DC Circuits: First-Order Circuitsfaraday.ee.emu.edu.tr/eeng223/Circuit...

EENG223: Circuit Theory I

DC Circuits: First-Order Circuits

Hasan Demirel


•  Introduc)on •  The Source-‐Free RC Circuit •  The Source-‐Free RL Circuit

•  Step Response of an RC Circuit •  Step Response of an RL Circuit

First-‐Order Circuits


•  A first-‐order circuit can only contain one energy storage element (a capacitor or an inductor).

•  The circuit will also contain resistance.

•  So there are two types of first-‐order circuits: §  RC circuit §  RL circuit

•  A first-‐order circuit is characterized by a first-‐order differen)al equa)on.

First-‐Order Circuits: Introduc)on


•  A source-‐free circuit is one where all independent sources have been disconnected from the circuit a6er some switch ac7on.

•  The voltages and currents in the circuit typically will have some transient response due to ini)al condi)ons (ini7al capacitor voltages and ini7al inductor currents).

•  We will begin by analyzing source-‐free circuits as they are

the simplest type. Later we will analyze circuits that also contain sources a6er the ini7al switch ac7on.

First-‐Order Circuits: The Source-‐Free Circuits


•  A source-‐free RC circuit occurs when its dc source is suddenly disconnected.

•  The energy already stored in the capacitor is released to the resistors.

First-‐Order Circuits: The Source-‐Free RC Circuits

V0

•  Since the capacitor is ini)ally charged, we can assume that at 7me t=0, the ini)al voltage is:

•  Then the energy stored:

•  Applying KCL at the top node: The image cannot be displayed. Your computer may not have enough memory to open the image, or the image may have been corrupted. Restart your computer,

•  By defini7on, iC =C dv/dt and iR = v/R. Thus,



V0

•  This is a first-‐order differen-al equa-on, since only the first deriva-ve of v is involved.

•  Rearranging the terms:

•  Integra7ng both sides:

•  ln A is the integra7on constant. Thus

•  Taking powers of e produces:

•  From the ini-al condi-ons: v(0)=A=V0

•  The natural response of a circuit refers to the behavior (in terms of voltages and currents) of the circuit itself, with no external sources of excita7on.


•  General form of the Differen7al Equa7ons (DE) and the response for a 1st-‐order source-‐free circuit:


§  In general, a first-‐order D.E. has the form:

00)(1≥=+ tfortx

dtdx

τ

§  Solving this DE (as we did with the RC circuit) yields:

0)0()( ≥=−

tforextxtτ

§  here τ= (Greek leRer “Tau”) = )me constant(in seconds)


•  Notes concerning τ: First-‐Order Circuits: The Source-‐Free RC Circuits

§  So, for an RC circuit: RC=τ

1) For the Source-‐Free RC circuit the DE is: 00)(1≥=+ tfortv

RCdtdv

2) τ is related to the rate of exponen)al decay in a circuit as shown below.

3) It is typically easier to sketch a response in terms of mul)ples of τ than to be concerning with scaling of the graph.


First-‐Order Circuits: The Source-‐Free RC Circuits Ex. 7.1: In Fig. 7.5, let vC(0)= 15 V. Find vC , vx and ix for t>0.

Solu)on

•  Equivalent Circuit for the above circuit can be generated:


First-‐Order Circuits: The Source-‐Free RC Circuits •  Equivalent Resistance seen by a Capacitor

§  For the RC circuit in the previous example, it was determined that τ= RC. But what value of R should be used in circuits with mul)ple resistors?

§  In general, a first-‐order RC circuit has the following )me constant:

§  where REQ is the Thevenin resistance seen by the capacitor.

§  More specifically, REQ = R (seen from the terminals of the capacitor for t>0 with independent sources killed.)

CREQ=τ


First-‐Order Circuits: The Source-‐Free RC Circuits Ex. : Refer to the circuit below. Let vC(0)= 45 V. DeterminevC , vx and io for t≥0.

Solu)on

•  Time constant τ :

•  Then:

•  Consider Req seen from the capacitor.

Ω=+×

= 12818612

eqR

sCReq 43112 =×==τ

V45)0()( 25.04 tt

CC eevtv −−

==

V154531)(

844)( 25.025.0 tt

Cx eetvtv −− ==+

=

V75.384515

8)()()( 25.0

25.025.0t

ttCx

o eeetvtvti −−−

−=−

=−

=


First-‐Order Circuits: The Source-‐Free RC Circuits Ex. 7.2: The switch in the circuit below has been closed for a long 7me, and it is opened at t= 0. Find v(t) for t≥0. Calculate the ini7al energy stored in the capacitor.

