8.1

CHAPTER 8 COMPLETE RESPONSE OF RC & RL CIRCUITS

The order of the circuit is defined by the number of irreducible storage elements in the circuit. In this chapter, we will find the complete response of a first-order circuit; a first-order circuit means there is only one storage element in the circuit, either an RC or RL circuit. A second-order circuit means there are two storage elements in the circuit; for example, RLC, LC, L2 or C2. This is the subject of Chapter 9.

There are two types of sources that will be applied to capacitors or inductors in this chapter:

Constant sources first-order response to a constant input source (easier)Time-dependent sources first-order response to a time varied input source (harder)

→

Response of a First-order Circuit to a Constant Input Regardless of how complicated the geometry of an RC or RL circuit is, we will always convert them into Thévenin and Norton equivalents:

RC circuit → convert into a Thévenin Equivalent circuit RL circuit → convert into a Norton Equivalent circuit

Why this approach? It turns out that this approach is more physically intuitive and easier to “see the answer” than other approaches that are more mathematical.

RC CIRCUIT A typical RC circuit will not be in Thévenin form. So we will convert any RC circuit into a Thévenin circuit by removing the load capacitor and using the standard tricks of the trade. Picturewise, we have

Our primary focus is analyzing the behavior of the capacitor’s voltage after a switch has been thrown. That is, we are most interested in determining its behavior vC(0 < t < 5τ) for t = 0+. Physically, this means the capacitor will either charge up or discharge from its initial to its final steady state value and vC(t) will describe this behavior completely. The sudden change in the circuit due to a switch being thrown, breaks the RC behavior into three distinct time periods.

• t < 0 (Circuit #1): Before the switch is closed, the voltage source has charged the capacitor to its initial steady state value vC(0−).

• 0 < t < 5τ (Circuit #2 & #3): The switch is thrown at t = 0 and the circuit is converted into a Thévenin equivalent. (This requires two circuits, (i) VOC and (ii) RTh). The capacitor’s voltage either charging or discharging from its initial value vC(0−) and this transient behavior is time dependent and exponential: vC(0 < t < ∞) ≡ vC(t).

• t = ∞: When sufficient time has elapsed, the capacitor has fully charged and has reached its final steady state value defined by vC(∞) = VTh = VOC.

• Solving the circuit problem (Circuit #4). If the capacitor’s voltage is not what is being solved for, one has to “unfold” the circuit to solve for the appropriate parameter of the problem.

8.2

In summary, there are two steady state values but it is what is in-between these two were the capacitor’s transient behavior is observed and interests us the most. When the capacitor's voltage is completely known it is called the complete response:

C C

C C

C oc

v (t 0) v (0 ) Initial steady-state valuev (t) Complete Response v (0 t 5 ) -state

v (t ) V Final steadyt

-ra

stnsient

ate value

− < ≡

≡ = < < τ = ∞ ≡

Looking ahead, we will find that this behavior is summarized in the following equation: Tht /R C

C C

Complete Response Forced Response Natural Response

oc oc v (t) V (v (0 ) V )e−−= + −

Let's start analyzing an RC circuit. Every time a switch is thrown, a new circuit is created and the capacitor naturally responding to its environment (how the resistors are organized) by either charging or discharging to the sudden change. Overall view of solving RC circuits with a constant source 1. BEFORE the switch is thrown, determine the initial steady state voltage vC(0−). 2. AFTER the switch is thrown, convert the RC circuit into a Thévenin circuit. 3. Solve the 1st order equation for the capacitor’s voltage vC(0 < t < ∞) ≡ vC(t). 4. Undo the Thévenin circuit and solve the circuit problem and interpret.

Step 1: BEFORE the switch is thrown, we solve the circuit for vC(0−). VDR leads the way and gives

3C S

1 2 3

Rv (0 ) VR R R

− =+ +

Step 2: AFTER the switch is thrown, the RC circuit is converted into a Thévenin equivalent using standard techniques. We remove the load capacitor and determine VOC and RTh:

The resulting Thévenin equivalent is

Step 3: To find the voltage vC(t), apply KVL to the Thévenin equivalent and solve for vC(t):

C CTh C C i Cdv /dtocV R i v (t)

== +

Rewriting it, we find

C C CTh C

Th Th

ococ

dv dv v V V R C v (t) dt dt R C R C

= + → + =

8.3

This first-order response equation is a form we will encounter again with the RL circuit, so it is useful to write this in a general form so that we can use it again. Define the parameters

CC C

ThTh Th

Th

oc

oc

x vdv v V dx xR C kdt R C R C dt

k V R C

= τ = → + = → + =

τ=

where τ is the time constant of the circuit and k is the effective source term since it is directly proportional to VOC. To integrate, I must isolate the x and t terms:

[ ]t / constant t/

dx (x k ) dx dt t exp Ln(x k ) exp integration constant dt (x k ) (x k ) e e Ae

− τ − τ

− τ = − → = − → − τ = − + τ − τ τ τ → − τ = ≡

∫ ∫

t / x(t) k Ae− τ→ = τ +

Applying the Initial Conditions (ICs) to determine A and k To determine the complete response x(t), constants A and k must be evaluated with the Initial Conditions (ICs) established by the circuit. These ICs are the initial x(0−) and final x(∞) steady state values of the circuit. Besides deriving the first order equation, the ICs become one of the most import steps in determining the complete response. Why? It is the ICs that "fit" the solution to the particular circuit you are solving and forces the solution to be realistic and finite. In other words, the general response has to match the initial and final steady states values determined by the circuit. In this process, we must setup two equations in order to solve all constants relative to x(0−) and x(∞). Of course, a Math 7 course cannot do this because these course cannot be expected to teach either physics or engineering while teaching the fundamentals of problem solving.

