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9.8 RL-Circuits
Earlier we have studied RC circuits, a circuit which consists of a resistor and a capacitor. In those circuits, we have seen that the current flowing through the circuit is not constant, and it changes exponentially.
Now we’re going to consider an electric circuit which consists of an inductor and a resistor. Let’s say we have an electromotive force. One terminal of this force battery is connected to a two way switch with a and b terminals, and then the switch is connected to a resistor with a resistance of R, and it is connected in a series to an inductor with an inductance of L. The circuit is completed and the branch b is an open branch initially.
Let’s say we turn the switch to a position, or throw the switch to a. If we do that, we’re going to have a closed loop for the charges to go, so we’re going to start with current i emerging, flowing through this path, through the resistor and through the inductor from positive towards the negative and of the power supply.
Now as soon as we turn the switch on, of course, we’re not going to reach to this current right away. It will start from 0 and increase towards its maximum value. As it makes this increase, then we will have a change in current, therefore, we’re going to end up with a self-induced EMF through this inductor. Because as this current flows through the inductor, as it increases, the associated magnetic field that it generates will also increase, and since the current is changing, we’re going to end up with a self-inducted electromotive force.
If we call this phase as “rise of current phase” and write down the loop equation by choosing path to triggers the circuit in a clockwise direction, the EMF arrow is pointing from negative to positive like this. If we start at any point in the circuit, let us say we just start right before the electromotive force, once we close the EMF in the direction of EMF arrow, then the potential increases by ε volts.
Then moving on, crossing the resistance in the direction of flow of current, the potential will decrease by i times R. Now, crossing the inductor, we’re going to end up with a self-induced EMF during the rise of current, and that self-induced EMF is going to show up such that it will try to oppose the increase of this current. Therefore, it’s going to behave as if it is opposing the original electromotive of force, or trying to generate a current flowing in opposite direction to the direction of this current. Therefore, it is going to cause a self-induced electromotive force, and that will be equal to minus, due to this position, L times di over dt, and that is the self-induced electromotive force.
Now the loop is completed. Algebraic sum of the change in electric potential will add up to 0. If we rearrange this expression, we will have L di over dt is going to be equal to ε minus iR. Again, when we look at this expression, like in the case of RC circuits, we’re ending up with a first order linear homogeneous differential equation. As we did in the RC circuit case, here also we do not need to know how to solve differential equations. Simply by rearranging the expression such that the dependent variable current is on one side and independent variable time is on the other side, we can easily solve the current as a function of time by taking the integral of both sides.
So if we leave out the i over dt alone and moving L di over dt to the other side, then it will be positive, and on the other side we will have ε minus iR. From here, if we make cross-multiplication, L di over ε minus iR will be equal to dt, or we can move inductance, L, to the other side also, then the i over ε minus iR will be equal to dt over L. So, doing this, the dependent variable current is left alone, or collected alone, on the left hand side of the equation, whereas the time independent variable is on the right hand side of the equation. Here, ε, electromotive force, resistance and inductance, these quantities are all constants.
Now, by taking the integral of both sides, we can calculate the current as a function of time. Here, though, we have to be careful with the initial conditions of the problem as time, t is equal to 0, this is the instant that we throw the switch to point a. The amount of current flowing through the circuit is 0. As the time proceeds to some t seconds, then we’re going to end up by a certain amount of current flowing through the circuit.
Well, now we’re going to take these integrals, for the left-hand side one, if we say let ε minus iR is equal to u, then –R di will be equal to du. So if we multiply both numerator and denominator by –R the ratio will not change, but we’re going to end up du in the numerator, and u in the denominator. So, if we multiply both numerator and denominator by –R, nothing will change, but –R di is going to be equal to du in terms of the new variable and denominator will be equal to u.
We can actually take this integral and let’s move –R to the other side of the equation, and in that case we will have –R over l times dt, and the integral of dt is going to give us just t. So we will end up with ln of u on the left hand side, and of course, since we changed the variable the boundaries too will change too, evaluate at u1 and u2. Again, we don’t need to calculate u1 and u2 because we will go back to the original variable of i current, and on the right-hand side we will have –R over L t.
Now, going back to the original variable, ln of u, which was ε minus iR, that will be evaluated at 0 and i, which will be equal to –R over L t. Substituting the boundaries, the left-hand side will be equal to ln of ε minus iR for the upper boundary, divided by, substituting in 0 for i, will make second term 0. Therefore we will only be left with ε in the denominator, and that will be equal to –R over L times t.
If we take the inverse natural logarithm of both sides, the left hand side will give us ε minus iR over ε, and on the right hand side we will have e to the minus R over L times t. And if we rearrange this expression, ε minus iR will be equal to ε times e to the –R over L t, leaving iR alone on one side of the equation. iR is going to be equal to ε minus ε e to the –R over L t, and from here, i of t will be equal to, we can take these two terms in ε parenthesis and we’re going to divide that by R, and inside of the parenthesis we will have 1 minus e to the –R over L t. So that expression is going to give us the current as a function of time.
