Study of DC Transients in R-L & R-C Circuits | Basic Electrical Technology - Electrical Engineering (EE) PDF Download

Objectives

• Definition of inductance and continuity condition for inductors.
• To understand the rise or fall of current in a simple series R − L circuit excited with dc source.
• Meaning of 'Time Constamt (τ)' for R − L circuit and explain its relationship to the performance of the circuit.
• Energy stored in an inductor
• Definition of capacitance and Continuity condition for capacitors.
• To understand the rise or fall voltage across the capacitor in a simple series R−C circuit excited with dc source.
• Meaning of 'Time Constamt (τ)' for R−C circuit and explain its relationship to the performance of the circuit.
• Energy stored in a capacitor

Introduction

So far we have considered dc resistive network in which currents and voltages were independent of time. More specifically, Voltage (cause → input) and current (effect → output) responses displayed simultaneously except for a constant multiplicative factor R (V = R×I). Two basic passive elements namely, inductor (L) and capacitor (C) are introduced in the dc network. Automatically, the question will arise whether or not the methods developed in lesson-3 to lesson-8 for resistive circuit analysis are still valid. The voltage/current relationship for these two passive elements are defined by the derivative (voltage across the inductor current flowing through the inductor ; current through the capacitor  voltage across the capacitor) or in integral form as  rather than the algebraic equation (V = IR) for all resistors. One can still apply the KCL, KVL, Mesh-current method, Node-voltage method and all network theorems but they result in differential equations rather than the algebraic equations that we have considered in resistive networks (see Lession-3 to lesson-8).

An electric switch is turned on or off in some circuit (for example in a circuit consisting of resistance and inductance), transient currents or voltages (quickly changing current or voltage) will occur for a short period after these switching actions. After the transient has ended, the current or voltage in question returns to its steady state situation (or normal steady value). Duration of transient phenomena are over after only a few micro or milliseconds, or few seconds or more depending on the values of circuit parameters (like R, , L and C ).The situation relating to the sudden application of dc voltage to circuits possessing resistance (R), inductance (L), and capacitance (C) will now be investigated in this lesson. We will continue our discussion on transients occurring in a dc circuit. It is needless to mention that transients also occur in ac circuit but they are not included in this lesson.

Significance of Inductance of a coil and dc transients in a simple R-L circuit

Fig.10.1 shows a coil of wire forming an inductance and its behavior is to resist any change of electric current through the coil. When an inductor carries current, it produces a certain amount of magnetic flux (Φ) in the core or space around it. The product of the magnetic flux (Φ) and the number of turns of a coil (an inductor) is called the ‘flux linkage’ of the coil. 

Considering the physical fact that the voltage across the coil is directly proportional to the rate of change of current through the inductor and it is expressed by the equation 

where L is the constant of proportionality called inductance of coil and it is measured in   The direction of induced emf is opposite to that of current increases or decreases (Lenz’s Law)

Let us assume that the coil of wire has 'N' turns and the core material has a relatively high permeability (or magnetic path reluctance is very low), so that the magnetic flux (Φ) produced due to current flowing through the coil is concentrated within the core area. The basic fundamental principle according to Faraday, the changing flux through the coil creates an induced emf (e) and it is expressed as 

In words, Faraday’s law states that the voltage induced in a coil (inductor) is proportional to the number of turns that the coil has, and also to the rate of change of the magnetic flux passing through its coils. From equations (10.2) and (10.3), one can write the following relation

The inductance of a coil can also be defined as flux (Φ) linkage per unit of current flowing through the coil and it is illustrated through numerical example.

Example- Consider two coils having the same number of turns ‘N’. One coil is wrapped in a nonmagnetic core (say, air) and the other is placed on a core of magnetic material as shown in fig.10.2. Calculate the inductances of both coils for same amount of current flowing through them.

Case-A: Nonmagnetic material

Inductance of nonmagnetic material 

Case-B: Magnetic material

Inductance of magnetic material 

Inductance calculation from physical dimension of coil

A general formula for the inductance of a coil can be found by using an equivalent Ohm`s law for magnetic circuit and the formula for reluctance. This topic will be discussed in detail in Lesson-21. Consider a solenoid-type electromagnet/toroid with a length much greater than its diameter (at least the length is ten times as great as its diameter). This will produce an uniform magnetic field inside the toroid. The length ‘l’of a toroid is the distance around the center axis of its core , as indicated in fig.10.3 by dotted line. Its area ‘A’ is the cross-sectional area of the toroid, also indicated in that figure.

Appling ampere-circuital law for magnetic circuit (see Lesson-21) one can write the following relation 

We know, flux is always given by the product of flux density (B) and area (A) through which flux density ids uniform. That is,

where B = μ H and H is the uniform field intensity around the mean magnetic path length ‘l’. Substituting the equation (10.6) into the defining equation for inductance, equation (10.4) gives

Remark-1: The expression (10.7) is derived for long solenoids and toroids, computation of inductance is valid only for those types.