• Before the switch is thrown, the capacitor’s voltage has reached its initial steady-state value of x(0−). Clearly, the general response has to satisfy this value at t = 0− and we force this IC onto the general response:

x(0 ) k A 1 A x(0 ) k− −= τ + ⋅ → = − τ

• After the switch is thrown, the capacitor’s voltage has evolved and reaches its final steady-state value x(∞). Once again, we force the general response to satisfy this IC at t = ∞:

t /x( ) k Ae− τ∞ = τ + =x( ) k

kA x(0 ) x( )−

∞ = ττ → = − ∞

So the application of the ICs on the general response places strong conditions on its shape:

A x(0 ) k A x(0 ) x( )k x( ) k x( )

− −= − τ = − ∞→

τ = ∞ τ = ∞

Combining all the terms together, the first-order response takes on the form of

t / t /A x(0) x( )k x( )

complete response forced response natural response

x(t) k Ae x(t) x( ) [x(0 ) x( )]e− τ + − τ= − ∞

τ= ∞= τ + → = ∞ + − ∞

Now translating this to the complete response of the capacitor’s voltage gives

Th

C

t /R CC CC C

Th Th x(t ) v (t ) complete response forced response natural response

ococ oc

dv v V v (t) V [v (0 ) V ]e dt R C R C

−+

=

+ = → = + −

8.4

Physical interpretation 1. Mathematically, the complete response satisfies the ICs:

C ocv (0 ) V− = C oc[v (0) V+ −

0C

C Coc oc

] e v (0)

v ( ) V [v (0) V ]

⋅ =

∞ = + − oce V−∞

⋅ =

2. Charging/Discharging of a capacitor. What does it mean for a capacitor to charge up? It means that vC(0−) is less than vC(∞) = VOC: VOC > vC(0−). Mathematically, the complete response reads

t / t /

C C

negative

C

oc oc oc

oc

v (t) V [v (0 ) V ] e V Ve

where V |v (0 ) V |

+ − τ − τ

+

= + − ≡ − ∆

∆ = −

That is, the minus sign indicates that the capacitor is charging to a higher voltage value. Physically (and in the words of Deane), when negative work is done on the capacitor it increases the electrical potential of the capacitor since the capacitor moves up higher on the electrical incline. What does it mean for a capacitor to discharge? It means that VOC is less than vC(0−): vC(0−) > VOC. Mathematically, the complete response tells us

t / t /

C Coc oc oc

positive

v (t) V [v (0 ) V ] e V Ve+ − τ − τ= + − ≡ + ∆

That is, the plus sign indicates that the capacitor is discharging to a lower voltage value. Physically, when positive work is done it decreases the electrical potential of the capacitor and moves lower down the electrical incline.

3. Time Constant

Another way to interpret the complete response and the role that the time constant plays, is to rewrite the complete response as

( ) ( ) t / t / t / t / t /

C C C initial finaloc oc ocv (t) V [v (0 ) V ]e v (0 )e V 1 e V e V 1 e+ − τ + − τ − τ − τ − τ= + − = + − ≡ + − These two terms show that

( )t /

initial

t /final

V e discharges from initial steady-state value

V 1 e charges to the final steady-state value

− τ

− τ

→

− →

If we only focus on the exponential behavior according to the table below,

• the capacitor has discharged from its initial voltage vC(0−) to essentially zero (0.7%

of its original value) after 5 time constants (5τ). • the capacitor has charged to its final steady state value of VOC after 5τ.

( )5 5C initial final initial final final ocv (5 ) V e V 1 e 0.007 V 0.993 V V V− −τ = + − = ⋅ + ⋅ ≈ ≡

8.5

The charging/discharging depends ONLY on the circuit elements: RTh and C.

Problem Solving Strategies for First Order Circuits with constant sources The four-step process for solving for the capacitor voltage is 1. Determine the capacitor’s initial steady state voltage vC(0−) before the switch is thrown. 2. After the switch is thrown, determine the Thévenin parameters (VOC & RTh) and the

inverse time constant 1/τ. 3. Put the voltage into solution form and interpret the solution:

Tht /R CC Coc ocv (t) V [v (0) V ]e−= + −

4. To solve the circuit problem for the appropriate parameter, go the circuit before the RC circuit was converted into a Thévenin circuit and relate unknown parameter to vC(t) using KCL or KVL.

Let’s apply this procedure to the following two examples.

Example 8.1 The circuit is in steady state before the switch closes at t = 0 s. a. Describe the behavior of the capacitor for all time? Is it charging or

discharging after the switch is closed? b. (i) Determine the capacitor voltage vC(t) for t > 0 s. (ii) Plot vC(t) vs. t

for -∞ < t < +∞.

Solution I will follow the problem-solving strategies from above.

Step 1: Determine vC(0−) = vC(0+) To determine the initial voltage vC(0−) before the switch is thrown, we replace the capacitor with its steady state equivalent and use whatever method is most convenient to get vC(0−). Since is circuit is completely open, the voltage across the capacitor is the same as the 3V-source: vC(0−) = 3V since v3Ω = v6Ω = 0.

Step 2: Determine the Thévenin parameters (VOC & RTh) and 1/τ

The switch is thrown and the RC circuit must be converted into a Thévenin equivalent. There are two circuits needed to find VOC & ISC:

• To determine VOC, we see that v6Ω = VOC use VDR to solve for it:

6 oc6v 3V 2V V

6 3Ω = ⋅ = =+

• To determine RTH, since there are only independent sources, we can short out the voltage source and solve for RTH:

ThR 6 6 3 8= + = Ω

Determine the time constant 1/τ

8.6

( )Th1 1 1R C 8 0.05 0.4 s 2.5 Hz

0.4τ = = ⋅ = → = = =

τ τ

Step 3: Put the voltage vC(t) into solution form and interpret the solution ( )

t / t /0.4 2.5tC Coc ocv (t) V [v (0 ) V ]e 2 3 2 e 2 e V+ − τ − −= + − = + − = +

Putting this together, the capacitor voltage behaves as

C 2.5t

3V t 0v (t)

2 e V t 0−

<=

+ ≥

Interpretation: the capacitor’s voltage starts off at 3V and discharges to 2V. After 5τ, the capacitor has almost arrived at its final voltage of 3V. The open circuit voltage acts like the terminal velocity of an object falling in air: it does not change it value again. Does the capacitor behave as it is supposed to? That is, is vC(0−) = vC(0+) and iC(0−) ≠ iC(0+)?

0C C

0CC C

v (0 ) v (0 ) 3V 2 e 3V

dv (0 )i (0 ) i (0 ) 0 C 0.05(0 2.5e ) 125 mAdt

− +

+− +

= → = + =

≠ → ≠ = − =

So, when the switch was thrown, the capacitor’s voltage doesn’t change instantly but the capacitor’s current does since it jumped from 0A to 125 mA; there is a discontinuity!

