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**Example- 6.14 RC-Circuits**

All right. Let’s go ahead and consider an example related to the RC circuits. Let’s see. Consider the following RC circuit. In this circuit, we have an electromotive force which provides *ε* volts of potential difference and its positive terminal is connected to a resistor with a resistance of *R*. Let’s see, we have a switch here and that switch is connected to a junction point which leads to another resistor with a resistance of *R*, and also it leads to, again, another resistance with resistance of *R*, which is connected in series to a capacitor with a capacitance of *C*.

Then, the circuit is completed by connecting all these leads to the other terminal of the power supply. First, we’d like to determine the current through each resistor at time *t* is equal to 0. Therefore, in part a), let’s say, we’d like to figure out the current, determine the current through each resistor at time *t* is equal to 0.

Well, if we consider the circuits at time *t* is equal to 0, which is the instant that we close the switch, and at that instant also, let’s say at *t* is equal to 0, the switch is closed and the capacitor is fully uncharged. Therefore, at this moment, when we turn the switch to on position, we can treat the capacitor, since it is not charged at all, we can treat the capacitor like a piece of wire over here. In other words, if we just place an ammeter before this resistance and ammeter after the capacitor, we will read exactly the same current value flowing through the circuit.

Of course, that current value for the branch in which the capacitor is located will decrease eventually as the capacitor is getting charged. But at time *t* is equal to 0, the capacitor over here will behave as if there is no capacitor over there, that we just have a straight wire here. In that case, we can redraw the circuit by saying that at time *t* is equal to 0, at that moment, the circuit is equivalent to this form. Here we have, again, *ε* volts provided by the EMF and resistance *R* here, *R* here, *R* there.

Let’s assume that, at this instant, current *i* emerges from the power supply and flows through the resistance *R* over here. At this point, it will see two available paths through this junction point. Some fraction will go through this *R*, let’s say it will go down, as *i*1, and then the remaining fraction will go through this branch as *i*2. These two currents will join together at this junction point, and they will become *i* again. So therefore, we will have these currents flowing through these appropriate branches.

If we choose clockwise direction to traverse these loops, we can therefore write down the loop equations for the loop 1 and loop 2 as, starting just right before the electromotive force, crossing it from negative to positive direction, in other words, in the direction of EMF arrow, then we will experience an increase in potential by *ε* volts. Then we will cross this resistance as we go in clockwise direction, in the direction of flow of current, therefore the potential will decrease by –*i* times *R*. Now, going down along this branch, we’re going to be crossing this resistance in the direction of flow of current, therefore the potential will further decrease by –*i* times *R*. Moving on in clockwise direction, the loop will be completed. The sum of these algebraic changes in potential will add up to 0.

For the second loop, we can write down similar type of equation. Starting at a specific point here, let’s say if we start from this junction point and moving on, we will cross this resistance in the direction of flow of current as we cross it in clockwise direction. Therefore the potential will decrease by –*i*2 times *R*. Then we will cross this resistance, now, in opposite direction to the direction of flow of current. Therefore, the potential will increase by *i*1 times *R*, and loop will be completed. Sum of the changes in potential, therefore, will add up to 0 here.

Now we will express the junction equations, and we can consider either this point or the other one, they will both give us the same equation. Here, the current going into the junction is *i*, and currents which are coming out of that junction are the *i*1 and *i*2. Therefore, *i* will be equal to *i*1 plus *i*2.

Okay. Let’s give some number to these equation, number 1 and number 2 and number 3. If we substitute equation number 3 in equation number 1, let’s see, from 3 and 1, we will have *ε* minus, for *i* we will substitute *i*1 plus *i*2, times *R*, plus *i*1*R* will be equal to 0. Moving on one more step, here we will have . . . actually, here, this is minus *i*1*R*. If we open the parentheses, we will have minus *i*1*R* here, another minus *i*1*R* will makes 2 minus 2 *i*1*R* minus *i*2*R* is equal to 0.