Continuity condition of Inductors

The current that flows through a linear inductor must always be a continuous. From the expression (10.1), the voltage across the inductor is not proportional to the current flowing through it but to the rate of change of the current with respect to time,   The voltage across the inductor (v_L) is zero when the current flowing through an inductor does not change with time.

This observation implies that the inductor acts as a short circuit under steady state dc current. In other words, under the steady state condition, the inductor terminals are shorted through a conducting wire. Alternating current (ac), on the other hand, is constantly changing; therefore, an inductor will create an opposition voltage polarity that tends to limit the changing current. If current changes very rapidly with time, then inductor causes a large opposition voltage across its terminals. If current changes through the inductor from one level to another level instantaneously i.e. in dt = 0 sec., then the voltage across it would become infinite and this would require infinite power at the terminals of the inductor. Thus, instantaneous changes in the current through an inductor are not possible at all in practice.

Remark-2: (i) The current flowing through the inductor cannot change instantaneously (i.e. i(0^-) just right before the change of current = − i(0⁺) just right after the change of current). However, the voltage across an inductor can change abruptly. (ii) The inductor acts as a short circuit (i.e. inductor terminals are shorted with a conducting wire) when the current flowing through the inductor does not change (constant). (iii) These properties of inductor are important since they will be used to determine “boundary conditions”.

Study of dc transients and steady state response of a series R-L circuit.

Ideal Inductor: Fig.10.4 shows an ideal inductor, like an ideal voltage source, has no resistance and it is excited by a dc voltage source V_s.

The switch ‘S’ is closed at time ‘t = 0’ and assumed that the initial current flowing through the ideal inductor just before closing the switch is equal to zero. To find the system response (i(t) - vs- t), one can apply KVL around the closed path.

KVL 

Equation (10.9) implies that the current through inductor increases with increase in time and theoretically it approaches to infinity as  but in practice, this is not really the case. 

Real or Practical inductor:

Fig.10.5 shows a real or practical inductor has some resistance and it is exactly equal to the resistance of the wire used to wind the coil.  

Let us consider a practical inductor is connected in series with an external resistance R₁ and this circuit is excited with a dc voltage V_s as shown in fig.10.6(a)

Our problem is to study the growth of current in the circuit through two stages, namely; (i) dc transient response (ii) steady state response of the system.

D.C Transients: The behavior of the current (i(t)) ; charge (q(t)) and the voltage in the circuit (like 
(v(t)) R − L ;R− C: R − L − C circuit) from the time (t (0⁺)) switch is closed until it reaches its final value is called dc transient response of the concerned circuit. The response of a circuit (containing resistances, inductances, capacitors and switches) due to sudden application of 

voltage or current is called transient response. The most common instance of a transient response in a circuit occurs when a switch is turned on or off – a rather common event in an electric circuit. 

Growth or Rise of current in R-L circuit

To find the current expression (response) for the circuit shown in fig. 10.6(a), we can write the KVL equation around the circuit

where V_s is the applied voltage or forcing function ,R_L  is the resistance of the coil, R₁ is the external resistance. One can combine the resistance of coil R_L to the external resistance R₁ in order to obtain a simplified form of differential equation. The circuit configuration shown in fig. 10.6(a) is redrawn equivalently in fig.10.6(b) for our convenience. The equation (10.10) is the standard first order differential equation and its solution can be obtained by classical method. The solution of first or second order differential equation is briefly discussed in Appendix (at the end of this lesson-10). The following relation gives the solution of equation (10.10)

Here, i_n (t) is the complementary solution/natural solution of differential equation (10.10). It is also sometimes called as transient response of system (i.e. the first part of response is due to an initial condition of the system or force free response). The second part i_f(t) of eq. (10.11) is the particular integral solution/force response or steady state
response of the system due to the forcing function (f (t) = VS) or input signal to the series R - L circuit. It may be noted the term A provide us the steady state solution of the first order differential equation while the forcing function (or input to the system) is step function (or constant input). More specifically, for a linear system, the steady state solution of any order differential equation is the same nature of forcing function (f (t)) or input signal but different in magnitude. We have listed in tabular form the nature of steady state solution of any order differential equation for various types of forcing functions (see in Appendix). To get the complete solution of eq. (10.10), the values of α, A₁ and A are to be computed following the steps given below:

Step-1: How to find the value of α ?

 we get a characteristic equation that will provide us the numerical value of α. This in turn, gives us the transient response of the system provided the constant A₁ is known to us.

The Characteristic equation of (10.10) is 

Step-2: How to obtain the constants A₁ and A ?

It may be noted that the differential eq. (10.10) must be satisfied by the particular integral solution or steady state solution i_f (t) .The value of i_f (t) at steady state condition (i.e.t →∞) can be found out using the eq.(10.11) and it is given below.

Using final condition (t → ∞)

 (note: at steady state (t → ∞) i_f (t) = A = constant from eq. (1U.11))

Using initial condition (t = 0)

Case-A: Assume current flowing through the inductor just before closing the switch ‘S” (at t = 0^-) is − = i(0^-) = 0.