Example 8.2 a. Describe the behavior of the capacitor’s voltage: (i) initial and final steady states, (ii)

charging or discharging? Assume steady state at t = 0−.

b. (i) Determine the current ix(t) for t > 0 s. (ii) Plot ix(t) vs. t for -∞ < t < +∞.

Solution Does the capacitor charge up or discharge when the switch is closed? I would claim that the capacitor discharges because before the switch is thrown, both sources will be contributing to the capacitor’s voltage. However, when the switch is closed, it cuts out the 4V-source and therefore, the capacitor sees less “source” adding to it.

Step 1: Determine vC(0−) With the circuit at initial steady state, the resulting circuit is shown on the right. To solve for vC(0−), I identify the nodes and note that there are two loops but one of the loops contains an isolated 4mA-source. This leads me to choose Simple Circuits in solving for vC(0−). Setting the ground at the bottom, I am going to follow the current starting with the 4mA-source. I see that the 4mA-source is in series with the right most 2kΩ-resistor. Since the 4V-source is connected to the ground, and I know the voltage across the 2kΩ-resistor, the node voltage for the positive terminal of vC(0−) can be calculated:

C 2kv ( ) 4V v 4 2k 4m 12VΩ+ = + = + ⋅ = To find the negative terminal of vC(−), I continue to follow the current through the ground and see that at the bottom node, the current is split evenly between the two branches since both branches have 4kΩ resistors. So 2mA goes up the 2kΩ, giving

8.7

Cv ( ) 2k 2m 4V− = − ⋅ = − Therefore, vC(0−) is

C C C Cv (0 ) v ( ) v ( ) 12 ( 4) 16V v (0 )− −= + − − = − − = =

Step 2: Find VOC, RTh and 1/τ The switch has now been thrown and the circuit now looks like

Converting this RC circuit into a Thévenin equivalent, the VOC- and RTh-circuits are below.

Determine VOC Using similar thinking as when we determine vC(0−), I am going to label the nodes and follow the current. The positive node VOC(+) is grounded and the 4mA-source splits between the two 4kΩ resistors such that half of the current (2mA) split at the bottom node again. The negative VOC(−) terminal of the open circuit is equal to v2kΩ. From Ohm's law, the voltage v2kΩ is -4V and VOC is

oc oc oc oc( ) ( ) 0 ( 4V) 4VV V V V= + − − = − − = = Determine RTh Opening the current source, the Thévenin resistance is

Th Th32R 6 2 k R= = Ω =

Determine 1/τ

The time constant is Th

1 1 2 1 20 1HzR C 3k 100 3

= = ⋅ = =τ µ τ

Step 3: Writing out the solution vC(t), we have

( ) t / 20t /3 20t /3

C Coc ocv (t) V [v (0 ) V ]e 4 16 4 e 4 12e V+ − τ − −= + − = + − = +

Putting this together, the capacitor voltage behaves as

C 20t/3

16 V t 0v (t)

4 12e V t 0−

<=

+ ≥

Does this agree with our initial assumptions whether the capacitor was going to either discharge or charge up? Because of the plus-sign on the transient solution, the capacitor discharges from 16V to 4V, as expected.

Step 4: Solving the circuit problem for ix(t), we need to determine (i) ix(0−) and (ii) ix(t). If we go back to the calculation for vC(0−), we find that ix(0−) was opposite in direction to the current: ix(0−) = −2mA. To determine ix(t), we to go the circuit before we converted the RC circuit into a Thévenin circuit and relate ix(t) to vC(t). In doing so, we see that vC(t) is “antiparallel” (opposite polarity) to vx(t); vx = −vC. Using Ohm’s law to determine ix(t), the current is

8.8

x 20t /3C

2mA t 0i (t) v (t) 2 6e mA t 0

2k−

− ≤=

− = − − >

Does this make physical sense? If we redraw the circuit and focus on the capacitor and the 2kΩ in question, it becomes clear that they are opposite in polarity and the reason why the current ix(t) is negative. So why is there a sudden discontinuity jump in the 2kΩ resistor? Since the capacitor is being forced to discharge, the current ix(t) is also being force to go to a lower (more negative) value as well. Checking on the capacitor’s current at t = 0+, we find that

4CC

dv (0 ) 20i (0 ) C 10 12 8mAdt 3

++ −= = ⋅ ⋅ =

As expected, we expected a discontinuity jump on the capacitor’s current from 0mA to 8mA. Furthermore, as the 14mA-source takes, the current reaches its final steady state value of ix(5τ = 0.75 s) = −2mA. RL Circuit Having now studied the RC circuit in detail, similar type of thinking occurs with the RL circuit and so of instead of repeating these steps, I will refer to the RC circuit for details. An RL circuit is converted into a Norton equivalent to find the current through an inductor:

Our primary focus is analyzing the behavior of the inductor’s current after a switch has been thrown: iL(0 < t < 5τ). That is, we want to answer the question of whether the inductor either stores up current or decays it from some initial steady state value in this parallel RL circuit. Similar to the RC behavior, three distinct time periods.

• t < 0 (Circuit #1): Before the switch is closed, the source stores up current in the inductor and establishes its initial steady state value iL(0−).

• 0 < t < 5τ (Circuit #2 & #3): The switch is thrown at t = 0+ and the circuit is converted into a Norton equivalent. (This requires two circuits, (i) ISC and (ii) RTh). Here, the inductor’s current either stores current or decays it from its initial value iL(0−). This transient behavior of the inductor’s current is time dependent and exponential: iL(0 < t < ∞) ≡ iL(t).

• t = ∞: When sufficient time has elapsed, the inductor is saturated and has reached its final steady state value defined by iL(∞) = IN (= ISC).

• Solving the circuit problem (Circuit #4). If the inductor’s current is not what is being solved for, one has to “unfold” the circuit to solve for the appropriate parameter of the problem.