The other equation we have is the equation number 2, and that is *i*1*R* is equal to *i*2*R*. So, from equation number 2, we can say that *i*1*R* is equal to *i*2*R*. Therefore, we can replace *i*2*R* with *i*1*R*, then *ε* minus 2*i*1*R* minus *i*1*R* will be equal to 0, and that is going to give us *ε* minus 3*i*1*R* is equal to 0. Finally, solving for *i*1, we’re going to end up with *ε* over 3*R*.

All right. Once we obtain the *i*1, we see that since *i*1 is equal to *i*2 from this equation number 2, *i*2 will also be equal to *ε* over 3*R*, and since *i* is equal to *i*1 plus *i*2, that is going to give us, if you add these two, 2*ε* over 3*R*. So these are the current values at time *t* is equal to 0, in other words, at the instant that we turn the switch in on position.

Now, in part b) of this problem, let’s try to determine the current through each resistor as *t* goes to infinity. To do that, let’s go back to our original circuit over here. When we turn the switch on position at time *t* is equal to 0, at that instant, as I mentioned, the capacitor will behave as if it is a piece of wire over here. But at the time proceeds, capacitor is going to started to get charged. In other words, the upper plate will get charged positively, whereas the lower plate is going to get charged negatively. So therefore, we will end up with positive charge accumulation over here, and the negative charge accumulation at the lower plate.

As the charge density increases in the plates of this capacitor, then they will generate increasing repulsive force to the incoming charges. As the time goes to infinity, capacitor is going to get charged fully. At that time, these charged plates will generate high enough repulsive force to the incoming charges. Therefore, the charge movement will stop. In other words, current will drop to 0 at that instant. So, once the capacitor is fully charged, then the current flowing through this branch will drop to 0. Therefore, as *t* goes to infinity, capacitor will get fully charged and the current through the branch of capacitor will drop to 0.

In that case, if we write down the loop equations for this case, none of the current will go through this branch for that moment. All the current will go through the first loop. So, all current will flow through the first loop. Therefore, our circuit is going to look like, something like this. This is the case as *t* goes to infinity, and we have resistances *R* for both resistors, the current *i* is going to be emerging from here, and the same *i* will be flowing through the circuit. In this case, if you write down the loop equation, we’re going to end up with *ε* minus *i* times *R* minus *i* times *R* is equal to 0, or *ε* minus 2*i* times *R* is equal to 0, and from there, solving for *i*, we will end up with *ε* over 2*R*.

Therefore, this will be the net current flowing through the circuit, and the current flowing through the branch in which the capacitor is located. For that part, the current will be 0. So, again, if you just make a note over here, current through the resistance, *R*, where the capacitor is located is going to be equal to 0 as *t* goes to infinity.

All right. Now, let’s look at the current in this circuit while the capacitor is getting charged. In other words, we turn the switch on, we don’t wait until *t* goes to infinity. In other words, the charging process continues while causing the current along this branch to decrease exponentially, and let’s try to obtain the current values at this arbitrary time. Let’s say, part c), current through each resistor at time *t*.

For that, if we redraw our circuit again, after we close the switch on, the current is going to be, *i* will be emerging from the power supply, which is generating *ε* volts of potential difference. It will flow through this first resistance, *R*, and then, some fraction will go through the second resistor as *i*1 through this *R*, and the remaining fraction is going to flow through this resistance, *R*, as, let’s say, *i*2. In the meantime, the capacitor is going to be getting charged. Eventually, *i*1 and *i*2 will join at this junction point, making current *i* again.

For this case, if we consider loop 1 and loop 2, again, by choosing a direction of clockwise to traverse these loops, we can write down the loop equations, starting with the loop 1 first. Here, again, crossing *ε* in the direction of EMF arrow, we will experience +*ε* volts of increasing potential, going through first resistance *R* over here, the potential will decrease by minus *i* times *R*, and going through this resistance, the potential will decrease *i*1 times *R*, since we will be crossing, again, in the direction of flow of current, the loop will be completed. The sum of the changes in potential will add up to 0.