Solu)on •  For t<0 the switch is closed; the capacitor is an open circuit to dc, as represented in Fig. (a).

•  For t>0 the switch is opened, and we have the RC circuit shown in Fig. (b).


•  Then:

•  The ini)al energy stored in the capacitor:


First-‐Order Circuits: The Source-‐Free RC Circuits Ex. : If the switch in Fig. below opens at t= 0, find v(t) for t≥0 and wC(0).

Solu)on

•  For t>0 the switch is opened, and we have the RC circuit shown in Fig. (b).


•  Then:

•  The ini)al energy stored in the capacitor:

•  For t<0 the switch is closed; the capacitor is an open circuit to dc as shown in Fig. (a).

V8)0(

0V824633)(

0 ==

<=+

=

Vv

tfortv

C

C

(a)

(b)

Ω=×

= 316412

eqR

sCReq 5.0613 =×==τ

V8)0()( 25.0 tt

C eevtv −−

==The image cannot be displayed. Your computer may not have enough memory to open the image, or the image may have been corrupted. Restart your computer, and then open the file again. If the red x still appears, you may have to delete the image and then insert it again.


First-‐Order Circuits: The Source-‐Free RC Circuits Ex. : The switch in the circuit shown had been closed for a long 7me and then opened at 7me t = 0.

a)  Determine an expression for v(t). b)  Graph v(t) versus t. c)  How long will it take for the capacitor to

completely discharge? d)  Determine the capacitor voltage at 7me

t=100ms. e)  Determine the 7me at which the capacitor

voltage is 10V.


•  A source-‐free RL circuit occurs when its dc source is suddenly disconnected.

•  The energy already stored in the inductor is released to the resistors.

First-‐Order Circuits: The Source-‐Free RL Circuits

•  At 7me, t=0 , the intuctor has the ini)al current:

•  Then the energy stored:

•  We can apply KVL around the loop above :

•  By defini7on, vL =L di/dt and vR = Ri. Thus,

I0 t=0



•  This is a first-‐order differen-al equa-on, since only the first deriva-ve of i is involved.

•  Rearranging the terms and integra7ng:

•  Then:

•  Taking powers of e produces:

•  Time constant for RL circuit becomes:

The natural response of the RL circuit is an exponen7al decay of the ini7al current.

I0 t=0


•  General form of the Differen7al Equa7ons (DE) and the response for a 1st-‐order source-‐free circuit:


§  In general, a first-‐order D.E. has the form:

00)(1≥=+ tfortx

dtdx

τ

§  Solving this DE (as we did with the RL circuit) yields:

0)0()( ≥=−

tforextxtτ

§  Then:

§  Where:

0)0()( 0 ≥==−−

tforeIeitittττ

RL

=τ


First-‐Order Circuits: The Source-‐Free RL Circuits •  Equivalent Resistance seen by an Inductor

§  For the RL circuit , it was determined that τ= L/R. As with the RC circuit, the value of R should actually be the equivalent (or Thevenin) resistance seen by the inductor.

§  In general, a first-‐order RL circuit has the following )me constant:

§  where REQ is the Thevenin resistance seen by the inductor.

§  More specifically, REQ = R (seen from the terminals of the capacitor for t>0 with independent sources killed.)

EQRL

=τ


First-‐Order Circuits: The Source-‐Free RL Circuits Ex. 7.3: Assuming that i(0) =10 A, calculate i(t) and ix(t) in the circuit below.

Solu)on

•  Subs7tu7ng Eq. (2) into Eq. (1) gives.

•  Thevenin resistance at the inductor terminals. we insert a voltage source with v0=1 V. Applying KVL to the two loops results (1)

(2)


First-‐Order Circuits: The Source-‐Free RL Circuits Ex. 7.3: Assuming that i(0) =10 A, calculate i(t) and ix(t) in the circuit below.

Solu)on

•  Time constant is:

•  Hence,

•  Thus, the current through the inductor is:


First-‐Order Circuits: The Source-‐Free RL Circuits Ex. 7.4: The switch in the circuit below has been closed for a long 7me. At t=0 the switch is opened. Calculate i(t) for t>0.

Solu)on

•  When t<0 the switch is closed, and the inductor acts as a short circuit to dc,

•  Using current division:

•  Current through an inductor cannot change instantaneously,


First-‐Order Circuits: The Source-‐Free RL Circuits Ex. 7.4: The switch in the circuit below has been closed for a long 7me. At t=0 the switch is opened. Calculate i(t) for t>0.

