EKT101 Electric Circuit Theory Chapter 5 First-Order and Second-Order Circuits

First-Order and Second Circuits Chapter 5 5.1 Natural response of RL and RC Circuit 5.2 Force response of RL and RC Circuit 5.3 Solution of natural response and force response in RL and RC Circuit 5.4 Natural and force response in series RLC Circuit 5.5 Natural and force response in parallel RLC Circuit

5.1 Natural response of RC circuit (1) A first-order circuit is characterized by a first-order differential equation. By KCL Ohms law Capacitor law

first-order circuit Two are 2 types of first-order circuits (RC and RL), because there are two ways to excite the circuits. The first way is by initial conditions of the storage elements in the circuits. In these so-called 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. The second way of exciting first-order circuits is by independent sources.  DC Source

5.1 Natural response of RC circuit (2) 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 excitation. Time constant Decays more slowly Decays faster the voltage decreases is expressed in terms of the time constant, denoted by t, the lowercase Greek letter tau. V decays faster for small t and slower for large t.

5.1 Natural response of RC circuit (3) The key to working with a source-free RC circuit is finding: voltage response of the RC circuit where KEY POINT: The initial voltage v(0) = V0 across the capacitor. The time constant  = RC.

5.1 Natural response of RC circuit (4) Example 1 Refer to the circuit below, determine vC, vx, and io for t ≥ 0. Assume that vC(0) = 60 V.

vx = 20e–0.25t Solution 1 4 𝑜ℎ𝑚 –40e–0.25t V io = –5e–0.25t A vC, vx, and io for t ≥ 0. Assume that vC(0) = 60 V. 4 𝑜ℎ𝑚 v0 where where vx = 20e–0.25t –40e–0.25t V io = –5e–0.25t A

5.1 Natural response of RC circuit (5) Example 2 The switch in circuit below is opened at t = 0, find v(t) for t ≥ 0.

When t < 0, the switch is closed as shown in Fig. (a). Solution 2 When t < 0, the switch is closed as shown in Fig. (a). When t > 0, the switch is open as shown in Fig. (b).

5.1 Natural response of RL circuit (6) A first-order RL circuit consists of a inductor L (or its equivalent) and a resistor (or its equivalent) By KVL Inductors law Ohms law

5.1 Natural response of RL circuit (7) A general form representing a RL where The time constant  of a circuit is the time required for the response to decay by a factor of 1/e or 36.8% of its initial value. i(t) decays faster for small t and slower for large t. The general form is very similar to a RC source-free circuit.

5.1 Natural response of RL circuit(8) Comparison between a RL and RC circuit A RL source-free circuit where A RC source-free circuit where

5.1 Natural response of RL circuit(9) The key to working with a source-free RL circuit is finding: where The initial voltage i(0) = I0 through the inductor. The time constant  = L/R.

5.1 Natural response of RL circuit(10) Example Find i and vx in the circuit. Assume that i(0) = 12 A.

Applying mesh analysis gives: Step 1: Find RTH at the inductor terminals by inserting a voltage source. i1 i2 i3 = i1 - i2 i3 Applying mesh analysis gives: Loop 1: 𝑤ℎ𝑒𝑟𝑒: 𝑣 𝑥 =1 𝑖 𝑖 −1+ 𝑣 𝑥 +2 𝑖 1 −i2 +2 𝑣 𝑥 = 0 −1+ 𝑖 1 +2 𝑖 1 −2 𝑖 2 +2 𝑖 1 = 0 Loop 2: −2 𝑣 𝑥 −2 𝑖 1 −i2 +6 𝑖 2 = 0 −2 𝑖 1 −2 𝑖 1 +2 𝑖 2 +6 𝑖 2 = 0

5.1 Natural response of RL circuit(11) Example 4 The switch in the circuit has been closed for a long time. At for t = 0 the switch is opened. Calculate i(t) for t > 0.