For the second loop, we’re going to have, again, starting at a specific point, starting from this junction point, let’s say, going in clockwise direction, we will have a decrease in potential as we cross resistance *R*, which will be equal to minus *i*2 times *R*. Now, as the capacitor is getting charged, the upper plate will be charged positively, the lower plate will be charged negatively, so, when we cross this in clockwise direction, we will go from positive to negative plate, therefore, we will experience a decrease in potential, and that will be equal to amount of charge on the capacitor at that instant divided by the capacitance.

Then, moving on, we will cross the other resistance in opposite direction to the EMF arrow. Therefore, the potential will increase by *i*1 times *R*, and the algebraic sum of the changes in potential from the complete traversal of this loop 2 will add up to 0. Finally, we will write down the junction equations, and the current into the junction is *i* and that should be equal to the sum of the currents coming out of junction, and which are *i*1 plus *i*2.

Again, let’s number these equations. This is equation number 1 and this is equation number 2. Again, and let’s say this one is equation number 3, substitute equation number 3 into equation number 1. Doing that, we will end up *ε* minus *i*1 plus *i*2 times *R* minus *i*1*R* is equal to 0, which will eventually then give us again *ε* minus 2*i*1*R* and minus *i*2*R* is equal to 0.

Here, let’s solve this equation for *i*1. Let’s see. Solving for *i*1. If you do that, first, let’s leave *ε* minus *i*2*R* alone here. That will be equal to 2*i*1 times *R*, and therefore *i*1 in terms of *i*2 will be equal to *ε* minus *i*2*R*, divided by 2*R*. Once you obtain *i*1 in terms of *i*2, let’s go ahead and substitute this into equation number 2. Doing that, we will have minus *i*2 times *R*, minus *q* over *C* plus, ow for *i*1 here, we will write down, it’s equivalent in terms of *i*2, which is *ε* minus *i*2*R* divided by 2*R* times *R*. That will be equal to 0.

Here, we can cancel this *R* with that *R*, and let’s try to simplify this equation. Let’s see. We have minus *i*2 times *R* here, and we have minus *i*2 times *R* over 2 here. If we have common denominator, that’s going to give us then, *ε* over 2 minus 3*i*2*R* over 2. And then, minus *q* over *C* is equal to 0. The *i*2 is the current flowing through the branch where this resistance and capacitor are connected in series, and *q* is the amount of charge at that moment in the capacitor. That *q* is related to *i*2 such that *i*2 is equal to *dq* over *dt*.

Moving on, then, we can express this as *ε* over 2 minus 3 over 2 *dq* over *dt* times *R*, and minus *q* over *C* is equal to 0. Again, as we have seen earlier in the RC circuits, we ended up with an equation for that loop 2, which, what we called as differential equation. Again, here, we have a first-order linear differential equation with constant coefficients. The dependent variable is the charge, *q*. Independent variable is the time. So, we will apply exactly the same procedure that we applied when we were analyzing the RC circuits. In other words, what we’re going to try to do is, we will try to arrange this equation such that the dependent variable *q* is on one side of the equation, and independent variable time is on the other side of the equation.

To do that, first, I’m going to move this differential term to the other side of the equation, and let me write it as 3 over 2*R* times *dq* over *dt* is equal to . . . and, on the other side, we will be left only with *ε* over 2 minus *q* over *C*. Now, we can make cross multiplication and rewrite this expression as *dq* divided by *ε* over 2 minus *q* over *C* is equal to *dt*, move this to the other side, and move also 3 over 2 to the other side, and therefore, we will have *dt* over 3 over 2*R*.

In doing this, we ended up with the dependent variable charge on the left-hand side of the equation, and the independent variable, time, on the right-hand side of the equation. Now, if we integrate both sides, then we will end up with charge as a function of time. Here again, we have to be careful with the initial conditions. Over here, again, at time *t* is equal to 0, the charge on the capacitor was 0. As time proceeds to some *t* seconds, then we will end up with a certain amount of charge stored in the capacitor.