In summary, there are two steady state values and it is in-between these two steady-states that the inductor’s transient current behavior is what we will be solving for:

I

L L

L L

L SC

i (t 0) i (0 ) Initial steady-state valuei (t) Complete Response i (0 t ) -state

i (t ) tFinal steady-state valransient

ue

− < ≡

≡ = < < ∞ = ∞ ≡

8.9

Looking ahead, we will find that this behavior is summarizes as I I Tht /(L/R )

L SC L SC

Complete Response Forced Response Natural Response

i (t) ( i (0 ) )e−−= + −

Skipping ahead (assuming we’ve already determine iL(0−)), we immediately go to the Norton circuit and solve for the inductor’s current iL(t) by applying KCL:

IThSC R Li i (t)= +

Since parallel elements have equal voltages, we use of Ohm’s law to rewrite the current through RTh

Th Th

L L LR L R

Th Th

di v (t) diLv v (t) L i (t)dt R R dt

= = → = =

Inserting this in the KCL equation above, gives the first-order response equation: I

ITh

SCL L LSC R L L

Th Th Th

di di iLi i (t) i (t) R dt dt L / R L / R

= + = + ⇒ + =

Since this has the same form as that of the Thévenin RC circuit differential, it clearly have the same solution form:

Lt /

L

dx x k x(t) x( ) [x(0 ) x( )]edt

− τ−+ = → = ∞ + − ∞τ

Translating this for the inductor, the completer response of the inductor’s current is I I

Lt /L SC L SC Th

Complete Response Forced Response Natural Response

i (t) [i (0 ) ]e where L/R− τ+= + − τ =

Because of the first order equation is the same, the Thévenin RC and Norton RL circuit have all the same physical traits and behaviors. I will only highlight how the inductor stores/decays the current.

Physical interpretation – Storing/Decaying of an inductor’s current What does it mean for an inductor to store up current? It means that iL(0−) is less than iL(∞) = ISC: ISC > iL(0−). Mathematically, the complete response reads

I I I I

I I

t / t /

L SC L SC SC

negative

L SC

i (t) [i (0 ) ] e e

where |i (0 ) |

+ − τ − τ

+

= + − ≡ − ∆

∆ = −

That is, the minus sign indicates that the inductor is storing a higher current value.

What does it mean for an inductor to decay current? It means that iL(0−) > ISC and now reads

I I I I t / t /

L SC L SC SC

positive

i (t) [i (0 ) ] e e + − τ − τ= + − ≡ + ∆

That is, the plus sign indicates that the inductor is decaying from a higher to a lower current value.

Example 8.3 a. Describe the behavior of the inductor’s current: (i)

initial and final steady states, (ii) storing or decaying current? Assume steady state at t = 0−.

b. (iii) Determine the voltage vx(t) for t > 0 s. (iv) Plot vx(t) vs. t for -∞ < t < +∞.

Solution

8.10

Is the inductor storing current or decaying it? At the initial steady state, the inductor shorts out everything left of the inductor. When the switch is thrown, it kills of the 500mA-source and now the inductor has to share the current with the rest of the resistors. So I expect a decaying current! We want to determine the voltage across the 600Ω resistor, not the current of the inductor. However, the inductor controls the current flow in the circuit and we first solve for the inductor current and then relate iL(t) to vx(t) to solve the circuit problem.

Step 1: Determine iL(0−) At the initial steady state, the inductor shorts out the resistors on the left. Therefore, current of the inductor and the current source are the same:

Li (0 ) 0.5A− =

Step 2: Determine ISC, RTh and 1/τ When the switch is thrown, we now convert the RL circuit

into a Norton equivalent. The two resulting circuit will give us ISC and RTh.

Determine ISC Replacing the inductor with a short circuit, the current through the i400Ω = ISC. Using VDR and Ohm's law to get i400Ω, we get

( )

ISC

400

400400

600 400 100V 37.5V600 400 400

93.8mA400

v

vi

Ω

ΩΩ

= ⋅ =+

→ = = =Ω

Determine RTh By shorting out the voltage source

Th ThR 400 400 600 400 240 640 R= + = + = Ω =

Determine 1/τL

The inverse time constant is Th

L L

R1 640 16400 HzL 0.1

= = = =τ τ

Step 3: Writing out the solution to the current iL(t) gives us

L 6400t

500 mA t 0i (t)

93.8 406.2e mA t 0−

<=

+ ≥

Interpretation: the inductor’s current starts off at 500mA and decays to a lower final steady-state value of 93.8mA (at 5τ = 0.78 ms) since it now “shares” the current with the other resistors when the 100V-source takes over. This is what we expected.

Step 4: Solving the circuit problem for v600Ω We need to determine (i) v600Ω (0−) and (ii) v600Ω(t). If we go back to the calculation for iL(0−), we immediately note that the 600Ω resistor was shorted out: v600Ω(0−) = 0. To determine v600Ω(t), we to go the circuit before we converted the RL circuit into a Norton circuit and relate v600Ω(t) to iL(t). Applying KVL to the right-handed loop in the circuit, we get

8.11

L600 400 L L

110

div v v (t) 400i (t)dtΩ Ω= + = +

Substituting in the inductor’s current iL(t) into KVL gives

( ) ( ) ( )

6400t 6400t600

6400t 6400t600

110v 400 93.8 406.2 e mA 406.2 6400 e mA

37.5 (162.5 260)e V 37.5 97.5e v (t)

− −Ω

− −Ω

= ⋅ + + ⋅ −

= + − = − =

Does this make physical sense? Since the 600Ω was initially shorted out, it starts off with no voltage drop across it. However, at t = 0+, we have a huge discontinuity:

( )0600v (0 ) 37.5 97.5e V 60V+

Ω = − = − Why is there a sudden large discontinuity jump from 0V to −60V? First note that one expects a large voltage drop across the inductor since that is when we expect the back emf to be the largest. Calculating the inductor’s voltage at t = 0+, we find that

3LL

110

di (0 )v (0 ) L 406.2 6400 10 260Vdt

++ −−= = ⋅ × = −

So at t = 0+, the inductor dominates the circuit and gives vx(0+) a polarity opposite to the 100V-source. However, the inductor cannot continue to sustain this voltage drop and dies off exponentially to 0V. The 100V-source takes over and charges the 600Ω resistor with the opposite polarity until it reaches its final steady state value of vx(5τ = 0.78 ms) = 37.5V.

As one can see from this example problem, solving these types of problems will become time consuming simply due to the fact that there are many steps in-between.

SEQUENTIAL SWITCHING When a series of sequential switches are thrown at various times, a series of sequential charge-ups or discharges will occur with the capacitor. Of course, every time a new circuit is created with the throwing of a switch, it creates new time constants and new complete responses.