Using the values of α , A₁ and A in equation (10.11), we get the current expression as

The table shows how the current i(t) builds up in a R-L circuit.

Note: Theoretically at time (t → ∞) the current in inductor reaches its steady state value but in practice the inductor current reaches 99.3% of its steady state value at time t = 5 τ (sec.).

The expression for voltage across the external resistance R₁ (see Fig. 10.6(a))

The expression for voltage across the inductor or coil

Graphical representation of equations (10.15)-(10.17) are shown in Fig.10.7 for different choices of circuit parameters (i.e., L & R)

Case-B: Assume current flowing through the inductor just before closing the switch ‘S” (at t = 0^-) is i (0^-)- i₀ ≠0.

Using equations (10.13) and (10.14), we get the values of 

Using these values in equation (10.11), the expression for current flowing through the circuit is given by 

The second part of the right hand side of the expression (10.18) indicates the current flowing to the circuit due to initial current i₀ of inductor and the first part due to the forcing function V_s applied to the circuit. This means that the complete response of the circuit is the algebraic sum of two outputs due to two inputs; namely (i) due to forcing function V_s (ii) due to initial current i₀ through the inductor. This implies that the superposition theorem is also valid for such type of linear circuit. Fig.10.8 shows the response of inductor current when the circuit is excited with a constant voltage source V_s and the initial current through inductor is i₀.

Remark-3:  One can also solve this differential equation by separating the variables and integrating. 

Time constant (τ) for exponential growth response (τ): We have seen that the current through inductor is represented by  

when a series R − L circuit is excited by a constant voltage source (V_s) and an initial current through the inductor i₀ is assumed to be zero. Further it may be noted that the current through the inductor (see fig.10.7) increases as time increases. The shape of growing current before it reaches to a steady state value   entirely depends on the parameters of R − L circuit (i.e. R & L) that associated with the exponential term  grows larger and larger the transient, because of its negative exponential factor, diminishes and disappears, leaving only the steady state.

Definition of Time Constant (τ) of R − L Circuit: It is the time required for any variable or signal (in our case either current (i (t)) or voltage (v_R1 or v_L)) to reach 63.2% (i.e the time at which the factor   becomes  (l-e^-1) x100 = 63.2% of its final value. It is possible to write an exact mathematical expression to calculate the time constant (τ) of any tirst-order ditterential equation.
Let ‘t’ is the time required to reach 63.2% of steady-state value of inductor current (see fig. 10.6(a)) and the corresponding time ‘t’ expression can be obtained as

The behavior of all circuit responses (for first-order differential equation) is fixed by a single time constant  and it provides information about the speed of response or in other words, it indicates how first or slow the system response reaches its steady state from the instant of switching the circuit. Observe the equation (10.15) that the smaller the time constant (τ), the more rapidly the current increases and subsequently it reaches the steady state (or final value) quickly. On the other hand, a circuit with a larger time constant (τ) provides a slow response because it takes longer time to reach steady state. These facts are illustrated in fig.10.7(a). In accordance with convenience, the time constant of an exponential term   is the reciprocal of the coefficient ‘a’ associated with the ‘t’ in the power of exponential term.

Remark-4: An interesting property of exponential term is shown in fig. 10.7(a). The time constant τ of a first order differential equation may be found graphically from the response curve.

It is necessary to draw a tangent to the exponential curve at time ‘t = 0’ and maintained the same slope until it intersects the steady state value of current curve at P point. A perpendicular is drawn from the point P to the time axis and it intersects the time axis at t =τ (see fig. 10.7(a)). Mathematically, this can be easily verified by considering the equation of a straight line tangent to the current curve at t = 0, given by y mt = where m is the slope of the straight line, expressed as

Here, we designated the value of time ‘t ’ required to reach y from   assuming a constant rate (slope) of growth. Thus, 

It is often convenient way of approximating the time constant (τ) of a circuit from the response curve (see fig.10.7(a) for curve-2).

Fall or Decay of current in a R-L circuit

Let us consider the circuit shown in fig. 10.9(a). In this circuit, the switch ‘S’ is closed sufficiently long duration in position ‘1’. This means that the current through the inductor reaching its steady-state value   and it acts, as a short circuit i.e. the voltage across the inductor is nearly equal to zero since resistance R_L << R₁. If the switch ‘S’ is opened at time ‘t’=0 and kept in position ‘2’ for as shown in fig. 10.9(b), this situation is referred to as a source free circuit.

