Electrical Engineering: Ch 8: RC & RL Circuits (13 of 43) Current=? in RL Circuit: Ex. 1

welcome to dialect online now let's take a look at an example of how to figure out the current through an inductor in an RL circuit well first of all what we have here is we have a 10 volt source connected to a 2 ohm resistor and a switch and the switch will open at time equals zero which means that the switch is closed before that moment in time and what we need to do then is look at the steady-state situation of this circuit when the switch is closed but other words when time is less than zero so at the moment that time is less than zero what does that circuit look like well the switch will be closed current will be flowing here it will split across the branch get to this branch point but in the steady-state situation a pure inductor does not offer any resistance or opposition to the current flow that means that the inductor acts like a short circuit and all the current that goes to the forum resistor will go through the inductor and none of it will go through the 16 ohm resistor so we can simply remove that 16 ohm resistor have that an open and make this a short and then let's see what the circuit that looks like we still have the 10 volt source we have the 2 ohm resistor we have the 4 ohm resistor then the 16 ohm resistor is gone the inductors are short and we still have the 12 volt resistor here like that so this is the 2 ohm resistor this here is a 12-1 resistor and the 4 ohm resistor now what we need to do is to figure out how much current flows through the circuit we need to simplify that even more and find the equivalent circuit of this notice that these are two resistors in parallel from this branch point to this branch point so the equivalent resistance of these two is the product over the sum so that would be 12 times 4 divided by 12 plus 4 which is equal to 48 divided by 16 which is equal to 3 these two resistors then form a single equivalent resistance of 3 ohms added to the 2 ohms here that gives us a 5 ohm resistance to the circuit for this 10 volt source so the equivalent circuit right here would be a 10 volt source and a single 5 ohm resistor and therefore the current leaving the 10 volt source in the circuit prior to opening up the switch while the inductor therefore is acting like a short will be simply using Ohm's law I equals V over R which is 10 volts over 5 ohms which is 2 amps so the initial current to the circuit as this will be 2 amps how much of that will flow to the forum resistor and a much of it will flow to the 12 ohm resistor after all we want to know how much current flows through the inductor here that's what we're trying to find is a function of time is equal to question mark for the range that contains the inductor so what we're going to do is we're going to say that the current I to the 4 ohm resistor is equal to the AI from the source times the ratio of the resistance of the other branch divided by the total 4 plus 12 and so that means that I to the 4 ohm resistor is equal to 2 amps times the ratio of 12 over 16 which is 3/4 and therefore I to the 4 ohm resistor is equal to 1.5 amps and that will be a current getting to this branch point right here now what we're going to do is we're going to open up the switch here at time equals 0 so what happens now when time is greater than zero well we're going to redraw the circuit when the switch is open this part of the circuit no longer plays a role because we don't have a continuous path here we only have to draw this part of the circuit so while we have Dennis we have the following circuits we have a 12-1 resistor 4 ohm resistor 16 ohm and then we have the inductor 12 ohms here that is 4 and here we have 16 and to Henry's alright now remember the initial current flowing through here let's call that I initial that's equal to 1.5 amps right here that's this current right here to the 4 ohm resistor now what happens when this situation happens I'll notice that all of that current initially went through the inductor because the inductor acts as a short after the curtain has been flowing for a while because the inductor only opposes a change in the current and none of it will have been going to the 16 ohm resistor so the initial current through the inductor will indeed be to 1.5 amps now we're going to redraw that circuit as a single equivalent circuit of an inductor and a single equivalent resistance so when we do that we get the following circuit notice that these two are combined into a single 16 ohm resistor in parallel to this 16 ohm resistor and of course when two resistors are in parallel and they have the same value they can be replaced by a single resistor of half the value so the equivalent resistance then would be 8 ohms and here we have the single 2 Henry inductor that would be the equivalent resistance of the circuit after the switch is opened from that we should be able to find our time constant the time constant is defined as the inductance divided by the equivalent resistance so this gauge that would be to Henry's divided by 8 ohms which is equal to 0.25 seconds and then finally we can find the equation for the current I as a function of time for the current going to the inductor is equal to the initial current times e to the minus T over tau now the initial current as we realize is going to be an initial current here going to the 4 ohm resistor because all of it goes to inductor none of it initially through the 16 ohm resistor so therefore this would be equal to 1.5 amps times e to the minus T over tau and tau is 0.25 seconds if we leave off the units and want to simplify that equation we could say that as a function of time is equal to 1.5 amps times e to the minus 4t and that would be the simplified form of the equation telling us what the current is through the inductor so again the methodology is you take your initial circuit you look at it before the switch is opened at time being less than zero you draw the equivalent circuit with a switch closed which means that this becomes an a short circuit and this becomes an open circuit you find the current from that current you figure out what portion the current goes to the 44 ohm resistor the rest goes to the 12 ohm resistor then you draw the equivalent circuit here after the switch is opened which means this part of circuit is removed you find the time constant using the equivalent circuit and then you use the general equation of the current once you find the initial current and you find the time constant you find the equation describing the current through the inductor and that's how it's done

Channel: Michel van Biezen

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Keywords: ilectureonline, ilectureonline.com, Mike, Mike van Biezen, van Biezen, ilecture, ilecture online, Electrical Engineering, RC & RL Circuits, Ch 8, Chapter 6, Introduction, First Order, Voltage, Current, Differential Equation, Capacitor, Resistor, Inductor, General Equation, Initial Charge, Voltage Divider, Parallel, Equilvalent, Circuit, Kirchhoff, Voltage Law, Ohm's Law, L/R Time Constant, RL Circuit, Finding the Current, Current=?

Id: mGINjDi3c_Q

Length: 7min 58sec (478 seconds)

Published: Sun Sep 25 2016