Mathematically what do we expect? Circuit 1: The switch is closed at t = 0− and steady state is established. The complete response of circuit-1 is

1t /C1 1 Coc1 oc1v (0 t t ) V (v (0 ) V )e− τ−< < = + −

Circuit 2: When the switched is thrown at t = t1, a second RC circuit is created that requires its own complete response and new time constant τ2. However, there are two new things that have to be accounted for. • The discharging of the capacitor starts at t = t1 and therefore, its complete response is

time shifted: t is replaced by t−t1. • The capacitor’s initial voltage is the final steady state value of the previous circuit-1 at

time t = t1. With these modifications, the discharging behavior of the capacitor then looks like

1

2 1

3 2

Initial steady state t 0 Circuit-1 with t 0Circuit-2 with t tCircuit-3 with t t

−

+

=

τ = τ = τ =

8.12

1 2(t t )/C2 1 2 C1 1oc2 oc2v (t t t ) V (v (t ) V )e− − τ< < = + −

Circuit 3: Following the same ideas as with circuit-2’s complete response, we write

C3 2 oc3v (t t ) V< < ∞ = C2 2 oc3(v (t ) V+ − 1 2 3 1 2 3( t t t )/ ( t t t )/C2 2)e v (t )e− − − τ − − − τ=

Let’s apply these ideas to the following example circuit.

Example 8.4 a. Describe the behavior of the inductor’s current: (i) initial and final steady states, (ii)

storing or decaying current? Assume steady state at t = 0−.

b. (iii) Determine the current i12Ω(t) for t > 0. (iv) Plot i12Ω (t) vs. t for -∞ < t < +∞.

Solution Since there are two switches, there are three-time intervals: t = 0−, 0 < t < 51ms, and t > 51ms.

Circuit #0: Determine iL1(0−) Note that i2Ω(0−) = iL1(0−). Using VDR to determine v2Ω(0−) and then apply Ohm’s law to calculate i2Ω(0−), we get

2 2 L14V2

2 6v 52V 4V i (0 ) 2A i (0 )2 6 (6 12)

− −Ω Ω

= = → = = =+ +

Determining the current i12Ω(0−) through the 12Ω resistor, we get

12 2 1232V12

12v 52V 32V i (0 ) 2.67A i (0 )2 6 (6 12)

− −Ω ΩΩ

= = → = = =+ +

Circuit #1: Determine iL1(0 < t < 0.051s) At t = 0+, the left switch is thrown close and shorts out the 52V-source, leaving the circuit without a source. Clearly, the inductor will decay it current.

• Determine the Norton parameters (ISC1, RTh1, τL1) for 0 < t < 0.051s.

Since there is no source supplying the circuit, we immediately know that ISC1 = 0. The Thévenin resistance is

Th1R 2 12 6 6= + = Ω . so that the inverse time constant is

Th1

1 1

61H

R1 16 HzL

Ω= = = =

τ τ

• Putting this into solution form: IL1 SC1i (0 t 0.051s)< < = IL SC1i (0 )−+ −( ) 1t / 6te 2e A− τ −=

• Determining the current i12Ω(0 < t < 0.051s), we can use CDR to determine it: 6t

12 L1 126 2i (0 t 0.051s) i e A i (0 t 0.051s)

6 12 3−

Ω Ω< < = = = < <+

Circuit #2: Determine iL2(t > 0.051s)

The second switch is thrown before the inductor has had the time to completely decay away its current.

8.13

So the complete response after the second switch is thrown has to be time shifted to t = t = 51ms and is given by

( )I I 2( t 0.051)/L2 SC2 L1 SC2i (t 0.051s) i (0.051s) e− − τ> = + −

• Determine the initial steady state value iL1(0.051s) The inductor’s final steady state value before the switch was thrown at t = 0.051 s becomes the initial steady state value after the switch is thrown:

6tL1 t 0.051si (t 0.051s) 2e 1.473A−

== = =

• Determine the Norton parameters (ISC2 , RTh2, τL2) for t > 0.051s With no supplying source in the circuit again, ISC2 = 0. The Thévenin resistance is simply those two series resistors and we get

ISC2 Th20 and R 2 12 14= = + = Ω The inverse time constant is

Th2

2 2

141H

R1 114 HzL

Ω= = = =

τ τ

• Putting this into solution form:

IL2 SC2i (t 0.051s)> = IL1 SC2i (0.051s)+ −( ) 2( t 0.051)/ 14(t 0.051)e 1.473e A− − τ − −=

Since the inductor and the 12Ω resistor are in series, they have the same current:

14(t 0.051)L2 12i (t 0.051s) i (t 0.051s) 1.473 e A− −

Ω> = > =

Summarizing the two-time intervals, we get

6t212 3

14(t 0.051)

2.67A 0 ti (t) e A 0 t 0.051s

1.473 e A 0 t 0.051s

−Ω

− −

>

= < < < <

By looking at the time constants for the two-time intervals, note that τ1 = 1/6 s while τ2 = 1/14 s. The inductor decayed at a much faster rate after the second switch was thrown.

UNIT STEP SOURCE and PULSE SOURCE In order to simplify switches in circuits, we now imbed switches in with the sources using unit step functions. A unit step function is defined as a mathematical function that has an instantaneous jump from zero to one at time t = t0:

00

0

0 t t Unit Step function u(t t )

1 t t<

≡ − = >

So how do we use this unit step function with sources? Consider the circuit

8.14

So instead of using a complicated set of switches in order to turn on a source, we will replace it with the unit step function (it easier to do it this way). Physically, this appears like the turning on of a light switch in your kitchen. There is an instantaneous voltage increases to V0 from zero volts.

To produce a pulse square voltage source of finite duration (or finite duty), a superposition of these two sources with unit step functions are used at different times.

RESPONSE of FIRST-ORDER CIRCUITS to NONCONSTANT SOURCES Up to now, we have only had constant sources. We now introduce time dependent sources so that the first order equation has a right-handed side that is not constant but time dependent.

k constant f(t) constant

dx x dx xk vs. f(t) dt dt

= ≠

+ = + =τ τ

There will be two significant changes how one derives the complete response for the time dependent circuit.