Since the current through an inductor cannot change instantaneously, the current through the inductor just before (i (0^-) and after ( − i(0⁺) opening the switch ‘S’ must be same. Because there is no source to sustain the current flow in inductor, the magnetic field in inductor starts to collapse and this, in turn, will induce a voltage across the inductor. The polarity of this induced voltage across the inductor is just in reverse direction compared to the situation that occurred during the growth of current in inductor (i.e. when the switch ‘S’ is kept in position ‘1’). This is illustrated in fig. 10.9(b), where the voltage induced in inductor is positive at the bottom of the inductor terminal and negative at the top. This implies that the current through inductor will still flow in the same direction, but with a magnitude decaying toward zero. Appling KVL around the closed circuit in fig. 10.9(b), we obtain

The solution of the homogeneous (input free), first-order differential equation with constant coefficients subject to the initial (boundary) inductor current (initial condition,

where α can be found from the characteristic equation of eq.(10.21) described by

At time t = 0 , the initial condition  is used in equation (10.22) to compute the constant A₁ and it is given below.

Using the values of A₁ and α in equation (10.22), we get final expression as

A sketch of i (t) for t > 0 is shown in fig.10.10. Here, transient has ended and steady state has been reached when both current in inductor and voltage across the inductor including its internal resistance are zero.

Time Constant (τ) for exponential decay response: For the source free circuit, it is the time τ by which the current falls to 36.8 percent of its initial value. The initial condition in this case (see fig. 10.9(a) is considered to be the value of inductor’s current at the moment the switch S is opened and kept in position ‘2’. Mathematically, τ is computed as

Alternatively, the time constant for an exponential decay response of a circuit may be computed graphically by adopting the steps (see equations (10.19) and (10.20)) as discussed before. In fig.10.10, a tangent is drawn to the exponential decay curve at time ‘ t = 0 ’ and maintained the same slope until the straight line intercepts time axis at time t =τ . Approximately, the value ofτ can thus be found directly from graphical representation of exponential decay curve.

Energy stored in an inductor

Let us turn our attention to power and energy consideration for an inductor. The instantaneous power absorbed by the inductor is expressed by product of the current through inductor i(t) and the voltage across it v(t).

Since the energy is the product of power and time, the energy absorbed by an inductor over a period is expressed as

where we select the current through inductor at time ‘ t₀ = − ∞’ is i(- ∞). Then, we have  

and from this relation we see that the energy stored in an inductor is always non-negative. At any consequent time at which the current is zero, no energy is stored in the inductor. The ideal inductor (R_L = 0Ω) never dissipates energy, but only stores. In true sense, a physical or practical inductor dissipates a very small amount of stored energy due to its small series resistance.

Example- Fig.10.11 shows the plot of current i (t) through a series R − L circuit when a constant forcing function of magnitude V_S = 50 V is applied to it. Calculate the values of resistance R and inductance L .

Solution: From fig.10.11 one can easily see that the steady state current flowing through the circuit is 10 A and the time constant of the circuit τ = 0.3 sec. The following relationships can be written as

Example-L.10.3 For the circuit shown in Fig.10.12, the switch ‘ ’ has been closed for a long time and then opens at t = 0.

Solution: When the switch S was in closed position for a long time, the circuit reached in steady state condition i.e. the current through inductor is constant and hence, the voltage across the inductor terminals a abd b is zero or in other words, inductor acts as short circuit i.e., (i) V_ab (0^-) = 0V. It can be seen that the no current is flowing through 6 Ω resistor. The following are the currents through different branches just before the switch ‘S’ is opened i.e., at t = 0^-.

 and the current through 10Ω resistor,   

The algebraic sum of these two currents is flowing through the inductor i.e., (ii) i_L (0^-) = 2 + 4 = 6 A.

When the switch ‘ S ’ is in open position The current through inductor at time t = 0⁺ is same as that of current i_L (0^-), since inductor cannot change its current instantaneously .Therefore, the current through i_x (0+) is given by

i_x (0⁺) = i_L (0⁺) = 6A .

Applying KVL around the closed loop at t=0⁺ we get,

The negative sign indicates that inductor terminal ‘b’ as +ve terminal and it acts as a source of energy or mathematically. V_ba (0⁺) 10V.

At steady state condition (t → ∞) the current through inductor is constant and hence inductor acts as a short circuit. This establishes the following relations:

The circuit expression  can be obtained using the KVL around the closed path (see fig.10.12).

KVL equation:

The solution of first order differential equation due to forcing function and initial condition is given by 

Initial and final conditions are:   current through inductor  Using initial and final conditions equation (10.30) we get,

From equation (10.30), we get the final expression as

Example: The switch ‘S’ is closed in position ‘1’ sufficiently long time and then it is kept in position ‘2’ as shown in fig.10.13. Compute the value of v_L and i_L (i) the instant just prior to the switch changing; (ii) the instant just after the switch changes. Find also the rate of change of current through the inductor at time  

Solution: We assume that the circuit has reached at steady state condition when the switch was in position ‘1’. Note, at steady state the inductor acts as short circuit and voltage across the inductor is zero.