Solution: When t<0 the switch is closed, and the inductor acts as a short circuit to dc. The resistor is short-circuited; the resulting circuit is shown in Fig. (a). 1 𝑅 𝑒𝑞 = 1 24 + 1 12 + 1 8 =4 𝑜ℎ𝑚 𝑉=4×15=60𝑉 Since the circuit is in parallel, hence each of the voltage is 60V 𝑖(𝑡)= 60 12 =5𝐴

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

Unit-Step Function (1) When the dc source of an RC or RL circuit is suddenly applied, the voltage or current source can be modeled as a step function, and the response is known as a step response. The step response is the response of the circuit due to a sudden application of a dc voltage or current source. The unit step function u(t) is 0 for negative values of t and 1 for positive values of t.

5.2 Force response of RC Circuit (1) The step response of a circuit is its behavior when the excitation is the step function, which may be a voltage or a current source. Initial condition: v(0-) = v(0+) = V0 v(0-):voltage across the capacitor just before switching and v(0+) is its voltage immediately after switching Applying KCL, or Where u(t) is the unit-step function

5.2 Force response of RC Circuit(2) Three steps to find out the step response of an RC circuit: The initial capacitor voltage v(0). When t<0 The final capacitor voltage v() — DC voltage across C. (final or steady-state value) When t>0 The time constant . When t>0 Note: The above method is a short-cut method. You may also determine the solution by setting up the circuit formula directly using KCL, KVL , ohms law, capacitor and inductor VI laws.

5.2 Force response of RC Circuit(3) Example 5 Find v(t) for t > 0 in the circuit in below. Assume the switch has been open for a long time and is closed at t = 0. Calculate v(t) at t = 0.5.

For the switch t<0 (the switch has been open) For the switch t<0 (the switch has been open). The capacitor acts like an open circuit to dc. Using the fact that the capacitor voltage cannot change instantaneously, 𝑣 0 − =𝑣 0 + =𝑣 0 =10

For t > 0, Switch is closed (use nodal analysis) 𝑣 ∞ −10 2 + 𝑣 ∞ −(−50) 6 =0 6𝑣 ∞ −60+2𝑣 ∞ +100=0 8𝑣 ∞ =−40 𝑣 ∞ =−5 𝑣 0.5 =−5+5.518=0.518V

Solution Ex 5 1 -The initial capacitor voltage v(0). 2- final capacitor voltage v() Use KCL at nod to get v(∞) 3- time constant . Use Rth =RC

5.2 Force response of RL Circuit(1) The step response of a circuit is its behavior when the excitation is the step function, which may be a voltage or a current source. Initial current i(0-) = i(0+) = Io Final inductor current i(∞) = Vs/R Time constant t = L/R

5.2 Force response of RL Circuit(2) Three steps to find out the step response of an RL circuit: The initial inductor current i(0) at t = 0+. The final inductor current i(). The time constant . Note: The above method is a short-cut method. You may also determine the solution by setting up the circuit formula directly using KCL, KVL , ohms law, capacitor and inductor VI laws.

5.2 Force response of RL Circuit(3) Example 6 The switch in the circuit shown below has been closed for a long time. It opens at t = 0. Find i(t) for t > 0.

Example 6 (solution) −→ 𝑖 𝑡 =2+(3−2) 𝑒 −10𝑡 1. Apply source transformation 30V At t < 0, the switch is closed so that the 5 ohm resistor is short circuited: 30 10 =3𝐴 𝑖 ∞ = 30 5+10 =2𝐴 −→ 𝑖 𝑡 =2+(3−2) 𝑒 −10𝑡

5.3 Solution of natural and force response in RC and RL Circuit Integrating both sides and considering the initial conditions, the solution of the equation is: Final value at t -> ∞ Initial value at t = 0 Source-free Response Complete Response = Natural response + Forced Response (stored energy) (independent source) = V0e–t/τ + Vs(1–e–t/τ)