Transitories in RC Circuits theory and study

This chapter consists of the items below. If you want to go directly to an item, click on it.

2. - RC Circuit click here!

3. - Solution of the Problems click here!

4. - RC Circuit Initial Values click here!

5. - Thévenin Equivalent click here!

6. - Electric Current in a Capacitor click here!

7. - Electric Voltage in a Capacitor click here!

In this chapter we will study resistor and capacitor associations and what is the behavior these circuits assume when subjected to an electrical voltage of type Direct Current (DC).

Let's repeat here two reminders that were used in chapter 3 as they are very important for understanding how a circuit works RC.

2. RC Circuit

See the Figure 22-01 for a typical circuit using a R resistor and a C capacitor in the serial configuration, and are connected to a V voltage source via an electric switch S.

In this circuit we have a switch S, which allows to turn the voltage source on and off that feeds the circuit. When closed, it applies a voltage from the voltage source V in the circuit formed by the resistor in series with the capacitor. In the technical literature represents the closing time of the S switch, as the time equal to t = 0⁺.

At the time of closing the key S, at t = 0 ⁺, as the capacitor is initially discharged (q = 0 and V_c = 0), its behavior will be short-circuited. So the voltage on the capacitor, V_c, will be equal to zero , and therefore, all source voltage will be applied over the R resistor. So we can calculate the current circulating through the R resistor at the instant t = 0⁺. To do this, simply apply Ohm's law, that is, I = V / R.

In the instant immediately after t = 0⁺, the capacitor then begins to be electrically charged by the electric current I. As the capacitor now has an electric charge, so it must also have a certain electrical voltage. This electrical voltage on the capacitor rises exponentially. The speed at which the capacitor acquires electric charge depends on the capacitor capacitance and resistor electrical resistance values that are in series with the capacitor. The values of these two components determine the so-called time constant of the circuit and is represented by the Greek letter τ (tau). Then we can write that:

τ = R C

With this information, we will present the equation that allows us to determine the current in the capacitor at any time.

eq. 22-01

Note that for t = 0, the exponential is raised to power zero, and as we know, any number raised to power zero equals ONE. So we conclude that we will have a current circulating through the circuit equal to i_c = V / R, as said before. Over time, that is, t > 0, the exponential is raised to a negative power, causing the circuit current to drop exponentially. And when t → ∞ the current will tend to zero.

It is easy to see that when the electric current tends to zero, the voltage over the capacitor increases exponentially until it reaches the voltage of the voltage source. So we can write the equation that determines this load, or:

eq. 22-02

Knowing these equations we can present graphs that show the behavior of a circuit RC.

We can see in the Figure 22-02 the graph of how the capacitor gets its electric charge over time. Note in the figure, that for a time equal to a time constant , capacitor acquires 63.2 % of your total charge. After two time constant, it reaches 86.5% of its total charge. In practice, we consider that after five time constant, the capacitor reaches its maximum electrical charge.

When the capacitor reaches its maximum electrical charge, we say that the circuit has reached the state of permanent regime. This means that if the circuit does not suffer any electrical disturbance later, the circuit tends to remain in this state indefinitely.

Now, be aware that as the voltage on capacitor increases , obviously the voltage on resistor decreases , since the voltage source has a fixed value. (constant). So the sum V_c + V_R must be equal to V. Therefore, when the capacitor acquires its maximum load, the voltage on the capacitor is V and of course on the resistor we have V_R = 0. Note that all these situations are perfectly in accordance with the equations presented above.

In the Figure 22-03, we have the graph of the electric current through the capacitor. Since resistor and capacitor form a series circuit, so this electric current is the same as circulating through the resistor. Thus, we conclude that the voltage graph on the resistor looks the same as the electrical current graph on the capacitor.

Note that as the electrical voltage on the capacitor increases (see previous figure), simultaneously the electrical voltage on the resistor decreases . In addition, this graph also represents the voltage drop across the resistor by simply replacing it on the vertical axis i_c by V_R.

3. Problem Solving

As stated earlier, let's look at the behavior of RC circuits through mathematical equations that lead us to solve the problems. Often we should use differential equations to solve them.