At t = 0^- , the current through and the voltage across the inductor are   respectively. When the switch is kept in position ‘2’, current through the inductor cannot change instantaneously but this is not true for the voltage across the inductor. At t = 0⁺ = , one can write the following expressions:

 (‘b’ is more + ve potential than ‘a’ terminal). Note that the stored energy in inductor is dissipated in the resistors. Now, the rate of change of current through inductor at time t = 0⁺ is obtained as

Example: L.10.5 Fig. 10.14(a) shows that a switch ‘S’ has been in position ‘1’ for a long time and is moved in position ‘2’ at time ‘t = 0’. Find the expression v(t) for t ≥ 0 .

Solution: When the switch ‘S’ is in position ‘1’, the current through inductor (using the fundamental property of inductor currents) at steady state condition (see fig.10.14(b)) is  

The circuit for the switch ‘S’ is in position ‘2’ is shown in fig.10.14 (c). The current in inductor can be computed using following two different methods.

Method-1: Using Thevenin’s theorem

Convert the part of a circuit containing independent sources and resistances into an equivalent Thevenin’s voltage source as shown in fig.10.14.(d).

Using the KVL around the closed path is 

The solution of the above equation is given by

where, i_n (t) = complementary/natural/transient solution of eq.(10.32)

i_f (t) = particular/ steady state/final solution of eq.(10.32)

The constants A₁ and A are computed using the initial and final conditions of the circuit when the switch is kept in position ‘2

At time t = 0,

At time t →∞, the current in inductor reached its steady state condition and acts as a short circuit in a dc source network. The current through inductor is

Using the above two equations in (10.33), one can obtain the final voltage expression for voltage v(t) across the terminals ‘a’ and ‘b’ as

Method-2: Mesh current method

Assign the loop currents in clockwise directions and redrawn the circuit as shown in Fig. 10.14(e). The voltage across the terminals ‘a’ and ‘b’ can be obtained by solving the following loop equations.

Loop-1:

Using the value of i₁(t) in equation (10.37) , we get

To solve the above first order differential equation we must know inductor’s initial and final conditions and their values are already known  and

The solution of differential equation (10.38) provides an expression of current i₂ (t) and this, in turn, will give us the expression of i₁(t) . The voltage across the terminals ‘a’ and ‘b ’ is given by

where,  i₂ (t), i₁(t) can be obtained by solving equations (10.38) and (10.36). The expressions for i₂ (t) and hence i₁ (t) arte given below:

Capacitor and its behavior

Fig.10.15 shows a capacitor consists of two pieces of metal (the plates) separated from each other by a good insulator (the dielectric), with two wires (the leads) attached to the metal plates.

A battery is connected across the capacitor to transport charge from one plate to the other until the capacitor charge voltage buildup is equal to the battery voltage V . The voltage across the capacitor depends on how much charge was deposited on the plates and also how much capacitance the capacitor has. In other words, there is a relationship between the voltage (V), charge (Q) and capacitance (C), they are related with a mathematical expression as 

where Q = magnitude of charge stored on each plate, V = voltage applied to the plates and the unit of capacitance is in Farad. Although the capacitance  of a capacitor is the ratio of charge per plate to the applied voltage but it mainly depends on the physical dimension of the capacitor. If the area of the plates is larger, the more would be the amount of charge stored over the surface of the plates, resulting higher value of capacitance. On the other hand, if the spacing ‘ ’ between the plates is closer, accumulates more charge over the parallel plates and thus increases the value of the capacitance. The quality of dielectric material has an effect on capacitance between the plates. The good quality of dielectric material indicates that higher the permittivity, resulting greater the capacitance. The value of capacitance can be expressed in terms physical parameters of capacitor as  where A is the area of each plate, d is the distance between the plates,   is the permittivity of free-space ε_r = relative permittivity of dielectric material and C is the capacitance in Farad. It is important to note that when the applied voltage across the capacitor exceeds a certain value the dielectric material breaks down and loses it insulation property.

Continuity condition of capacitors

To find the current-voltage relationship of the capacitor, one can take the derivative of both sides of Eq.(10.39)

The voltage-current relation can also be represented by another form as

 is voltage across the capacitor at time ‘t₀’. It can be seen that when the voltage across a capacitor is not changing with time, or, in other words, the capacitor is fully charged and the current through the capacitor is zero (see Eq.10.40). This means that the capacitor resembles as an open circuit and blocks the flow of current through the capacitor. Equation (10.40) shows that an instantaneous (Δt = 0) change in capacitance voltage must be accompanied by an infinite current that requiring an infinite power source. In practice, this situation will not occur in any circuits containing energy storing elements. Thus, the voltage across the capacitor (or electric charge q(t)) cannot change instantaneously (ie.,Δt = 0) = , that is we cannot have any discontinuity in voltage across the capacitor.

Remark-5

(i) The voltage across and charge on a capacitor cannot change instantaneously (i.e. v_c(0^-) just right before the change of voltage = v_c(0⁺) just right after the change of voltage). However, current through a capacitor can change abruptly. (ii) The capacitor acts as an open circuit (i.e., when the capacitor is fully charged) when voltage across the capacitor does not change (constant). (iii) These properties of capacitor are important since they will be used to determine “boundary conditions”.

