to you.

If you have more questions, feel free to ask them on the forum.

An electrical installation is fed by an effective (RMS) voltage \(U_e = 200\; V\).

It consumes power \(P = 12\; kW\).

The frequency is \(f = 50\; Hz\) and effective (RMS) intensity \(I_e=80\; A\).

Question

Knowing that this installation is of the inductive type, calculate the resistance \(R\) and inductance \(L\) which, placed in series and with the same supply, would be equivalent to the installation.

Solution

We can write the equations :

\({U_e} = \sqrt {{R^2} + {{(L\omega )}^2}} \;{I_e}\)

And :

\(P = {U_e}{I_e}\cos \varphi = RI_e^2\)

We can deduce :

\(R = \frac{P}{{I_e^2}} = 1,875\;\Omega\)

Then :

\(L = \frac{{\sqrt {{{({U_e}/{I_e})}^2} - {R^2}} }}{\omega } = 5,26\;mH\)

Question

Calculate the capacitance \(C\) to place in parallel to the installation to meet the power factor value \(0,9\).

Solution

Equivalent admittance is calculated, in complex notation :

\(\;{\underline y _{eq}} = jC\omega + \frac{1}{{R + jL\omega }} = \frac{R}{{({R^2} + {L^2}{\omega ^2})}} + j\left[ {C\omega - \frac{{L\omega }}{{({R^2} + {L^2}{\omega ^2})}}} \right]\)

Can be calculated :

\(\tan \varphi = \frac{{{R^2} + {L^2}{\omega ^2}}}{R}C\omega - \frac{{L\omega }}{R}\)

Then the value of \(C\) :

\(C = \frac{{Rtan\varphi + L\omega }}{{{R^2} + {L^2}{\omega ^2}}}\)

With \(cos(\varphi)=0,9\), it comes :

\(C=1,3\;mF\) or \(C=0,38\;mF\)