Solu)on

When t>0 the switch is open and the voltage source is disconnected. We now have the source-‐free RL circuit in Fig. (b).

•  The 7me constant is :

•  Thus,


First-‐Order Circuits: The Source-‐Free RL Circuits Ex. : Determine an expression for i(t). Sketch i(t) versus t.


First-‐Order Circuits: Step Response of an RC Circuit •  Step Response (DC forcing func)ons) •  Consider circuits having DC forcing func)ons for t > 0 (i.e., circuits that have

independent DC sources for t > 0). •  The general solu)on to a differen)al equa)on has two parts: •  x(t) = xh+ xp = homogeneous solu)on + par)cular solu)on •  or •  x(t) = xn+ xf = natural solu)on + forced solu)on

•  xn is due to the ini)al condi)ons in the circuit •  and xf is due to the forcing func)ons (independent voltage and current

sources for t > 0).

•  xf in general take on the “form” of the forcing func)ons, •  So DC sources imply that the forced response func)on will be a constant(DC), •  Sinusoidal sources imply that the forced response will be sinusoidal, etc.


First-‐Order Circuits: Step Response of an RC Circuit •  Step Response (DC forcing func)ons) •  Since we are only considering DC forcing func)ons in this chapter, we assume

that : xf = B (constant). •  Recall that a 1st-‐order source-‐free circuit had the form Ae-t/τ. Note that there

was a natural response only since there were no forcing func)ons (sources) for t > 0. So the natural response was

0/ >= − tforAex tn

τ

•  The complete response for 1st-‐order circuit with DC forcing func)ons therefore will have the form: x(t) = xf + xn

τ/)( tAeBtx −+=

•  The “Shortcut Method”: An easy way to find the constants B and A is to evaluate x(t) at 2 points. Two convenient points at t = 0 and t = ∞ since the circuit is under dc condi)ons at these two points. This approach is some)mes called the “shortcut method.”


First-‐Order Circuits: Step Response of an RC Circuit •  Step Response (DC forcing func)ons) •  The “Shortcut Method” :

§  So, x(0) = B + Ae0= B + A §  And x(∞) = B + Ae-∞= B

•  Complete response yields the following expression:

τ/)]()0([)()( texxxtx −∞−+∞=

•  The Shortcut Method-‐ Procedure: The shortcut method will be the key method used to analyze 1st-‐order circuit with DC forcing func)ons:

1.   Analyze the circuit at t = 0-: Find x(0-) = x(0+) 2.   Analyze the circuit at t = ∞: Find x(∞) 3.   Find τ= REQC or τ= L/REQ 4.   Assume that x(t) has the form x(t) = x(∞)+[x(0) –x(∞)] e-t/τ using x(0) and

x(∞)


First-‐Order Circuits: Step Response of an RC Circuit •  Step Response (DC forcing func)ons) Notes: The “shortcut method” also works for source-‐free circuits, but x(∞) = B=0 since the circuit is dead at t = ∞. If variables other than vC or iL are needed, it is generally easiest to solve for vC or iL first and then use the result to find the desired variable.


First-‐Order Circuits: Step Response of an RC Circuit Ex. 7.10: The switch in Fig. Below has been in posi7on A for a long 7me. At t=0 the switch moves to B. Determine v(t) for t>0 and calculate its value at t =1 s and 4 s.

Solu)on •  Voltage across the capacitor just before t=0. Capacitor is open circuit under dc condi7ons:

•  Capacitor voltage cannot change instantaneously:

•  For t>0 (switch to B). Thevenin Resistance connected to the capacitor:

•  Time constant:


First-‐Order Circuits: Step Response of an RC Circuit Ex. 7.10: The switch in Fig. below has been in posi7on A for a long 7me. At t=0 the switch moves to B. Determine v(t) for t>0 and calculate its value at t 1 s and 4 s.

Solu)on •  Since the capacitor acts like an open circuit to dc at steady state, v(∞) = 30 V. Thus,


HW#3 Ex. 1 : Find v(t) and i(t) for t ≥0.

Ex. 2 : Find v(t) and i(t) for t ≥0.


First-‐Order Circuits: Step Response of an RL Circuit Ex. 3: Find v(t) and i(t) for t ≥0.

DC Circuits: First-Order Circuitsfaraday.ee.emu.edu.tr/eeng223/Circuit...

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