Study of dc transients and steady state response of a series R-C circuit.

Ideal and real capacitors: An ideal capacitor has an infinite dielectric resistance and plates (made of metals) that have zero resistance. However, an ideal capacitor does not exist as all dielectrics have some leakage current and all capacitor plates have some resistance. A capacitor’s leakage resistance is a measure of how much charge (current) it will allow to leak through the dielectric medium. Ideally, a charged capacitor is not supposed to allow leaking any current through the dielectric medium and also assumed not to dissipate any power loss in capacitor plates resistance. Under this situation, the model as shown in fig.10.16(a) represents the ideal capacitor. However, all real or practical capacitor leaks current to some extend due to leakage resistance of dielectric medium. This leakage resistance can be visualized as a resistance connected in parallel with the capacitor and power loss in capacitor plates can be realized with a resistance connected in series with capacitor. The model of a real capacitor is shown in fig. 10.16(b).

In present discussion, an ideal capacitor is considered to study the behavior of dc transients in R-C circuit.

Charging of a capacitor or Growth of a capacitor voltage in dc circuits

Let us consider a simple series R − Ccircuit shown in fig. 10.17(a) is connected through a switch ‘S’ to a constant voltage source V_s.

The switch ‘S’ is closed at time ‘t = 0’ (see fig. 10.7(a)). It is assumed that the capacitor is initially charged with a voltage v_c (0) = v₀ and the current flowing through the circuit at any instant of time ‘t’ after closing the switch is i (t).

The KVL equation around the loop can be written as  

The solution of the above first-order differential equation (10.41) due to forcing function V_s is given by  (natural response/transient response) + v_cf (t) (steady-state response)

The constants A₁ , and A are computed using the initial and boundary conditions. The value of α is obtained from the characteristic equation given by (see in detail in Appendix)

At steady state, the voltage across the capacitor is    which satisfy the original differential equation (10.41). i.e.,

Using the initial condition (at t = 0) in equation (10.43), we get

The values of A₁ , A, and Eq. (10.43) together will give us the final expression for capacitor voltage as 

Response of capacitor voltage with time is shown in fig. 10.18.

Special Case: Assume initial voltage across the capacitor at time ‘ t = 0’ is zero i.e., v_c (0) = v₀ = 0 . The voltage expression for capacitor at any instant of time can be written from Eq.(10.44) with v_c (0) = v₀ = 0 .

Voltage across the capacitance  

Voltage across the resistance  

Charge accumulated on either plate of capacitor at any instant of time is given by

where Q is the final charge accumulated in the capacitor at steady state ( i.e., t → ∞). Once the voltage across the capacitor vc (t) is known, the other quantities (like, v_R (t), i (t), and q (t)) can easily be computed using the above expressions. Fig. 10.19(a) shows growth of capacitor voltage v_c (t) for different choices of circuit parameters (assumed that the capacitor is initially not charged). A sketch for q (t) and i (t) is shown in fig. 10.19(b).

Following the definition given in section L.10.3.1, time constant of each of the exponential expressions described in Eqs. 10.45 to 10.48 may be found as τ = R C (for RC circuit). 

Following the definition given in section L.10.3.1, time constant of each of the exponential expressions described in Eqs. 10.45 to 10.48 may be found as τ = R C (for RC circuit). 

Discharging of a capacitor or Fall of a capacitor voltage in dc circuits 

Fig. 10.17(b) shows that the switch ‘S’ is closed at position ‘1’ for sufficiently long time and the circuit has reached in steady-state condition. At ‘ t = 0’ the switch’ S ’ is opened and kept in position ‘2’ and remains there. Our job is to find the expressions for (i) voltage across the capacitor (v_c) (ii) voltage across the resistance (v_R) (iii) current (i (t)) through the capacitor (discharging current) (iv) discharge of charge (q(t)) through the circuit.

Solution: For t < 0, the switch ‘S’ in position 1. The capacitor acts like an open circuit to dc, but the voltage across the capacitor is same as the supply voltage V_S. Since, the capacitor voltage cannot change instantaneously, this implies that v_c (0^-) = v_c (0⁺) = Vs
When the switch is closed in position ‘2’, the current i(t) will flow through the circuit until capacitor is completely discharged through the resistance R. In other words, the discharging cycle will start at t = 0. Now applying KVL around the loop, we get

The solution of input free differential equation (10.49) is given by

v_c (t) = A₁e^αt

where the value of α is obtained from the characteristic equation and it is equal to     The constant A₁ is obtained using the initial condition of the circuit in Eq.(10.50). Note, at ‘t = 0’ (when the switch is just closed in position ‘2’) the voltage across the capacitor v_c (t) = V_s. Using this condition in Eq.(10.50), we get  

Now the following expressions are written as 

Voltage across the capacitance  

Voltage across the resistance

Voltage across the resistance  

An inspection of the above exponential terms of equations from (10.51) to (10.53) reveals that the time constant of RC circuit is given by τ = RC (sec.)

