INC 111 Basic Circuit Analysis Week 8 RL circuits

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Transcript of INC 111 Basic Circuit Analysis Week 8 RL circuits

  • INC 111 Basic Circuit AnalysisWeek 8RL circuits

  • Non-periodic SignalThere are infinite number of non-periodic signal.

    This course will cover only the most basic one, a step.

    A step is a result from on/off switches, which is common in our daily life.0V9VOn switch

  • AC

    1V

    1

    I

    AC

    2V

    1

    I

  • VoltageCurrenttimetime1V2V1A2A

  • AC

    1V

    1

    L

    I

    AC

    2V

    1

    L

    I

  • VoltageCurrenttimetime1V2V1A2A

  • I am holding a ball with a rope attached, what is the movement of the ball ifI move my hand to another point?Movements

    Oscillation

    Forced position changePendulum Example

  • Transient Response or Natural Response (e.g. oscillation, position change temporarily)

    Fade over timeResist changes

    Forced Response (e.g. position change permanently)

    Follows inputIndependent of time passed

  • Forced responseNatural responseat different timeMechanical systems are similar to electrical system

  • Transient Response RL Circuit

    RC Circuit

    RLC Circuit

  • RL CircuitKVL

    AC

    u(t)

    +

    -

    L

    R

    i(t)

  • Assume that i(t) = g(t) make this equation true. considerHowever, g(t) alone may be incomplete. The complete answer isi(t) = f(t) + g(t)where f(t) is the answer of the equation

  • Proof: Answer has two partsf(t) is the answer ofthis equationtherefore------------------(1)g(t) is the answer ofthis equationtherefore------------------(2)

  • f(t)+g(t) is alsothe answer ofthis equationthereforeIfmust be true

  • Transient ResponseForced Responsei(t) consists of two partsTherefore, we will study source-free RL circuit first

  • Source-free RL CircuitInductor L has energy stored so thatthe initial current is I0Compare this with a pendulumwith some height (potential energy) left.height

    L

    R

    +

    -

    -

    +

    i(t)

  • There are 2 ways to solve first-order differential equations

    L

    R

    +

    -

    -

    +

    i(t)

  • Method 1: Assume solutionwhere A and s is the parameters that we want to solve forSubstitutein the equationThe term that can be 0 is (s+R/L) , thereforeThe answer is in format

  • Initial conditionfromSubstitute t=0, i(t=0)=0We got

  • Method 2: Direct integration

  • ti(t)I0Approach zeroNatural Response only

  • Time ConstantRatio L/R is called time constant, symbol Time constant is defined as the amount of time usedfor changing from the maximum value (100%) to 36.8%. Unit: second

  • ti(t)2ANatural Response + Forced Response1AForced response = 1A comes from voltage source 1VApproach 1A

  • SwitchClose at t =0Open at t =03-way switch

    t=0

    t=0

    t=0

    t=0

    t=0

  • SwitchClose at t =0Open at t =03-way switcht > 0

    t=0

    t=0

    t=0

    t=0

    t=0

  • tv(t)1Vtv(t)1V0V0VStep function (unit)

    R

    R

    1V

    1V

    t=0

    t=0

  • Will divide the analysis into two parts: t0When t0, the current start changing. The inductor discharges energy.Using KVL, we can write an equation of current with constantpower supply = 1V with initial condition (current) = 2A

    R=1

    2V

    t=0

    1V

    L

  • For t>0

  • We can find c2 from initial conditioni(0) = 2 ASubstitute t = 0, i(0) = 2Therefore, we haveSubstitute V=1, R=1

  • RL Circuit Conclusion Force Response of a step input is a step

    Natural Response is in the form where k1 is a constant, whose value depends on the initial condition.

  • ResponsetimePeriod 1How to Solve Problems?Period 2Period 3Divide in to several periods (3 periods as shown below)Period 1, 3 have constant V, I -> Use DC circuit analysisPeriod 2 is transient.

  • Calculate Transient (period 2) Start by finding the current of the inductor L first

    Assume the response that we want to find is in form of Find the time constant (may use Thevenins)

    Solve for k1, k2 using initial conditions and status at the stable point

    From the current of L, find other values that the problem ask

  • ExampleThe switch is at this position for a long timebefore t=0 , Find i(t)Time constant = 1 sec

    R=1

    2V

    t=0

    1V

    L=1H

    i(t)

  • At t=0, i(0) = 2 AAt t = , i() = 1 ATherefore, k1 = 1, k2 = 1The answer is

  • 2A1A

  • ExampleL has an initial current of 5A at t=0Find i2(t)The current L is in form ofTime constant = R/L, find ReqTime constant(Thevenins)

    R1

    L

    R2

    R3

    R4

    i(0)=5

    +-

    i2(t)

  • Find k1, k2 using i(0) = 5, i() = 0At t=0, i(0) = 5 AAt t = , i() = 0 ATherefore, k1=0, k2 = 5i2(t) comes from current divider of the inductor currentGraph?

  • ExampleL stores no energy at t=0Find v1(t)Find iL(t) first

    1

    1H

    1V

    2

    t=0

    2

    +v1(t)-

  • Find k1, k2 using i(0) = 0, i() = 0.25At t=0, i(0) = 0 AAt t = , i() = 0.25 ATherefore, k1=0.25, k2 = -0.25v1(t) = iL(t) RGraph?

  • *