Be the circuit show in the Figure 22-04 where we put the S switch in position 1, allowing the capacitor acquires electrical charge through the resistor of 10 ohms . We know that, initially the voltage on the capacitor is V_C = 0 volt. So the capacitor starts to charge and its voltage increases exponentially with a speed which depends on the time constant of the circuit formed by the resistor of 10 ohms and the capacitor of 10 µF.

As we know the value of the components, we can calculate the time constant, τ, given by:

^-6

^-4

With this value, we can write the capacitor charge current equation i₁, or:

i₁ = 50/10 e^{-(t/100 )µs} = 5 e^{- 10 000 t}

Similarly, we can write the equation that expresses the electric voltage over the capacitor, that is:

^{-(t/100 )µs}

^{- 10 000 t}

With this, we establish the mathematical behavior of this circuit. Realize that now we can calculate the value of i₁ and V_C anytime.

Let's suppose we want to know the value of V_C in the instant t = 250 µs. Just replace this value of t in the above equation and we will have the value, or:

^{- (250/100)}

^{- 2.5}

In other words: after 2.5 times the circuit time constant, the capacitor is already at 91.8% the maximum voltage it can reach. This is what we call instant value. It is clear in this example, we can predict at what time we want the capacitor to acquire determined voltage simply by choosing the appropriate values of R and C.

We see in the Figure22-05 a graph showing how the value of R modifies the circuit time constant by changing the capacitor charging time. The higher R, the more slowly the capacitor charge rises, and vice versa. Here we assume that the capacitor value has not changed. Of course we can keep the R value fixed and vary C. Keeping the reasoning, we conclude that the higher C, the more slowly the capacitor charge rises, and vice versa.

Now let's look at the situation with key S at position 2

Let's assume that the S key was left 10 time constants in position 1. This ensures that the capacitor has a maximum electrical voltage of 50 volts . If we move the switch S to position 2, the capacitor will be discharged by the resistor 40 ohms. The new time constant will be:

τ = R C = 40 (Ω ) x 10 x 10^-6 (F ) = 40^-4 s = 400 µs

Notice that as the resistance value quadrupled, the time constant also quadrupled. That way we can now write the equations for the electric current i₂ and for the electrical voltage on the capacitor, V_C.

i₂ = 50/40 e^{-(t/400 )µs} = 1.25 e^{- 2 500 t}

V_C = 50 e^{-(t/400 )µs} = 50 e^{- 2 500 t}

4. RC Circuit Initial Values

So far we have studied circuits where the capacitor was initially discharged, that is, V_C = 0. From this moment we will analyze circuits where the capacitor has an initial charge, that is, V_C ≠ 0. This voltage that appears at the capacitor terminals is called the initial value. As soon as we close the key, the transitional phase begins and for practical purposes we can consider that it only ends after five time constants. After that we enter into the so-called steady state, to determine the values of the transient phase we need a equation that allows us to reach the desired values. This equation is shown below.

eq. 22-03

In this equation, the meaning of the variables are:

V_C - capacitor voltage at any time t
V_i - initial voltage on capacitor
V_f - final voltage on capacitor
t - time we want to calculate the voltage V_C
τ - circuit time constant

If you're interested in knowing how we got to this equation, Click Here!

To see a practical application of using this technique, Click Here!

5. Thévenin Equivalent

Many electrical circuits do not have the simple form we have studied so far. In this item we will study more complex cases and so we should resort to the known Thevenin theorem. So we have to find the Thévenin resistance. of the circuit to calculate the time constant.

Example - Let's look at an example. that appears on the page 291 (example 10.10) from the book of Robert Boylestad [3]. The circuit is shown in the Figure 22-06.

Item a - Let's determine the mathematical expression for the transient behavior of voltage V_c and of the current i_c as a function of time after key closure (position 1 in t = 0 s).

Considerations - Note that we have three resistors in the circuit when the S switch is in position 1. To calculate the time constant, we must reduce to a single resistor, which will be the Thévenin resistance. Let's apply the concepts we learned in chapter 15 to calculate it. We know that we must short circuit the voltage source. See the Figure 22-07 for how the modified circuit for calculating Thévenin resistance.

