AC Through Series RL Circuit – Problem 1 – AC Circuits – Basic Electrical Engineering

AC Through Series RL Circuit – Problem 1 – AC Circuits – Basic Electrical Engineering


hello friends in this video we are going to see how series RL circuit works by taking the simple problem so I have taken a problem based on series RL circuit so in this problem we are supposed to calculate the impedance of the circuit given below in a circuit what are the things given current flowing through the circuit is given as Phi ampere voltage across resistance is 16 volt and voltage across inductor is 12 volt and we are supposed to find out the impedance of this circuit so in series RL circuit first I will draw a voltage triangle so voltage triangle for series other circuit will be like this its a V R along the positive x axis perpendicular to we are taken in anti-clockwise direction I will have V L and resultant of V R and V L is nothing but voltage applied that will make angle Phi with V R so from this I can say voltage applied is nothing but root of V R square plus V L square equal to root of 16 square plus 12 square if I solve I will get voltage applied for the circuit as 20 volt but this is just a magnitude we need to calculate angle Phi also so from this I can say tan inverse of V L upon VR equal to 5 so 5 I will get as tan universe off 12/16 which is nothing but 36.87 degree now I can say voltage applied to this circuit is 20 at an angle 36.87 degree volt its a series RL circuit so voltage leads current by angle Phi therefore I consider current angle as 0 or reference quantity hence voltage applied is 20 at an angle 36.87
42
00:03:02,060 –>00:03:08,840
degree now getting impedance is very easy so impedance Z of a circuit is voltage applied divided by current flowing through the circuit so it is equal to 20 at an angle 36.87 degree divided by current given is 5 ampere 5 at an angle 0 as I said earlier and considering this as a reference phasor so angle is 0 degree so impedance I am getting as 4 at an angle 36.87 degree ohm I can get a separate values of R and XL by converting this into a rectangle form so in rectangle form z equal to 3.2
57
00:03:54,919 –>00:04:10,359 plus J 2 point 4 ohm I need to compare this value of Z with r plus J Excel so that I will get resistance of the circuit as 3.2 ohm and inductive reactance XL as 2.40 ohm if i know the frequency of supply I can get L but right now they have not mentioned anything about the frequency so lets keep our answer up to this only thank you

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