circuit with resistor and inductor

[logistic_guy]

Determine \(\displaystyle \bold{I}_s\) in the circuit shown if the voltage source supplies \(\displaystyle 2.5 \ \text{kW}\) and \(\displaystyle 0.4 \ \text{kVAR}\) (leading).

 

[khansaheb]

Determine \(\displaystyle \bold{I}_s\) in the circuit shown if the voltage source supplies \(\displaystyle 2.5 \ \text{kW}\) and \(\displaystyle 0.4 \ \text{kVAR}\) (leading).

View attachment 39054

show us your effort/s to solve this problem.

 

[barbsabbs]

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[logistic_guy]

For now it's obvious that both currents: the current coming from the current source and the current coming from the voltage source, are entering the the resistor and the inductor.

Let \(\displaystyle \bold{I}_v\) be the current of the voltage source.

And

Let \(\displaystyle \bold{I}_z\) be the current in the resistor and inductor.

Then

\(\displaystyle \bold{I}_z = \bold{I}_s + \bold{I}_v\)

We have three unknowns and we have to figure out a way to find two of them!

Thinking....

 

[logistic_guy]

If you solved simple \(\displaystyle \text{DC}\) circuits, I am sure that you would be familiar with ohm's law:

\(\displaystyle V = I \times R\)

Now even if we have \(\displaystyle \text{AC}\) circuits and combinations of resistors, inductors, and capacitors, the law stays the same, but instead of \(\displaystyle R\), we write \(\displaystyle \bold{Z}\).

where \(\displaystyle \bold{Z}\) is called impedance. The impedance is just the combination of \(\displaystyle R\) and \(\displaystyle L\) (inductor) in our current circuit.

So, we have this formula:

\(\displaystyle \bold{V} = \bold{I}_z \times \bold{Z}\)

This formula will help us solve for \(\displaystyle \bold{I}_z\).

\(\displaystyle \bold{I}_z = \frac{\bold{V}}{\bold{Z}} = \frac{120}{8 + j12} \ \text{A}\)

In the next post, we will try to figure out a way to find the current, \(\displaystyle \bold{I}_v\).

 

[logistic_guy]

The problem has already given us the complex power of the voltage source which is:

\(\displaystyle \bold{S} = P + jQ = 2500 - j400\)

We made \(\displaystyle Q\) negative because in the problem we were given that the reactive power is leading. When the reactive power \(\displaystyle Q\) is leading it is less than zero.

Since \(\displaystyle \bold{S}\) is power it still holds the formula of power which is:

\(\displaystyle \bold{S} = \bold{V}\bold{I}_v^{*}\)

where \(\displaystyle \bold{I}_v^{*}\) is the conjugate of \(\displaystyle \bold{I}_v\).

Then,

\(\displaystyle \bold{I}_v^{*} = \frac{\bold{S}}{\bold{V}} = \frac{2500 - j400}{120\angle 0^{\circ}} = \frac{\sqrt{2500^2 + (-400)^2}\angle\tan^{-1}\frac{-400}{2500}}{120\angle 0^{\circ}} = \frac{2531.8\angle -9.09^{\circ}}{120\angle 0^{\circ}} = 21.1\angle -9.09^{\circ} \ \text{A}\)

\(\displaystyle \bold{I}_v\) is the same as \(\displaystyle \bold{I}_v^{*}\), only the angle has the opposite sign, so:

\(\displaystyle \bold{I}_v = 21.1\angle 9.09^{\circ} \ \text{A}\)

Finally, we have:

\(\displaystyle \bold{I}_s = \bold{I}_z - \bold{I}_v = \frac{120}{8 + j12} - 21.1\angle 9.09^{\circ} = \frac{120\angle 0^{\circ}}{\sqrt{8^2 + 12^2}\angle \tan^{-1}\frac{12}{8}} - 21.1\angle 9.09^{\circ}\)


\(\displaystyle = \frac{120\angle 0^{\circ}}{14.422\angle 56.31^{\circ}} - 21.1\angle 9.09^{\circ} = 8.321\angle -56.31^{\circ} - 21.1\angle 9.09^{\circ}\)


\(\displaystyle = 8.321[\cos -56.31^{\circ} + j\sin -56.31^{\circ}] - 21.1[\cos 9.09^{\circ} + j\sin 9.09^{\circ}]\)


\(\displaystyle = 4.616 - j6.924 - 20.835 - j3.334 = -16.219 - j10.258\)


\(\displaystyle = \sqrt{(-16.219)^2+(-10.258)^2}\angle\tan^{-1}\frac{-10.258}{-16.219} - \pi = 19.19\angle -147.69^{\circ} \ \text{A}\)