1. For a constant source, we converted our RC or RL circuit into a Thévenin or Norton circuit. As a result, the solution always had the same form of

t/x(t) x( ) x(0 ) x( ) e− − τ = ∞ + − ∞ For time dependent sources, we will find that it is more time effective to not convert the circuit into a Thévenin or Norton circuit, but “grab it” by the horns and derive a new first order equation every time. We will only encounter two types of time dependent sources in this chapter: exponential and sinusoidal.

at1 2

1 2

K K ef(t) effective source

K K sin t (cos t)

− +≡ = + ω ω

2. Since we will not have the same first order equation, the complete response will have to be derived every single time, and consequently, is more complicated. There is an additional minor complication that arises because there is a change in language (which is most unfortunate). Here is the language change:

force response natural response

completeresponse

natural response force response

t/

t/ t /

x( ) [x(0 ) x( )]e ) constant source

x(t) x(0 )e x( )(1 e ) nonconstant sour

−

−

− τ

− τ − τ

∞ + − ∞ →

=+ ∞ − →

ce

In my personal opinion, I prefer the bottom complete response form since I find it more “physical” in explaining the behavior of the circuit. The important thing to realize is that this is simply a change in language (and therefore, no “big deal”).

Interpretation of time dependent complete response When a constant source is driving the circuit, the storage element’s behavior is to exponentially decay from its initial state value x(0−) while forcing it towards its final state value x(∞) according to

complete response natural response force response

t/ t/ x(t) x(0 )e x( )(1 e ) − − τ − τ= + ∞ −

8.15

A time dependent source driving the circuit behaves exactly the same way but the big change is the force response is now time dependent

f n f complete response natural response force response

t/ x(t) x(0 )e x (t) x x− − τ= + ≡ +

Furthermore, we know a lot about the force response because physically, if the time dependent source is sinusoidal, then the force response must also be sinusoidal. In other words, the functional shape of the time dependent source will “force” the force response to have the exact functional form as the time dependent source.

There are two types of time dependent sources (exponential and sinusoidal) that we will encounter and therefore, two types of force responses (exponential and sinusoidal):

atat1 21 2

f1 2 31 2effective force

source response

B e B K K e f(t) x (t)

B sin t B cos t) BK K sin t (or c os t)

−− ++ ≡ → ≡ ω + ω ++ ω ω

A “worded brief overview” of the process is as follows: • Derive the first order equation using Simple Circuits, and do a simply check to see if

the final steady state agrees with your equation. • From the first order equation, you will be able to write the general complete response

(natural + force) down that contains an arbitrary constant A. In the language of differential equations, one will generate a homogeneous & particular solution to the first order equation.

• Derive the forcing response constants B1 and B2; Solve for the IC x(0−) and apply it to the general response to derive the arbitrary constant A.

• Write out the complete response (with all solved for constants) and solve the circuit problem

Problem Solving Strategies for First-order Circuits with Time Dependent Sources Capacitor equation: iC = C dvC/dt Inductor equation: vL = L diL/dt

Step 1: Derive the First Order Equation Use Simple Circuits to write an equation that contains x and dx/dt (i.e. vC & iC or iL & vL). Rewrite any resistor voltage or current in terms of (vC & iC or iL & vL). Apply the capacitor or inductor equation to convert the first order equation into a first order differential equation and write it into it “proper form”:

dx x f(t)dt

+ =τ

Do a simply check to see if the final steady state x(∞) agrees with your equation.

Step 2: Derive the general response equation The general response equation is of the form

-t /natural force 1 f1 2 f2 3x(t) x (t) x (t) Ae B x B x Bτ= + = + + +

Forcing Response xf(t) The forcing response has the same form as the effective source term f(t) in the first order equation:

at

1 21 2f

1 2 31 2

-at B e B K K e f(t) x (t)

B sin( t) B cos( t) B K K sin( t) (or cos( t))

− ++ = → = ω + ω ++ ω ω

Substitute the general force response into the first order equation and solve for constants B1 and B2:

f 1 f1 2 f2 3

f ff 1 f1 2 f2 3

x (t ) B x + B x B

dx x f(t) x (t) B x + B x Bdt = +

+ = → = +τ

Substitute the constants B1, B2 and B3 into the general response equation.

8.16

Natural Response xn(t) From the first order equation, pick-off the time constant 1/τ and write down the natural response:

nt/x Ae− τ=

Write the general complete response were the only unknown constant is A: t/ at

1 2two possiblet /1 f1 2 f 2 3 solutions t /

1 2 3

Ae B e Bx(t) Ae B x B x B

Ae B sin t B cos t B

− τ −− τ

− τ

+ += + + + → + ω + ω +

Step 3: Solve the IC for x(0−) and the arbitrary constant A • Solve for the IC [vC(0−) or iL(0−)] • Apply the IC to the general complete response equation and solve for constant A. • Write out the complete response

Step 4: To solve the circuit problem for the appropriate parameter, go to the circuit at t = 0+ and relate the unknown parameter to x(t) using KCL or KVL or both.

Let's consider a straightforward example and go through the above steps to get an idea as to what to do.

Example Determine vC(t) for t > 0 when vC(0−) = 4V.

Solution What do we expect for the capacitor’s response vC(t) to the sine voltage source? Since the capacitor is initially charged to 4V, as soon as the sinusoidal voltage source turns on at t = 0, the voltage cannot change instantaneous and therefore, cannot have a discontinuity. However, the sinusoidal source will force the capacitor’s voltage to oscillate at an angular frequency of ω = 20 rad/s.

Step 1: Derive the first-order equation • Since this is a series circuit, apply KVL to write down an equation that contains both vC

and iC. Next, rewrite the resistor voltage in terms of iC since the 10Ω in series with the capacitor (using Ohm’s law):

R CS C R C Cv Ri

v v v v Ri=

= + = +

• Apply the capacitor equation to rewrite iC in terms of vC (this turns it into a first order differential equation):

CC

C CdvS c C C Ci Cdt

110

dv dvv v Ri v RC vdt dt=

= + = + = +

Write this equation into its proper form:

S

C C S CC

effective sourcev 10sin20t

dv v v dv 10v 100sin20tdt 1/10 1/10 dt=

+ = → + =

Check your equation. At steady state, the capacitor is in AC steady state that means that its response is sinusoidal. The details we cannot answer at the moment since this requires us to solve the first order equation.