To be able to take this integral, of course, the right-hand side is easy, because 3 over 2 is *R*, we can just simply take it outside of the integral. That part, the integral of *dt*, will give us just *t*. To be able to take the left-hand side’s integral, again, we will say that let *ε* over 2 minus *q* over *C* is equal to *u*. If that is the case, then minus *dq* over *C* will be equal to *du*. So, if we multiply both numerator and denominator by minus 1 over *C*, multiply and divided by, the ratio will not change.

But in this form, I’m going to have *du* in the numerator in terms of this new variable, and *u* in the denominator, which [inaudible 0:31:20] take the integral of *du* over *u*, which is ln of *u*. Therefore, the left-hand side is going to give us ln of *u*, where *u* is *ε* over 2 minus *q* over *C*, which will be evaluated at 0 and *q*. For the right-hand side, again, integral of *dt* will give us *t*. I’m going to move this minus 1 over *C* also to the right-hand side, and we can move this 2 to the numerator, so we’re going to end up with -2 over 3*RC* times *t*, which will be evaluated at 0 and *t*.

If we substitute the boundaries, we will have ln of, first substituting *q* for *q* here, we’re going to have *ε* over 2 minus *q* over *C*, and then substituting 0 for *q* will give us the second term as 0, so we will only have *ε* over 2. And will be equal to minus, if we substitute *t* for time, *t*, we will end up with 2*t* over 3 *RC*, and if you substitute 0, it’s going to give us just 0.

All right. By taking the inverse natural logarithm of both sides, left-hand side will give us *ε* over 2 minus *q* over C, divided by *ε* over 2, which will be equal to *e* to the -2*t* over 3*RC*, which will be equal to *ε* over 2 minus *q* over *C* is equal to *ε* over 2 times *e* to the -2*t* over 3 *RC*. If you leave *q* over *C* alone on one side of the equation, then we will have *ε* over 2, which is common. Therefore in *ε* over 2 parentheses 1 minus *e* to the -2*t* over 3*RC*. Leaving *q* alone, charge as a function of time will be equal to *C* times *ε* divided by 2 times 1 minus *e* to the -2*t* over 3*RC*. This is going to be the amount of charge which is getting stored into the capacitor as the time proceeds.

Well, knowing the charge as a function of time, we can easily calculate the current flowing through that branch, and that is simply *i*2, and that is equal to change in charge with respect to time. If you take the derivative of this with respect to time, we will have *Cε* over 2, coming from this product, derivative of the exponential term, will give us minus 2 over 3*RC* times the exponential term. That minus and this minus will make plus, therefore we will have 2 over 3*RC* *e* to the -2*t* over 3*RC*. There, these 2’s will cancel, and this *C* and that *C* will cancel, and therefore, the current flowing through the resistor, which is connected in series to the capacitor, will be equal to *ε* over 3*R* times *e* to the -2*t* over 3*RC*.

If you look at these expressions, earlier we said that the exponent of an exponential term should be dimensionless. Therefore this 2 over 3*RC* term has to have the dimensions of inverse time. As a matter of fact, for this circuit, the capacitive time constant for that second loop, becomes 3*RC* over 2. In other words, *τ**C*, for the loop 2 becomes equal to 3*RC* over 2 for this specific circuit.

We can see that the current along that branch is going to be decreasing exponentially, and we can easily see that at time *t* is equal to 0, *i*2 is, when we substitute 0 for time over here, if the 0 is 1, *i*2 is 3 over, actually, *ε* over 3*R*. If we compare this with the value we obtained in part a) of this problem, at time *t* is equal to 0, we found the current flowing through that branch, which we called that current as *i*2, is equal to *ε* over 3*R*. So it is exactly the same current, and it verifies our answer for part a).

