RL and RC circuits first- order response Electric circuits ENT 161/4

RL and RC circuit original response A first-order circuit is characterized by a first-order differential equation. This circuit contain resistor and capacitor or inductor in one close circuit.

The natural response of a circuit refers to the behaviour ( in terms of voltages and currents) of the circuit itself, with no external sources of excitation. RL circuit: circuit that have resistor and inductor. RC circuit: circuit that have resistor and capacitor.

Natural response RC circuit

Consider these three condition : 1. At initially, t=0 -, switch doesn’t change for some time 2. At initial, t=0 +, switch doesn’t change for some time 3. At final condition, t→∞, switch doesn’t change for some time

Known t ≤ 0, v(t) = V 0. voltage  Therefore t ≥ 0:

For t > 0,

Natural response RC circuit graph

This show that the voltage response of the RC circuit is an exponential decay of the initial voltage. constant, τ = RC

Constant τ define how fast voltage reach stable condition :

Natural response RL circuit

Consider these three condition : 1. At initially, t=0 -, switch doesn’t change for some time 2. At initial, t=0 +, switch doesn’t change for some time 3. At final condition, t→∞, switch doesn’t change for some time

Known at t ≤ 0, i(t) = I 0  Therefore t > 0, Current

For t > 0,

EXAMPLE Switch in circuit for some time before open at t=0. Calculate a) I L (t) at t ≥ 0 b) I 0 (t) at t ≥ 0+ c) V 0 (t) at t ≥ 0+ d) Total energy percentage that stored in inductor 2H that absorb by 10Ω resistor.

Answer a) Switch close for some time until t=0, known voltage at inductor should be zero at t = 0 -. Therefore, initial current at inductor was 20A at t = 0-. Thus i L (0 + ) also become 20A, because immidiate changes for current didn’t exist in inductor.

Equivalent resistance from inductor and constant time

Therefore, current i L (t)

b) Current at resistor 40Ω could be calculate by using current divider law,

This current was at t ≥ 0 + because i 0 = 0 at t = 0 -. Inductor will become close circuit when switch open immediately and produce changes immediately at current i 0. Therefore,

c) V 0 could be calculate by using Ohm’s Law,

d) Total power absorb by 10Ω resistor

Total energy absorb by 10Ω resistor

Initial energy stored at 2H inductor

Therefore, energy percentage that absorb by 10Ω resistor

Step response RC circuit The step response of a circuit is its behaviour when the excitation is the step function, which may be a voltage or a current source.

Consider these three condition : 1. At initially, t=0 -, switch doesn’t change for some time 2. At initial, t=0 +, switch doesn’t change for some time 3. At final condition, t→∞, switch doesn’t change for some time

Known at t ≤ 0, v(t)=V 0 For t > 0, voltan

Current for step response RC circuit

Then, for t >0 Where

V f = Force voltage or known as steady-state response V n = known as transient response is the circuit’s temporary response that will die out with time.

Step response RC circuit graph force Natural total

Step Response RL circuit

Consider these three condition : 1. At initially, t=0 -, switch doesn’t change for some time 2. At initial, t=0 +, switch doesn’t change for some time 3. At final condition, t→∞, switch doesn’t change for some time

known i(t)=I 0 at t ≤ 0.  For t > 0, Current

Finally,

Question Switch in those circuit was at x position for some time. At t=0, switch move to position y immediately. Calculate, (a) Vc(t) at t ≥ 0 (b) V 0 (t) at t ≥ 0+ (c) i 0 (t) at t ≥ 0+ (d) Total energy absorb by 60kΩ resistor.

Answer (a) Constant for circuit  V C (0)=100V  equivalent resistor = 80kΩ.

Then, V C (t) for t ≥ 0:

Answer (b) V 0 (t) could be calculate by using voltage divider law.

Answer (c) current i 0 (t) can be calculated by using ohm’s law

Answer (d) Power absorb by 60kΩ resistor

Total energy

Second-order RLC circuit RLC circuit : circuit that contain resistor, inductor and capacitor Second-order response : response from RLC circuit Type of RLC circuit: 1. RLC series circuit 2. RLC parallel circuit

Original response for parallel RLC circuit

Take total current flows out from node

differential of t,

Take

Characteristic equation known as zero :

The root of the characteristic equation are

Response for RLC parallel circuit

The root of the characteristic equation are where:

summarize ParameterTerminologyValue in natural response s 1, s 2 characteristic equation α frequency Neper resonant radian frequency

Roots solution s 1 and s 2 depend on α and Consider these cases saperately: 1. If < α, voltage response was overdamped 2. If > α, voltage response was underdamped 3. If = α, voltage response was critically damped

Overdamped voltage response Solution for overdamped voltage

constant A1 and A2 can be determined from the initial conditions v(0+) and Known,

Here v(0+) = V0 and initial value for dv/dt was

Solution for overdamped natural response, v(t) : 1. Calculate characteristic equation, s1 and s2, using R, L and C value. 2. Calculate v(0+) and using circuit analysis.

