All right, so far we have introduced capacitors and resistors as some of the basic components of simple electric circuits. We have studied their connections, and we have seen that, for circuits, that they consist of resistors only, the amount of current flowing through the circuit is constant.
Now we’re going to consider a circuit which will consist of a resistor and a capacitor. We call these type of circuits as RC circuits. Let’s say we have an electromotive force of ε volts, which is first connected to a two-way switch with one terminal a here, and the other terminal b is here, and the switch is connected to a resistor with a resistance of R, which is connected to a capacitor with a capacitance of C in serial form.
Then the circuit is completed by connecting the other terminal or other plate of the capacitor to the other terminal of the electromotive force. The other terminal of the two-way switch is connected to this branch in this form.
Let’s say first we throw the switch to position a. When we do this, then we generate a closed path for the charges to go. Now the positive charges, which are collected at the positive terminal of the power supply, that they’re continuously repelling one another. As soon as the switch is thrown to position a, they will move along this available path through the resistor R, and they will come and be collected on the upper plate of capacitor C.
In a similar way, the negative charges are continuously repelling one another at the negative terminal, and whenever they see this available path, they will move and get collected on the lower plate of the capacitor. And this charging mechanism will continue until the plates reach high enough charge density so that they will generate large enough repulsive force to the incoming charges. At that point, or at that instant, charging will stop, the capacitor will reach to its maximum charge, and therefore, the motion of the charges will stop.
If we look at this from the current point of view then, as soon as we turn the switch to position a, then we’re going to start a certain amount of current emerging from the power supply, flowing from positive towards the negative end of the circuit, and flowing through the resistance R and capacitor along this path. Of course, we will not end up any current through this branch, because at this moment, that branch is an open branch.
Well, of course, one interesting observation can be made in a circuit something like this, that when we look at the circuit carefully, we see that since a capacitor is two conducting plates separated by an insulating medium, the region between these plates is an insulator. Now, we have seen that an insulating medium does not conduct the electricity. In other words, it does not provide the charges to move freely around. But despite that, although we have an open circuit, if we introduce an ammeter just before the capacitor, an ammeter right after the capacitor, we will see that these two ammeters will read exactly the same current value at the same instant, indicating that this insulating medium is behaving like a conducting medium.
We will study and answer this question later on, once we introduce the concept of magnetic field. But for now, we will just analyze the circuit and convince ourselves that, indeed, the current flowing through the circuit is i, and of course, as the capacitor gets charged, then the amount of this current will get lesser and lesser. And once the capacitor is fully charged, then due to this high enough repulsive force generated by these charged plates, then the motion of charges will stop. In other words, the amount of current will drop to 0.
So we can easily say that the major difference from a circuit which has resistors only is that, while the current in those circuits was constant as it flows through the circuit, in a circuit like this, it is not constant anymore, but it is a function of time. It changes with time, and as the time proceeds, as the capacitor is getting charged, then it drops and eventually goes to 0 once the capacitor is fully charged. So we will now study this simple loop and we will call this phase, which is the case that we through the switch to point a, as charging phase of the capacitor, or simply charging phase, which is the case that the switch is thrown to point a.
Okay, let’s try to express the loop rule or loop equation for this circuit. In order to do that, we will choose first a direction to traverse the circuit, and let’s choose clockwise direction. Starting just right before the electromotive force, going in clockwise direction, we will be crossing this unit in the direction of EMF arrow, since the EMF arrow points from negative terminal of the power supply towards the positive terminal, giving us the direction in which the charges are gaining electrical potential energy.
So crossing them from negative to positive, we will experience an increase in potential, and that will be equal to +ε volts. Let’s say loop equation. And then, moving on in clockwise direction, we will be crossing the resistance in the direction of flow of current. Therefore, the potential will decrease by i times R. In other words, as much as the potential difference between the ends of this resistance.
Again, moving on in clockwise direction, we will be crossing the capacitor from its positive plate to its negative plate. In other words, we will be going from high potential region to low potential region. Therefore, the potential will decrease by as much as the potential difference between these plates, and that is minus q over C.
Again, moving along the clockwise direction, we will come back to the point that we started with. Therefore, the loop is completed. Algebraic sum of the changes in potential from the complete traversal of this closed loop will add up to 0.
If we look at this equation, we see that it’s a function of current, and as well as charge. But since current is related to charge as the change in charge with respect to time, we can express this equation in terms of charge only, as ε minus R times dq over dt, minus q over C is equal to 0.
Let’s rearrange this equation and express it as R dq over dt plus one over Cq is equal to ε. This is a new type of equation and most or all of you have not seen it before. Here in this equation, the charge is called “dependent variable” and the time is called “independent variable”. Here, resistance R is constant, and as well as capacitance of the circuit is constant, and also electromotive force is constant.