Well, let’s check whether it satisfies our initial conditions. At time t is equal to 0, we will have ε R 1 minus e to the 0 is 1, and indeed, the current in the circuit will be equal to 0, and as time t goes to infinity, in other words, if we wait long enough, in that case we know that the current will approach or go to its maximum value.
From there we can figure out the maximum value for the current, and that is going to be equal to limit as t goes to infinity of this term, ε over R, 1 minus e to the –R over L t. As t goes to infinity, exponential term will go to 0, and therefore, we’re going to end up with imax, the maximum value of current is equal to ε over R. And that’s what we expect because this is directly from the ohms law, the current is equal to voltage divided by the resistance.
And if we go back to our circuit diagram, once the current reaches its maximum value, in other words, if we wait long enough, then the current will reach to its maximum value. It’s not going to be changing anymore at that time, so the i over dt will be 0, and self-induced EMF will drop to 0, and in that case, the L over here is going to be nothing but a piece of wire, part of the circuit, then the potential difference for that moment across the resistor is going to be whatever the potential difference supplied by the power supply, ε volts. That ε volts divided by the current flowing through that resistor will give us the maximum current flowing through this circuit.
Here, again, we have an exponential function, and as I mentioned earlier, in physics, the arguments of these functions, sine, cosine, and exponentials, should be dimensionless, otherwise, there’s something wrong with our calculations. If we then look at the exponent of the exponential term, we know that the time has the dimensions, units of seconds, let’s say, therefore R over L has to have the units of, or dimensions of inverse time, so that they will cancel one another. We’re going to end up with a dimensionless quantity.
At this point we’re going to define a quantity called inductive time constant, and that is equal to, defined as ratio of inductance to the resistance in the LR circuit. If we check the units of τL, writing down the units of inductance and resistance in explicit form, the unit of inductance was henry, unit of resistance of ohm, in an SI unit system. In explicit form, inductance was magnetic flux divided by current. So that was weber per amps divided by, ohm was the ratio of voltage to the current amps. So here, these amps will cancel. We’re going to end up with weber per volt, and Weber in explicit form is tesla meters squared divided by volt.
Moving on now, tesla, the explicit form of tesla was, if we recall that from the magnetic force expression, qv cross B and so the tesla was equal to force, that is newtons, divided by coulombs and divided by velocity, which is meters per second. Therefore we will have seconds per meter over here in explicit form times meters square and divided by volts, and volt was energy per unit charge, joules per coulomb.
So here this meter and that meter squared will cancel and the coulombs will cancel. In the numerator we’re going to end up with newton meters divided by joules, and we also have seconds, but newton meter is nothing but work done and that is therefore nothing but joules in SI unit system. So, inductive time constant is going to be equal to joules per joules times seconds and joules will cancel. Indeed, the unit of L over R, inductance divided by resistance, will have the dimensions of time, which is going to be in seconds, an SI unit system.
Okay. Now, let’s go back to our circuit. In this case, throw the switch to point b. When we do that, we will basically take this electromotive force unit out of the circuit. In other words, we’re going to take this part out of the circuit. Our loop is going to be consisting of, through this path, passing through branch b like this. So in doing so, as soon as we turn the switch off, in other words, move it point b, the current is going to decrease from its maximum value towards 0, because now the power supply’s not existing in the circuit.
In terms of the loop equations, this is really equivalent, just taking ε 0. So the loop equation becomes i times R times L times di over dt is equal to 0. Let’s call this phrase as “decay of current”, and for that part loop equation becomes, ε becomes 0. Let’s say switch is thrown to point b and as a result of that ε becomes 0. Therefore our loop equation becomes equal to R times i plus L di over dt is equal to 0.
Again, applying the separation of variables, keeping L di over dt alone and moving R times i to the other side, and then making a cross multiplication, will give us di over i on the left-hand side, which will be equal to –R over L dt on the right-hand side. Again, by taking the integral of both sides, in this case our initial conditions are such that at time t‘s equal to 0, we’re starting with the maximum current flowing through the circuit. As the time proceeds to some t seconds, we’re going to end up with a certain amount of current left flowing through the circuit.
The left-hand side integral will give us ln of i evaluated at imax and i, and the right-hand side will give us –R over L times t. Again, substituting the boundaries, we will have ln of i over imax is equal to –R over L t. Taking the inverse natural logarithm of both sides, i over imax is going to be equal to e to the –R over L t, and solving for current, as a function of time, we will end up with imax times e to the –R over L t. So, the current behavior during the rise of current we found that i of t‘s equal to imax times 1 minus e to the –t over τL, inductive time constant, and for the decay of current part or phase we found that i of t‘s equal to imax times e to the –t over τL.
Okay. Again, one can easily calculate what happens for both rise and decay of current parts if a one time constant duration of time elapses. For the rise of current, if we wait for one time constant, inductive time constant of time, then we will see that we’re going to have imax times 1 minus e to the -1 and that is equal to 0.63. Therefore, if we wait one inductive time constant of time long, during the rise of the current, 63% of the maximum current will be achieved.
On the other hand, if we wait one time constant of time during the decay phase, then imax times e to the -1, which is going to be equal to 0.37 times imax, indicating that 37% of the maximum current will decay during that process.