It is perceived by the circuit, which R₁ and R₂ are in parallel. As we already know to calculate the parallel of two resistors, performing the calculation we find the value of 20 kΩ. And obviously this parallel is in series with R₃, therefore, we find for Thévenin resistance the value of:

_th

We are now able to calculate the circuit's time constant.

_th

^-6

Having the value of the Thévenin resistance, let us calculate the value of Thévenin Voltage. As we know, for this we must calculate the voltage open circuit, as shown in the figure below.

Looking at the circuit in the Figure 22-08 , we realize that there will be no electric current flowing through R₃, therefore to calculate the Thévenin Voltage is enough calculate the voltage on the resistor R₂. R₁ and R₂ form a resistive divider that is easy to solve. So:

_th

₂

₁

₂

And so, performing the calculation, we conclude that:

V_th = 7 V

It is now possible to build the equivalent circuit of Thévenin by replacing the capacitor in the circuit as it appears in the Figure 22-09.

To find the mathematical expression that defines the electrical voltage on the capacitor in the circuit, we need to use eq. 22-03, shown below.

eq. 22-03

Through the calculations performed, we define the initial and final values of the voltage on the capacitor, or:

V_i = 0 V and V_f = V_th = 7 V

Then, substituting these values in the above equation and grouping similar terms, we find:

V_c = 7 (1- e^{- (t/6 ) ms} )

In other words, to t = 0 we have V_c = 0 because 1 - e^-0 = 0. As t grows, e^{- (t/6 ) ms} tends to zero and therefore V_c will tend to the source voltage of 7 volts.

To determine the mathematical expression that defines the electric current in the capacitor, let's use eq. 22-01. For a better understanding, let's repeat it here.

eq. 22-01

See the circuit in the Figure 22-10.

In this particular case, V is the Thévenin voltage (7 volts), R is the Thévenin resistance (30 kΩ) and C is the capacitor value (0.2 µF) in the circuit. Replacing the above equation and performing the calculation we find the mathematical expression that was requested in the problem.

i_c

^{- (t/6 ) ms}

What this equation says is that when we turn on the key, that is, to t = 0, the current that passes through the capacitor is of 233 µA ( because e⁰ = 1). If t increases, the current tends to zero, as expected.

6. Electric Current in a Capacitor

As a capacitor is made up of two conductors separated by a dielectric or insulating material, it means that the electrical charge is not conducted through the capacitor. Although applying a voltage to the capacitor's terminals does not cause it to conduct charges through its dielectric, it can produce small displacements of a charge within it. As the voltage varies with time, this displacement also varies with time, causing the so-called displacement current.

At the terminals of a capacitor, the displacement current is indistinguishable from a conduction current. Therefore, we can state that:

"The current in a capacitor is proportional to the temporal variation of the voltage across it."

eq. 22-04

Note that the equation eq. 22-04 perfectly expresses the above definition. Therefore, two conclusions can be drawn:

If the voltage v at the terminals is constant, then the current in the capacitor is null, or i = 0.

If the voltage v on the capacitor changes instantaneously, then i = ∞. Physically, this is impossible, as it would require infinite power. This implies that a capacitor cannot undergo instantaneous voltage variations. In other words: we cannot have discontinuity in v(t).

The reason the current in the capacitor is zero when the voltage across it is constant is that a conduction current cannot be established in the dielectric material. Thus, the capacitor in the presence of a constant voltage behaves like an circuit or open loop.

7. Electrical Voltage in a Capacitor

Working algebraically the eq. 22-04 and integrating we obtain the voltage on the capacitor when we know the current flowing through it. See the eq. 22-05.

eq. 22-05

The time t_o is called initial time and the voltage v(t_o) is called initial condition. Most of the time we do t_o = 0, a convenient value.

Example

Let it be a capacitor, whose capacitance is unknown, connected directly to a current source of value given by:

^{- 1.2 t}

_-1

And the voltage across the capacitor is:

^{- 1,2 t}

_-1

We want to determine the value of the capacitor's capacitance. To solve this problem we will use eq. 22-05. So substituting for numerical values, we have:

^{- 1.2 t}

Equating the coefficients of e^{- 1.2 t} , we obtain:

C = 3.125/1.25 = 2.5 F

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