In calculus, we have a special name for these type of equations. They’re called differential equations, and specifically the one that we’re dealing over here is called first order, because the derivative term is only in first order, and it is called “first order linear differential equation”. And since the coefficients of the independent variable are constant, it is also called as with constant coefficients.
Later on, you’re going to take a whole semester course and study different types of differential equations and introduce the methods how to solve them. Of course, at this level, we’re not going to go into the details of the methods of solving the differential equations, but when we consider this first order linear differential equation with constant coefficients, this is the simplest form, we really don’t need to have a background of how to solve a differential equation. The only thing that we will do over here is that we will try to express the equation such that the dependent variable is on one side of the equation, and the independent variable is on the other side of the equation. We call this method as “separation of variables”.
In order to do that, let’s leave R dq over dt alone on the left-hand side of the equation. Then we will have ε minus q over C on the other side. Now let’s do cross-multiplication, which will give us R dq is equal to ε minus q over C times dt. And if you divide both sides by R times ε minus q over C, which will eventually give us dq over ε minus q over C is equal to dt over R.
Now we can easily see that we have collected the dependent variable charge terms on the left-hand side of the equation, whereas the independent variable time terms on the other side, on the right-hand side of the equation. If we take the integral of both sides now, we will end up with an expression for charge as a function of time.
Here we have to be careful with the initial conditions of the physical problem. We know that at the instant that we throw the switch to point a, the charge on the capacitor is 0, and if we wait for a little while, then we’re going to end up with certain amount of charge on the capacitor. And eventually, the capacitor will reach to its maximum charge of, let’s say q0 coulombs. So, at time t is equal to 0, we have the charge of 0 in the capacitor, and as time proceeds to some t seconds, then we will end up with some certain amount of charge to be stored in the place of the capacitor.
Okay, now the task becomes taking these integrals, and the right-hand side is obviously easy. For the left-hand side, we will do our usual transformation. We will say that let ε minus q over C be u. Then, since ε is constant, and as well as C is constant, if you take the derivative of both sides, we will have minus dq over C will be equal to du.
So to be able to express the integral in terms of the new variable u, we see that, in terms of this new variable, dq is equal to, leaving dq alone is going to be equal to –C times du. Or, if we multiply both numerator and denominator of this equation by -1 over C, nothing will change. So, in doing that, we’re going to have du at the numerator and u at the denominator. So our integral will take the form of du over u, which we know that the integral of du over u is ln of u.
Now we can move this term over here, -1 over C, to the other side, and therefore our integral becomes ln of u, and u is ε minus q over C on the left-hand side, which is going to be evaluated at 0 and q, will be equal to integral of dt is t, and R is constant, so we can take it outside of the integral. And moving this -1 over C to the other side, then we will have -1 over RC times t, which will be evaluated at 0 and t.
Now substituting the boundaries, we’re going to have ln of ε minus q over C, first by substituting q for q, and then minus ln of, substituting 0 for q, will give us just ln of ε, is equal to, on the right-hand side, substituting t for t, we’ll have -1 over RC times t. And if you substitute 0, that’s going to give us 0 for time.
We can also express the left-hand side as ln of ε minus q over C over ε is equal to -1 over RC times t. If we take the antilog or anti-natural logarithm of both sides, then the left-hand side is going to be equal to ε minus q over C over ε. And let me write this in explicit form so that you can remember. So inverse ln of this quantity is going to be equal to inverse ln of t over RC. The left-hand side, therefore, will give us just ε minus q over C over ε is equal to, and inverse ln of t over RC can be expressed as e to the –t over RC.
And from there, ε minus q over C is equal to ε times e to the –t over RC, Moving one step further, q over C, therefore, will be equal to ε minus ε e to the –t over RC, or q is going to be equal to C times ε minus ε e to the –t over RC.
And finally, we can express charge as a function of time as C times ε 1 minus e to the –t over RC, close parentheses. So that expression, therefore, gives us how the charge stored in the capacitor is varying with time. Here, we know that at t is equal to 0, we shall have 0 charge in the capacitor. As a matter of fact, if we substitute 0 for time here, we will have e to the 0, which is equal to 1. 1 minus 1 will make 0. Indeed, q at t is equal to 0 will be equal to 0, and that’s what we were expecting.
Furthermore, as t goes to infinity, in other words, if we wait long enough, then we expect the capacitor to be fully charged. So as t goes to infinity, the exponential term will go to 0. Then we will end up with q0. So q will go to q0, the maximum charge, and that will be equal to, since the exponential term will go to 0, we will just have 1 times Cε. Therefore, the maximum charge is going to be equal to capacitance times the potential difference at that instant between the plates of the capacitor.