Step 2: Derive the general response vC(t) = vn + vf By looking at the first order equation, the general response equation must have the form of

-t /C n f 1 2

general natural response general force response

v (t) v (t) v (t) Ae B sin20t B cos20tτ= + = + +

8.17

My job now is to determine the constants B1, B2, and 1/τ.

Determine the constants B1 and B2 To determine B1 and B2, we substitute the general force response into the first-order equation:

[ ] ( )ff 1 2 1 2

ff vdv /dt

dv 10v 20 B cos20t B sin20t 10 B sin20t B cos20t 100sin20t dt

+ = − + + =

Grouping together like sine and cosine terms:

[ ] [ ]2 1 1 220B 10B sin20t 20B 10B cos20t 100sin20t− + ⋅ + + ⋅ =

The only way this equation can be true is if all of the sine/cosine terms on the left side equals those on the right side of the equation.

[ ][ ]

1 2 1 2

1 21 2

10B 20B sin20t 100sin20t 10B 20B 100

20B 10B 0 20B 10B cos20t 0

− ⋅ = − = → + =+ ⋅ =

Solving this 2 × 2 matrix for B1 and B2:

11 2

2

B10 20 100 B 2, B 4

B20 10 0−

= → = = −

Determine 1/τ From the proper form of the first order equation, we pick-off the time constant 1/τ:

CC

dvdx x 1f(t) 10v 100sin20t 10 Hzdt dt

+ = → + = → =τ τ

Substituting in these constants into the general complete response gives

12

10tB 2C n f B 4

1/ 10

v (t) v v Ae 2sin20t 4cos20t−==−

τ=

= + = + −

Step 3: Solve vC(0−) and the arbitrary constant A

Solve for the IC vC(0−) Since this was given in the problem, we write vC(0−) = 4V.

Determine the constant A Apply vC(0−) to the general complete response equation and solve for constant A:

0Cv (0 ) 4 Ae 2sin0− = = + 4cos0 A 4 A 8− = − → =

The complete first-order response is

( )10tCv (t) 8e 2sin20t 4cos20t V−= + −

Interpret the solution Using PSpice, the complete response is plotted below.

8.18

• Is vC(0−) = 4V = vC(0+)? ( )0

C Cv (0 ) 4V v (0 ) 8e 2sin0 4cos0 V 8 4 4V− += = = + − = − =

• Does the capacitor’s voltage oscillate at a frequency of ω = 20 rad/s? 10 12 f 20 rads/s f 3.1847 Hz T 0.314 s

2 fω

ω = π = → = = = → = =π π

This is exactly the same period found on the plot. • What is the capacitor’s maximum voltage at steady state?

When the capacitor is at steady state, that means that the exponential decay of the natural response is undetectable and the force response dominates its behavior:

10tCt large

lim v (t) 8e−

→=

can beignored

2sin20t 4cos20t 2sin20t 4cos20t+ − ≈ −

Let’s check this. Use PSpice to pick-off the time value at one of the peak voltage values. Why? The equation that has to be solved to obtain the critical values is a transcendental equation, which is not trivial and a big “no thank you!” in trying to solve it. Picking off the time for one of the peaks, t = 2.0183 s, and plugging this time into the complete response it predicts the voltage

( )10 2.0183C

8

C

natual response force response

v (t 2.0183) 8e 2sin20 2.0183 4cos20 2.0183 V

1.37 10 0.914 ( 3.558)

4.472 V v (t 2.018 s)

− ⋅

−

= = + ⋅ − ⋅

= × + − −

= = =

As expected, the natural response will make no contribution to the complete response for large times. Furthermore, the theoretical prediction by the complete response compares very well with the PSpice value.

Example 8.5 a. Describe the behavior of the inductor’s current: (i) initial and final steady states, (ii)

storing or decaying current? Assume steady state at t = 0−.

b. (iii) Determine the voltage v24Ω(t) for t > 0. (iv) Plot v24Ω(t) vs. t for -∞ < t < +∞. Interpret the solution.

Solution What is the initial behavior of the inductor at t = 0−? By performing a source transformation on the 5A-source (this becomes a 30V-source), we will have two opposing voltage sources such that the role of the 5A-source is to “reduce” the inductor’s initial current. At t = 0+, the switch “kills off” the 5A-source and therefore, the inductor’s current increases or stores up

8.19

more current. The role of the exponential source at t = 0+ is to give a sudden burst of voltage to the circuit after the switch. When writing out the solution, we connect the inductor’s current to v24Ω(t) using Ohm’s law: v24Ω(t) = 24iL(t).

Step 1: Derive the first-order equation • Since this is a series circuit, apply KVL to write down an equation that contains both iL

and vL. Then rewrite the 24Ω resistor voltage in terms of the inductor’s current using Ohm’s law:

5t24 L L L Sv v 24i v (t) v 32 8e−

Ω + = + = = + • Apply the inductor equation to rewrite vL in terms of iL (turns it

into a first order differential equation) and write it into its proper form:

5t 5tL LL L8

di di24i 32 8e 3i 4 edt dt

− −+ = + → + = +

Check your equation. The final steady state value of the inductor’s current should be given by the first order equation. To determine this, set iL(∞) = constant and substitute this into the first order equation at t = ∞:

L

LL

( )

L

i

di d ( )3idt

idt∞

+ =∞

LL43 ( ) 4 e i ( ) A3

i −∞+ = + → ∞ =∞

The final steady state circuit will have the exponential 8e−5t-source decay away and the inductor is replaced with its steady state equivalent. Solving for iL(∞), we see that the current iL(∞) is given by Ohm’s law:

L32 4 i ( ) A24 3

∞ = =

So the first order equation behaves as expected.

Step 2: Derive the general response iL(t) = in + if.

By looking at the first order equation, the general response equation must have the form of

-t / 5tL n f 1 2

general force response

i (t) i (t) i (t) Ae B e B τ −= + = + +

My job now is to determine the constants B1, B2, and 1/τ.