This means that at time t =τ , the capacitor’s voltage v_c drops to 36.8% of its initial value (see fig. 10.20(a)). For all practical purposes, the dc transient is considered to end after a time span of 5 τ . At such time steady state condition is said to be reached. Plots of above equations as a function of time are depicted in fig. 10.20(a) and fig. 10.20(b) respectively.

Energy stored in a capacitor 

The ideal capacitor does not dissipate any of the energy supplied by the source. It stores energy in the form of an electric field between the conducting plates. Let us consider a voltage source VS is connected to a series R - C circuit and it is assumed that
the capacitor is initially uncharged. The capacitor voltage (v_c(t)) and current (i_c(t)) waveforms during the charging period are shown in fig.10.21 (see the expressions (10.45) and (10.47)) and instantaneous power (p_c (t) = v_c(t)xi(t)) supplied to the capacitor is also shown in the same figure.

Let us consider the instantaneous power supplied to the capacitor is given by 

P_c (t) = V_c (t) X i(t)

Now, the energy supplied to the capacitor in dt second is given by 

Total energy supplied to the capacitor in t seconds is expressed as

(Note initial voltage across capacitor is zero and q(t)t is the charge accumulated on each plate at a time t) .

When the capacitor is fully charged, its terminal voltage is equal to the source voltage V_s. The amount of energy stored in capacitor in the form of electric field is given by

where Q is the final charge accumulated on each plate of the capacitor at steady state ( i.e., t → ∞) i.e., when the capacitor is fully charged.

Example: L.10.6 The switch ‘S’ shown in fig.L.10.22 is kept open for a long time and then it is closed at time

(vi) find the time constants of the circuit before and after the switch is closed

Solution: As we know the voltage across the capacitor v_c (t) cannot change instantaneously due to the principle of conservation of charge. Therefore, the voltage across the capacitor just before the switch is closed vc (0^-) = voltage across the capacitor just after the switch is closed  vc (0⁺) = 40 V (note the terminal ‘a’ is positively charged. It may be noted that the capacitor current before the switch ‘ S’ is closed is i_c (0^-) = 0A.. On the other hand, at t = 0, the current through 10 Ω resistor is zero but the current through capacitor can be computed as 

(note, voltage across the capacitor cannot change instantaneously at instant of switching). The rate of change of capacitor voltage at time ‘t = 0’ is expressed as

Time constant of the circuit before the switch was closed    Time constant of the circuit after the switch is closed is  

(replace the part of the circuit than contains only independent sources and resistive elements by an equivalent Thevenin’s voltage source. In this case, we need only to find the Thevenin resistance R_Th).

Note: When the switch is kept in closed position, initially the capacitor will be in discharge state and subsequently its voltage will decrease with the increase in time. Finally, at steady state the capacitor is charged with a voltage   (theoretically, time required to reach the capacitor voltage at steady value is

Example: The circuit shown in Fig.10.23 has been established for a long time. The switch is closed at time t = 0 . Find the current (i)   (ii) at steady state the voltage across the capacitors,  

Solution: (i) At t = 0^- no current flowing through the circuit, so the voltage at points ‘b’ and ‘d’ are both equal to 50volt. When the switch ‘S’ closes the capacitor voltage remains constant and does not change its voltage instantaneously. The current i₁(0⁺) through a - b branch must then equal to zero, since voltage at terminal ‘b’ is equal to v_b (0⁺)= 50 volt., current through b - c is also zero. One can immediately find out the current through c − e equal

i₃(0⁺) = 1 A which is the only current flow at t = 0⁺ around the loop ‘ d - c - e - d . Note the capacitor across ‘d - e’ branch acts as a voltage source, the change of capacitor

(ii) at steady state the voltage across each capacitor is given

At steady state current delivered by the source to the different branches are given by  

Example: The circuit shown in fig. 10.24(a) is switched on at time t = 0 . How long it takes for the capacitor to attain 70 % of its final voltage? Assume the capacitor is initially not charged. Find also the time constant (τ) of the circuit after the switch is closed.

The circuit containing only resistive elements and independent current source (i.e., nontransient part of the circuit) is converted to an equivalent voltage source which is shown in fig.10.24(b).

Fig.10.24(c) shows the capacitor C is connected across the Thevenin’s voltage terminals ‘a’ and ‘b’ in series with Thevenin’s resistance R_Th . 

The parameters of Thevenin’s voltage source are computed below: 

Using KVL around the closed path, one can find the current through the capacitor and hence, the voltage across the capacitor.

The solution of the differential equation is given by

Using the initial and boundary conditions of the circuit, we obtain the final expression of voltage across the capacitor v_c(t) as

Let ‘t’ is the time required to reach the capacitor voltage 70% of its final (i.e., steady state) voltage.

Example: The switch ‘S’ of the circuit shown in fig.10.25(a) is closed at position ‘1’ at t = 0.

