41 - AC CIRCUITS PROBLEM 9 |Finding the Impedance and Power of an AC Circuit|

[@risingrookie7]

God bless you ❤🎉

[@SkanCityAcademy_SirJohn]

Amen Amen thanks so much

[@adumark5683]

Nice one ,good work

[@SkanCityAcademy_SirJohn]

Thank you so much for your comment

[@stanleynnajiofor9548]

Well explained..m

[@SkanCityAcademy_SirJohn]

Thanks so much, Stanley

[@richardobrasi7323]

And pls how will you know you have to use complex numbers to solve a particular question

[@SkanCityAcademy_SirJohn]

The impedance z is made up of real and imaginary component. The real component is the resistance and the imaginary component is the reactance as a result of the capacitance or inductance. Hence you need to use complex numbers

[@richardobrasi7323]

Pls what shows that is an RL series circuit

[@SkanCityAcademy_SirJohn]

RL series circuit has a resistor in series with an inductor

[@richardobrasi7323]

@@SkanCityAcademy_SirJohnpls can you show me where the question mentioned the present of an inductor

[@capableoldman3260]

Thank you❤

[@SkanCityAcademy_SirJohn]

Most welcome🤗

[@richardobrasi7323]

Pls can u solve it without using complex numbers

[@SkanCityAcademy_SirJohn]

No, ones there is a capacitance or inductance you cant do away with complex numbers.

[@richardobrasi7323]

But there was nothing like inductor in the question or …. If yes pls show me

[@SkanCityAcademy_SirJohn]

Positive reactance is inductive. Negative reactance is capacitive. So if z = 3 + j4, it means the reactance X is 4 ohms Since its positive it is inductive, Hence X = XL = 4ohms

[@SkanCityAcademy_SirJohn]

So first find the impedance z, then you will know if its positive or negative reactance

[@richardobrasi7323]

@@SkanCityAcademy_SirJohn thank you very much

[@richardobrasi7323]

@@SkanCityAcademy_SirJohn pls how do I find the actual power and reactive power of 1kVA load operating at 0.85 power factor leading

[@Easymoney911]

The noise behind your voice is disturbing

[@SkanCityAcademy_SirJohn]

I Will rectify that in subsequent videos...thank you

[@Easymoney911]

Welcome