3. Calulate A1 and A2 by solve those equation 4. Insert s1, s2, A1 and A2 value to calculate overdamped natural response for t ≥ 0.

Example for overdamped natural response for v(0) = 1V and i(0) = 0

Underdamped voltage response At > α2, root of the characteristic equation was complex number and those response called underdamped.

Therefore ωd : damped radian frequency

underdamped voltage response for RLC parallel circuit was

constant B1 and B2 was real number. Solve those two linear equation to calculate B1 and B2,

Example for underdamped voltage response for v(0) = 1V and i(0) = 0

Critically Damped Voltage Response Second-order circuit was critically damped when = α. When circuit was critically damped, two characterictic root equation was real and same,

Solution for voltage Linear equation to calculate D1 and D2 value

Example for critically damped voltage response at v(0) = 1V and i(0) = 0

Step response RLC parallel circuits

From Kirchhoff current law

Known Therefore

Have,

There are two solution to solve the equation, direct approach and indirect approach.

Indirect approach From Kirchhoff’s current law:

Differential

Depend on characteristic equation root :

Insert in Kirchhoff’s current law eq:

Direct approach It’s simple to calculate constant for the equation directly by using initial value response function.

Constant of the equation could be calculate from and

The solution for a second-order differential equation with a constant forcing function equals the forced response plus a response funtion identical in form to the natural response.

If and Vf represent the final value of the response function. The final value may be zero,

Natural response for RLC Series circuit The procedures for finding the natural or step responses of a series RLC circuit are the same as those used to find the natural or step responses of a parallel RLS circuit, because both circuits are described by differential equations that have the same form.

RLC series circuit

Summing the voltages around the closed path in the circuit,

differential

Characteristic equation for RLC series circuit

Characteristic equation

Neper frequency (α) for RLC series circuit and resonant radian frequency was,

Current response Overdamped Underdamped critically damped

Three kind of solution

Step response for RLC series circuit The procedures for finding the step responses of series RLC circuit are the same as those used to find the step response of a parallel RLC circuit.

RLC series circuit

Using Kirchhoff’s voltage law,

Current known as,

Differential for current

Insert in Voltage current law equation

Three solution that possibly for v C

Contoh 1 Tenaga awal yang disimpan oleh litar berikut adalah sifar. Pada t = 0, satu punca arus DC 24mA diberikan kepada litar. Nilai untuk perintang adalah 400Ω. 1. Apakah nilai awal untuk i L ? 2. Apakah nilai awal untuk ? 3. Apakah punca-punca persamaan ciri? 4. Apakah ungkapan numerik untuk iL(t) pada t ≥ 0?

Jawapan 1. Tiada tenaga yang disimpan dalam litar sebaik sahaja punca arus digunakan, maka arus awal bagi induktor adalah sifar. Induktor mencegah perubahan yang serta-merta pada arus induktor, oleh itu i L (0)=0 sebaik sahaja suis dibuka.

2. Nilai awal voltan kapasitor adalah sifar sebelum suis dibuka, oleh itu ia akan sifar sebaik sahaja suis dibuka. Didapati: maka

3. Dari elemen-elemen dalam litar, diperolehi

Oleh kerana, maka punca-punca persamaan ciri adalah nyata

4. sambutan arus induktor adalah overdamped dan persamaan penyelesaian adalah

Dua persamaan serentak:

Penyelesaian numerik:

Contoh 2 Tiada tenaga disimpan dalam inductor 100mH atau kapasitor 0.4µF apabila suis di dalam litar berikut ditutup. Dapatkan v C (t) untuk t ≥ 0.

Jawapan Punca-punca persamaan ciri:

Punca-punca adalah kompleks, maka sambutan voltan adalah underdamped. Oleh itu, diperolehi voltan v C :

Pada awalnya, tiada tenaga tersimpan dalam litar, maka:

Selesaikan untuk dan

penyelesaian untuk v C (t)