Find voltage v_c(t) and current i_c (t) expressions for t > 0 . Assume that the capacitor is initially fully uncharged (i.e., . v_c (0) = 0).
(i) find the mathematical expressions for v_c(t) and i_c (t) if the switch ‘S’ is thrown into position ‘2’ at t = t (sec.) of the charging phase.
(ii) plot the waveforms obtained in parts (i) to (ii) on the same time axis for the voltage v_c (t) and the current i_c (t) using the identified polarity of voltage and current directions.

Solution: (i) The current source is converted to an equivalent voltage source and it is redrawn in fig.10.25(b) when the switch ‘ S ’ is in position ‘1’.

The voltage expression across the capacitor using the initial and boundary conditions of the circuit, one can write v_c(t) as

Note that the time constant of the circuit in part (i) is τ = RC m =100 sec.

(ii) The switch ‘S’ is thrown into position ‘2’ at t =τ =0.1sec. and the corresponding circuit diagram is shown in fig.10.25(c).

Note, at time t = t = 0.1sec., the capacitor is charged with a voltage = v_c(t = 0.1) = 40(1-e ^-10x0.1) = 40x0.632 = 25.28V and at the same time (t = t = 0.1sec.) the current in capacitor is 4 x e ^-10t = 4x 0.368 = 1.472 (in mA) . Considering the fig.10.25(c), one can write KVL around the closed path

where R_eq = 4 + 6 =10kΩ and the capacitor is now in discharging phase.

The solution of Eq.(10.64) can be found using the initial and final voltage of the capacitor (initial voltage

and it is given by

Discharging current expression is given by (note, current direction is just opposite to the assigned direction and it is taken into account with a –ve sign)

(Note, the above two expressions are valid only for t ≥ τ)

The circuit responses for charging and discharging phases in (i) and (ii) are shown in fig.10.25 (d).

Remark-6 Note that the current through the capacitor (see fig. 10.25(d)) can change instantaneously like the voltage across the inductor.

Appendix-A  

L.10.A Solution of n^th order linear time invariant differential equation excited by forcing function. 

Let us consider a linear time invariant circuit having several energy source elements is described by the following dynamic equation. 

where a₁, a₂, a₃, ••• a_n-1 , a_n are constant coefficients associated in the differential equation and they are dependent on circuit parameters (like, R, L and C for electric circuit) but independent of time, f (t) is the forcing or driving function and x(t) is the solution of differential equation or response of the system. We shall discussion the solution of differential equation restricted to second order differential, say n = 2 in equation (10.A1).

The solution of this differential equation provides the response of circuit and it is given by

where x_n(t) is the natural response of circuit, obtained by setting f (t) = 0, and x_f (t) is the forced response that satisfies the original differential equation (10.A2).
By setting f (t) = 0 in equation (10.A2), as given in equation (10.A4), the force free equation is obtained

 (Homogeneous equation)

The solution of such differential equation (or homogeneous equation) is known as natural solution or complementary solution or transient solution and it is denoted by x_n (t). To get the natural solution x_n (t) of equation (10.A4) the following steps are considered.

Let us use the following operators 

in equation (10.A4) and results an equation given by

Since x ≠ 0 , the above equation can be written as

which is known as characteristic equation for a circuit whose force free equation is Eq.(10.A4). The natural or transient solution of Eq.(10.A4) is expressed by the exponential terms as given below.

where α₁ and α₂ are the roots of characteristic equation (10.A5). The roots of second order characteristic equation with real coefficients is either real or complex occur in conjugate pairs. The constants A₁ and A₂ are evaluated from initial or boundary conditions of circuit. The principles of continuity of inductance current and capacitance voltage are used to establish the required boundary conditions.

If x_n (t) is the natural or transient solution of unforced (or homogeneous) equation differential, it must satisfy its own differential equation  

Further, if   is the complete solution of given differential Eq.(10.A2), it must satisfy its own equation

Using the equation (10.A7) in Eq.(10.A8), we get

The above equation implies that x_f (t) is the forced solution or steady state solution of second order differential equation (10.A2). Steady state solution of some common forcing functions is listed in Table (assume  

Table: Steady state solution x_f (t) for any order differential equation excited by some common forcing function.

Coefficients involve in the steady state solution can be found out by using the boundary conditions of the circuit.

Remark-7

(i) Eq. (10.A2) is the differential equation description of a linear circuit, superposition may be used to find the complete solution of a forcing function which is sum of natural and steady state responses. (ii) Eq.(10.A6) is the natural solution of force-free linear differential equation. Note that the constants α₁ and α₂ are the roots of the characteristic equation (10.A5) and they are entirely depending on the circuit parameters. The roots of the characteristic equation may be classified as

Case-1: Real or Complex but distinct

The natural solution of homogeneous equation (10.A4) is given as

Case-2: Roots are repeated (i.e. α₁ = α₂ = α or multiplicity of roots of order 2) The natural solution of homogeneous equation (10.A4) is given as