E e 2415
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E E 2415. Lecture 7 Natural and Step Responses of RL and RC Circuits. Conservation of Charge (1/4). Energy transferred if v 10  v 20 Total system charge is conserved. Conservation of Charge (2/4). Initial stored energy:. At equilibrium:. Conservation of Charge (3/4). Initial Charge:.

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E E 2415

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E e 2415

E E2415

Lecture 7

Natural and Step Responses of RL and RC Circuits


Conservation of charge 1 4

Conservation of Charge (1/4)

  • Energy transferred if v10 v20

  • Total system charge is conserved


Conservation of charge 2 4

Conservation of Charge (2/4)

Initial stored energy:

At equilibrium:


Conservation of charge 3 4

Conservation of Charge (3/4)

Initial Charge:

Final Charge:

Since


Conservation of charge 4 4

Conservation of Charge (4/4)

Final stored energy:

Energy consumed in R:


Conservation of flux linkage 1 3

Conservation of Flux Linkage (1/3)

  • Energy transferred if i10 i20

  • Total system flux linkage is conserved.

Initial stored energy:

At equilibrium:


Conservation of flux linkage 2 3

Conservation of Flux Linkage (2/3)

Initial flux linkage:

Final flux linkage:

Since


Conservation of flux linkage 3 3

Conservation of Flux Linkage (3/3)

Final stored energy:

Energy consumed in R:


Natural rl response 1 2

Natural RL Response (1/2)

  • Inductor has initial current, io.

  • Switch opens at t = 0

  • Inductor current can’t change instantaneously


Natural rl response 2 2

Natural RL Response (2/2)

Separate

the

variables:

Integrate:

KVL:

Exponential of both

sides:


Natural rc response 1 2

Natural RC Response (1/2)

  • Capacitor has initial voltage, vo.

  • Switch closes at t = 0.

  • Capacitor voltage can’t change instantaneously

KCL:

Separate the variables:


Natural rc response 2 2

Natural RC Response (2/2)

Integrate:

Exponential of both

sides:


Rl step response 1 4

RL Step Response (1/4)

  • Make-before-break switch changes from position a to b at t = 0.

  • For t < 0, Io circulates unchanged through inductor.


Rl step response 2 4

RL Step Response (2/4)

  • For t > 0, circuit is as below.

  • Initial value of inductor current, i, is Io.

  • The KVL equation provides the differential equation.


Rl step response 3 4

RL Step Response (3/4)

Solution has two parts:

Steady State Response

Transient Response

Determine k by initial conditions:


Rl step response 4 4

RL Step Response (4/4)

  • Inductor behaves as a short circuit to DC in steady state mode


Rc step response 1 3

RC Step Response (1/3)

  • Switch closes at t = 0.

  • Capacitor has initial voltage, Vo.

v-i relationship:

By KVL & Ohm’s Law:


Rc step response 2 3

RC Step Response (2/3)

  • Response has two parts

    • steady state

    • transient

  • Use initial voltage to determine transient

Steady State Response

Transient Response


Rc step response 3 3

RC Step Response (3/3)

  • Capacitor becomes an open circuit to DC after the transient response has decayed.


Unbounded response 1 5

Unbounded Response (1/5)

  • Need Thévenin equivalent circuit from terminal pair connected to inductor

  • Let initial current = 0A in this example.


Unbounded response 2 5

Unbounded Response (2/5)

Voltage divider to get vx:

Then

Thévenin voltage


Unbounded response 3 5

Unbounded Response (3/5)

Therefore:


Unbounded response 4 5

Unbounded Response (4/5)

Steady state:

Transient:


Unbounded response 5 5

Unbounded Response (5/5)

Use initial conditions to determine k.

Complete response is unbounded:


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