We can also see that as *t* goes to infinity, then the exponential term will go to 0. Therefore *i*2, indeed, will go to 0. Now, let’s try to determine the current as a function, current *i*1, which is the amount of current flowing through the middle branch as a function of time, and, in order to do that, we will use the expression that we obtained, current *i*1 in terms of current *i*2, which is over here. *i*1 is equal to *ε* minus *i*2*R* over 2*R*. So *i*1 is equal to *ε* minus *i*1 times *R* over 2*R*.

Therefore, we will have *ε* minus, we found *i*1 . . . actually, this is *i*1, and this is *i*2. We found *i*2 as a function of time, as *ε* over 3*R* times *e* to the -2*t* over 3*RC* divided by 2*R*, and we’re multiplying *i*2 by *R* here, in the numerator of the equation. These *R*‘s will cancel, and if we have common denominator in the numerator, we will have *i*1 is equal to 3*ε* minus *εe* to the -2*t* over 3*RC* divided by 2*R*, and the common denominator 3 will move to the denominator also, so 3 times 2 will give us 6*R* in the denominator.

Here, *ε* is common in first and second term, therefore, we can write down *i*1 as *ε* over 6*R* times 3 minus *e* to the -2*t* over 3*RC*. Let’s check this expression with our initial condition at time *t* is equal to 0. Substituting 0 for time *t* over here, exponential term will go to 1, 3 minus 1 will give us 2, therefore *i*1 is going to be equal to 2*ε* over 6*R*. Canceling 2 and 6, we will end up with 3 in the denominator, therefore *i*1 is going to be equal to *ε* over 3*R*. Again, this is exactly the same result that we have obtained in part a) for the current flowing through the middle branch right here at time *t* is equal to 0, *ε* over 3*R*.

Now, let’s calculate the current flowing through the first branch as a function of time. Since that current *i* was equal to *i*1 plus *i*2, we can express *i* as *i*1 — we found that that is *ε* over 6*R* — times 3 minus *e* to the -2*t* over 3*RC* plus *i*2, and we have found *i*2 as *ε* over 3*R*, *e* to the -2*t* over 3*RC*. So that is *ε* over 3*R* times *e* to the minus 2*t* over 3*RC*.

We can express these in *ε* over 3*R*, parentheses as from the first term over here, we will have 3 over 2 and the second term will give us one-half times *e* to the -2*t* over 3*RC*. Plus the last term over here, which is *e* to the minus 2*t* over 3*RC*. Adding these last two terms to one another, we will have current as a function of time, which is equal to *ε* over 3*R*, 3 over 2, one-half plus 1 will give us just plus one-half *e* to the -2*t* over 3*RC*. We can write down now, *i* of *t* is equal to, let’s take 2 outside of the bracket also here, *ε* over 6*R* times 3 plus *e* to the -2*t* over *RC*.

This will give us the current, as a function of time, flowing through the first branch of the circuit. Again, if you test this for time *t* is equal to 0 case, at time *t* is equal to 0, exponential term will give us just 1. Therefore *i* of *t* will be equal to *ε* over 6*R* times, 3 plus *e* to the 0 will give us just 1. So *i* of *t* will be equal to, or I should write this as *i* at *t*, is equal to 0 is going to be equal to 3 plus 1 is 4, 4*ε* over 6*R*. Simplifying this, we will have 2 in the numerator and 3 in the denominator, therefore, *i*, the current at time *t* is equal to 0, will be 2*ε* over 3*R*.

Again, if we compare this value with the value that we obtain in part a) for the current *i*, we see that we have exactly the same result, verifying that it satisfies the condition for time *t* is equal to 0. Therefore, the current through the first branch as a function of time is equal to this quantity, and current through the middle branch as a function of time is equal to this quantity, and current through the last branch, where the resistance and the capacitor is located, is equal to, as a function of time, to this quantity. And in this circuit, the time constant, capacity of time constant, becomes equal to 3*RC* over 2.