That allows us to write the charge as a function of time as equal to the maximum charge stored in the capacitor, q0, times 1 minus e to the –t over RC. So this is how the charge is varying with time during charging phase in the capacitor. Okay. This is, therefore, a charging phase.
Now if we look at also why indeed q0, the maximum charge, is equal to Cε, and if we go back to our circuit diagram, now we know that we said that, once the capacitor is fully charged, then the charge movement stops means that the current drops to 0 at that instant.
So once the current flowing through the circuit is 0, and the capacitor is charged to its maximum charge of q0, then the potential difference between the plates of the capacitor becomes directly equal to the potential difference generated by the electromotive force, and that is ε. Therefore, at that instant, once the charge is q0, the potential difference between the plates of the capacitor is ε, and as you recall, C is equal to amount of charge stored in the capacitor divided by the potential difference between the plates of the capacitor.
So C becomes q0 over ε, and solving for q0, indeed the maximum charge becomes C times ε, so that verifies that what we obtain over here. Once the capacitor is fully charged, the current drops to 0 and the maximum charge in the capacitor at that instant is equal to capacitance times the potential difference between the plates of the capacitor for that moment.
Well, knowing the charge, we can easily obtain the current flow through this circuit, since current is equal to dq over dt. Therefore, it’s going to be equal to derivative of this quantity with respect to time, and if you take the derivative, the first term is q0 which is constant. The derivative of a constant is 0. That’s going to give us just 0.
Derivative of second term with respect to time will give us -1 over RC times this minus, which will make positive. So we’ll have q0 over RC times e to the –t over RC. Well, q0 was C times ε, and if we substitute that quantity here for q0, then the current becomes equal to Cε over RC times e to the –t over RC. The capacitances will cancel. Then the current is going to be equal to ε over R times e to the –t over RC.
Well, we know that when we throw the switch to position a, again, going back to our circuit diagram, at the moment that we throw the switch to point a, we will start with the maximum current, because at that time, there’s no charge in the capacitor, and therefore, the charges will not see any repulsive force generated by the charge on the plates of the capacitor.
So we will start with the maximum current, and as the time proceeds, the capacitor will get charged. Therefore, we’re going to end up with some repulsive force to those incoming charges. It means that the current will start to decrease, and once the capacitor is fully charged, then the current will go to 0.
So in that sense, then, we can say at time t is equal to 0, i is equal to i0, which is the maximum current. Okay, if we look at our equation at time t is equal to 0, exponential term will be equal to e to the 0, which will give us just 1. So i0 will be equal to ε over R, and that is the maximum current, and that occurs once we throw the switch to position a. Right after that moment, then the current starts to decrease exponentially. So i becomes i0 times e to the –t over RC, and that’s the behavior of current during the charging phase of the capacitor.
Okay, so obviously current is not constant in this circuit, and it is decreasing exponentially. As a matter of fact, if we plot current as a function of time, it is going to be a function. At time t is equal to 0, it will start with its maximum value, i0, and then it will decrease exponentially as the time goes to infinity.
Now, when we look at both charge and as well as current expression, we see this exponential behavior, and as I mentioned earlier, for a physical equation to make sense, in other words, to be correct, the argument of the exponent or exponential function, or if you’re dealing with trigonometric functions, the argument of those functions, should be dimensionless. In other words, if we end up with a dimensional quantity, then there has to be something wrong with our expression. In that sense, since we know that the exponent of the exponential function should be dimensionless, we can say that the product of resistance and capacitance should have to have the dimensions of time, so that the time and time will cancel. We’re going to end up with a dimensionless quantity.
Let’s check that out. R times C, by making a dimensional analysis, and let’s use SI unit systems. The resistance, by definition, is potential difference divided by current, and the capacitance is the amount of charge stored in the capacitor divided by the potential difference between the plates of the capacitor. The potential differences will cancel. Then we will end up RC is equal to charge divided by current.
If we express these in terms of their units, in SI unit system, charge is in terms of coulombs, and the current is in charge per unit time, which is coulombs per second. Coulombs will cancel, and indeed, we’re going to end up with 1 over 1 over second, which will give us seconds. So RC has the dimensions of time. Therefore, time divided by time in the exponential terms will give us dimensionless quantity.
This product, of course, will be unique for every RC circuit. In other words, every RC circuit will have a unique value for the resistance that they have and the capacitance that they have. So we have a special name for this product. We will denote this by τC, and it is called capacitive time constant. And τC is equal to product of resistance and the capacitance of the specific RC circuit.