Determine the constants B1 and B2 To determine B1 and B2, we will substitute the general force response into the first-order equation:

( )5t 5t 5t 5tff 1 2 1 2 1

di 3i 5B e 3 B B e 3B 2B e 4 e dt

− − − −+ = − + + = − = +

Equating the constant and e−5t terms on both sides of the equation, we solve for B1 and B2: 5t 5t4 1

2 2 1 13 23B 4 B and 2B e e B− −= → = − = → = −

Determine 1/τ From the proper form of the first order equation, we pick-off the time constant 1/τ:

5tff

di 13i 4 e 3 Hzdt

−+ = + → =τ

Substituting in the constants into the general complete response gives

12

3t 5tB 1/2L n f B 4/3

1/ 3

4 13 2i (t) i i Ae e− −=−

=τ=

= + = + −

Step 3: Solve iL(0−) and the arbitrary constant A

8.20

Solve for the IC iL(0−) Starting with the circuit at t(0−), the inductor is replaced with its steady state equivalent and the circuit looks like the one on the right. Using a source transformation to convert the 5A-source into a series of voltage source, Ohm’s law gives us

L L1

1532 30i (0 ) A i (0 )6 24

− +−= = + =

+

Determine the constant A Apply iL(0−) to the general complete response equation and solve for constant A

0 0L

234 13 2 30

1 4 115 3 2i (0) Ae e A A= = + − = + − → = −

Substituting this into the complete response gives

− −= − −5t 3tL

234 13 2 30i (t) e e

Interpret the solution Using PSpice, the complete response is plotted below.

• Is iL(0−) = 0.0667A = iL(0+)? ( )0 0

L L23 234 1 4 1

3 2 30 3 2 30 i (0 ) 0.06667A i (0 ) e e V 0.06667A− += = = − − = − − =

• Does the inductor’s current value at t = 2τ match PSpice value? ( )5 0.400 3 0.400

L

L

234 1 43 2 30 3 i (2 0.400 s) e e A 0.0677 0.2309

1.0347A i (2 )

− ⋅ − ⋅τ = = − − = − −

= = τ

• Does the inductor’s current value at t = ∞ match PSpice value? ( )L L

234 13 2 30 i ( ) e e A 1.333A i ( )−∞ −∞∞ = − − = = ∞

The theoretical prediction modeled by the complete response compares very well with the PSpice values.

Step 4: Solve for the voltage across the 24Ω resistor. Since the inductor and the 24Ω resistor are always in series, either before or after the switch (including the unit step function turn on), they will always have the same current. Therefore, the behavior of v24Ω(0−) and v24Ω(t) should be continuous at t = 0. Applying Ohm’s law, we get v24Ω(t):

( )− −Ω = = − −5t 3t

24 L925v (t) 24i (t) 32 12e e V

Since the voltage v24Ω is only scaling the inductor current by a factor of 24, the behavior of the 24Ω is exactly the same as iL(t). Therefore, I will not repeat the same analysis as I did for the inductor current.

8.21

Example 8.6 a. Describe the behavior of the capacitor’s voltage:

(i) initial and final steady states, (ii) charging or discharging? Assume steady state at t = 0−.

b. Determine vC(t) for t > 0.

Solution What is the behavior of the capacitor in this circuit? Initially, the capacitor will be charged by the 38.5V-source. However, when the switch is thrown open at t = 0+, there is a burst of energy into the circuit due to the 8e−5t-source but this dies off completely after 5τ = 1 s. So there is no source supply ix (no 2ix-source) and the capacitor will discharge to zero.

Step 1: Derive the first-order equation To solve for the capacitor's voltage vC(t) at t = 0+, not only will we have to use KVL but also include a dependent condition (DC) since there is dependent source in the circuit. • Write down an equation that contains both vC and iC. Applying

KVL around the outside loop of the circuit, it will include the 12Ω resistor. Next rewrite the v12Ω in terms of the dependent current ix:

5t 5t12 C X Cv 8e v 12i 8e v 0− −− + = − + =

Note that this equation does not contain both vC and iC. We need to bring in the DC to rewrite ix in terms of iC.

• To apply the DC, use KCL to the top node and write X X C X C

13i 2i i i i+ = → =

Substituting ix back into our equation, we now have an equation that contains both vC and iC:

1X C3

5t 5tX C C Ci i

1312i 8e v 12 i 8e v 0− −

=− + = ⋅ − + =

• Using the capacitor equation to rewrite iC in terms of vC (as well into its proper form), we get

CC

5t 5tCdvC C Cidt

136

dv4i v 0 8e 9v 72edt

− −=

+ = = → + =

Check your equation. At t = ∞, there are no active sources and the capacitor’s voltage vC(∞) should go to zero. Substituting vC(∞) = constant into our first order equation gives

C

CC C C

v ( )

dv 9v 0 9v ( ) 72e 0 v ( ) 0dt

−∞

∞

+ = + ∞ = = → ∞ =

So the first order equation behaves as expected.

Step 2: Derive the general response vC(t) = vn + vf By looking at the first order equation, the general response equation must have the form of

-t / 5tC n f

general force response

v (t) v (t) v (t) Ae Be τ −= + = +

My job now is to determine the constants B and 1/τ.

Determine the constants B To determine B, we will substitute the general force response into the first-order equation:

5tf

5t 5t 5t 5tff

v Be

dv 9v 5Be 9Be 4Be 72e B 18dt −

− − − −

=

+ = − + = = → =

Determine 1/τ From the proper form of the first order equation, we pick-off the time constant 1/τ:

8.22

5tCC

dv 19v 72e 9 Hzdt

−+ = → =τ

Substituting in the constants into the general complete response gives

9t 5tCv (t) Ae 18e− −= +

Step 3: Solve vC(0−) and the arbitrary constant A

Solve for the IC vC(0−) Starting with the circuit at t(0−), I use a source transformation to convert the 38.5V-source into a current source and redraw the circuit as

Since all of these elements are parallel to each other, the initial voltage vC(0−) can be calculated across any of the elements; I will focus on the 12Ω-resistor:

C 12 xv (0 ) v 12i−Ω= − = −

Since this clearly depends on the controlling current ix, I will apply CDR to the 12Ω-resistor

( )x x x3i 2i 12.83 i 4.28A

3 12= + → =

+

and the calculation of vC(0−) is

C x Cv (0 ) 12i 51.3 V v (0 )− −= − = − = Determine the constant A Apply vC(0−) to the general complete response equation and solve for constant A

0 0Cv (0 ) 51.3 Ae 18e A 18 A 69.3V− = − = + = + → = −

Substituting this into the complete response gives

5t 9tCv (t) 18e 69.3e− −= −

As expected, the capacitor's voltage goes to zero fairly quickly. The natural response dies off faster at e-9t than the force response at e-5t.