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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

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 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: