Blog #49 - Another Practice Problem

As part of my problem series in this video, I will be solving another practice problem.

Once I am finished with this series of problems I will be posting them in my Stan store, at this web address…https://stan.store/GVB

In this problem we are being asked to find the value of the inductance L if the impedance of the net work is real at 60 Hz.

In this context, the term "real" means that the impedance of the circuit has no imaginary part.  The term “real” indicates that the reactive impedance is at 0° and can be replaced by a short circuit. Let's look at this in terms of the impedance equation…

Z = 1 Ω and 2 Ω in parallel with the impedances of the inductor and the impedances of the capacitor which is zero… therefore, the reactance impedances must cancel each other, providing zero impedance… or a short circuit around the 2 Ω resistor… that means the term 2||[XL + XC] must be “0” … that can only happen when the term

[xL + XC] is zero.… plugging in the impedances values for the inductor and capacitor…we get… jωL + 1/jωC = 0… which means… jωL = - 1/jωC…in order to get rid of j operator. Let's multiply both sides of the equation by 1/j which gives us… ωL = 1/j(sqrd)ωC…but remember…

j(sqrd) is = -1… therefore… ωL = 1/ωC…dividing both sides of the equation by ω we get… L = 1/ω(sqrd)C. Plugging in actual known values. Remember that ω is in radians so and 60 Hz is equal to 377 radians…so we see that… the value of the inductor is 703.6 microHenries.

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Remember, this video has been brought to you by PSPT, where you will find electrical train training videos when you go to this web address…https://bit.ly/47YB3vh...which will also give you a free copy of my 50 page crib sheets that you can use while viewing any of the courses or just keep handy during your every day work.