What is the indication of this value? Let’s see what happens, how much charge is stored in an RC circuit into the capacitor when one time constant of time elapses. So in order to find that, we will calculate q at t is equal to τC. That’ll be equal to q0 times 1 minus e to the –τC over τC, which is going to give us q0 times 1 minus e to the -1. And if we calculate this quantity, that’ll be equal to 0.63. It means that if we wait 1 time constant of time, then 63% of the maximum charge will be stored in the capacitor.
In a similar way, if we look at the current during the same time interval, i at t is equal to τC will be equal to i0 times e to the minus τC over τC, which is going to be equal to i0 times e to the -1, which will give us 0.37i0. So it means that if we wait one time constant of time after we throw the switch to position a, 37% of the maximum current will be decayed.
All right. Now, let’s go back to our circuit and consider the case that we throw the switch to position b. In that case, let’s see what happens. So if we now throw the switch to position b, in doing that, we’re going to be literally taking out this unit out from the circuit. In other words, we will be taking the power supply, electromotive force, out of the circuit.
So our closed loop is going to be then in this form over here, and let’s assume that at time t is equal to 0 in this case, we have the maximum charge on the capacitor. Once we throw the switch to position b, then the capacitor is going to discharge through this path. In other words, through the resistance R.
And the loop equation, in this case, will be exactly identical to the previous case, except ε will be 0, since we’re just taking the ε outside of the circuit. Therefore, if we write down the loop equation for this case, and let’s call this phase as the discharging phase, which is the case that switch is thrown to point b while the capacitor is fully charged, then if we express the loop equation for this case, it will be equal to Ri plus q over C is equal to 0.
Again, using the fact that current is equal to change in charge with respect to time, we can express this equation as R dq over dt plus 1 over C times q is equal to 0. Here again, what we have is a first-order linear differential equation with constant coefficients. We apply the same technique that we applied for the first case, try to express this equation such that the dependent variable is on one side of the equation and the independent variable is on the other side of the equation. So R dq over dt will be equal to minus q over C, and making a cross-multiplication and rearranging the terms, we will have dq over q is equal to minus dt over RC.
Again, once we separate the dependent variable from the independent variable in this way, we can integrate both sides by considering the initial conditions, and they are such that, at time t is equal to 0, now we have the charge in the capacitor as the maximum charge. The capacitor is fully charged, therefore q0, and as the time proceeds to some t seconds, we end up with the amount of q left in the capacitor.
By taking the integrals of both sides, the left-hand side is going to give us ln of q, evaluated at q0 and q, will be equal to minus, again RC is constant. We can take it outside of the integral. Integral of dt will give us t, so we will have t over RC evaluated at 0 and t. Substituting the boundaries, we will have ln of q over q0 will be equal e to –t over RC. By taking the inverse logarithm of both sides, q over q0 will be equal to e to the –t over RC, or simply τC, since we called the time constant as product of resistance and the capacitance of the circuit.
Therefore, q is going to be equal to, as a function of time, q0 times e to the minus 2 over RC, or q0 times e to the minus t over τC. And that expression, therefore, gives us the behavior of charge as a function of time during discharging period or discharging phase.
Again, since current is dq over dt, rate of change of charge, and from here, we can calculate the current by taking the derivative of this term, which will be equal to minus q0 over RC times e to the –t over RC. Again, recall that maximum charge was equal to ε times C, and substituting this for q0, the current will be equal to ε over R e to the –t over RC, or i of t is equal to ε over R, was simply the maximum current, therefore, i of t can be expressed as i0 times e to the –t over RC. Of course, we have a negative sign in front of this equation.
As you can see, the current as a function of time is identical to the charging phase, except with a negative sign over here. And as you’ll remember from the Kirchoff rules, we said that the current’s sign is directly associated with the direction of flow of current. So in the first part, we have assumed, or we said, as a matter of fact, the current flows from positive end towards the negative end, in clockwise direction, for the charging phase.
For the discharging phase, once we throw the switch to this point, then the associated current is going to be flowing from positive plate towards the negative plate of the capacitor. Therefore, in the discharging phase, current is going to be flowing in counterclockwise direction, i, which is the case for discharging phase. That is just the opposite flow direction relative to the first case, and that’s also why the negative sign is appearing in front of the current expression.
But the behavior of current is identical to the charging phase. Again, we can calculate the amount of charge left in the capacitor after one time constant of time for the decay of current part or discharging phase. That’ll be q0 times e to the -1, which is going to be 0.37 times q0, indicating that the 37% of q0 will be decayed, or the 37% of the maximum charge will be discharged once one time constant of time elapses.
In a similar way, the amount of current after one time constant of time will be equal to i0 times e to the -1, and that will also give us that this 37% of the maximum current is going to be decayed if we wait one time constant of time once we throw the switch to position b.