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DC Power

*DC TRANSIENT ANALYSISCHAPTER 5

*ObjectivesInvestigate the behavior of currents and voltages when energy is either released or acquired by inductors and capacitors when there is an abrupt change in dc current or voltage source.To do an analysis of natural response and step response of RL and RC circuit.

*Lectures contents5-1 NATURAL RESPONSE OF RL CIRCUIT5-2 NATURAL RESPONSE OF RC CIRCUIT5-3 STEP RESPONSE OF RL CIRCUIT5-4 STEP RESPONSE OF RC CIRCUIT

*First Order CircuitA circuit that contains only sources, resistor and inductor is called and RL circuit.A circuit that contains only sources, resistor and capacitor is called an RC circuit.RL and RC circuits are called first order circuits because their voltages and currents are describe by first order differential equations.

*Review (conceptual)Any first order circuit can be reduced to a Thvenin (or Norton) equivalent connected to either a single equivalent inductor or capacitor.

In steady state, an inductor behave like a short circuit.In steady state, a capacitor behaves like an open circuit.

*The natural response of an RL and RC circuit is its behavior (i.e., current and voltage ) when stored energy in the inductor or capacitor is released to the resistive part of the network (containing no independent sources)

The steps response of an RL and RC circuits is its behavior when a voltage or current source step is applied to the circuit, or immediately after a switch state is changed.

- *5-1 Natural Response of an RL circuitConsider the following circuit, for which the switch is closed for t
*Solving the circuitFor t 0, i(t) = IoFor t 0, the circuit reduce to

At t = 0, the inductor has initial current Io, hence i(0) = IoThe initial energy stored in the inductor is,

*Cont.Applying KVL to the circuit:

From equation (4), let say;

*Cont.

Integrate both sides of equation (5);

Therefore,

hence, the current is

*Cont.From the Ohms law, the voltage across the resistor R is:

And the power dissipated in the resistor is:

Energy absorb by the resistor is:

*Time Constant, for RL circuitTime constant, determines the rate at which the current or voltage approaches zero.

The time constant of a circuit is the time required for the response to decay to a factor of 1/e or 36.8% of its initial current

Natural response of the RL circuit is an exponential decay of the initial current. The current response is shown in Fig. 5-1

Time constant for RL circuit is

And the unit is in seconds. Figure 5-1

*The expressions for current, voltage, power and energy using time constant concept:

*Switching timeFor all transient cases, the following instants of switching times are considered.t = 0- , this is the time of switching between - to 0 or time before.t = 0+ , this is the time of switching at the instant just after time t = 0s (taken as initial value)t = , this is the time of switching between t = 0+ to (taken as final value for step response)The illustration of the different instance of switching times is:

*Example 1For the circuit below, find the expression of io(t) and Vo(t). The switch was closed for a long time, and at t = 0, the switch was opened.

*Solution :When t < 0, switch is closed and the inductor is short circuit.

When t > 0, the switch is open and the circuit become;Therefore; iL(0-) = 20AHence; iL(0+) = iL(0-) = 20ACurrent through the inductor remains the same (continuous)

*RT = (2+10//40) = 10

So, time constant, sec

By using current division, the current in the 40 resistor is:

Using ohms Law, the Vo is,

Hence:

*Example 2The switch in the circuit below has been closed for a long time.At t = 0, the switch is opened. Calculate i(t) for t > 0.

*Solution :When t < 0, the switch is closed and the inductor is short circuit to dc. The 16 resistor is short circuit too.

calculate i1;

using current division, calculate i(0-)Hence; i(0) = i (0-) = 6ASince the current through an inductor cannot change instantaneously

*When t > 0, the switch is open and the voltage source is disconnect.i(0+) = i(0) = i (0-) = 6Ahence;RT = Req = (12 + 4)// 16 = 8Time constant, Thus, Because current through the inductor is continuously

*5-2 Natural Response of an RC CircuitThe natural response of RC circuit occurs when its dc source is suddenly disconnected. The energy already stored in the capacitor, C is released to the resistors, R.Consider the following circuit, for which the switch is closed for t < 0, and then opened at t = 0:

*Solving the circuitFor t 0, v(t) = VoFor t > 0, the circuit reduces to

At t = 0, the initial voltage v(0) = VoThe initial value of the energy stored is

*Cont. Applying KCL to the RC circuit:(1)(3)(4)(5)(2)

*Cont. From equation (5), let say:

Integrate both sides of equation (6):

Therefore:

(6)(8)(7)

*Cont.Hence, The voltage is:

Using Ohms law, the current is:

The power dissipated in the resistor is:

The energy absorb by the resistor is:

*Time Constant, for RC circuitThe time constant for the RC circuit equal the product of the resistance and capacitance,Time constant, sec

The natural response of RC circuit illustrated graphically in Fig 5.2

Figure 5.2

*The expressions for voltage, current, power and energy using time constant concept:

*Example 3The switch has been in position a for a long time. At time t = 0,the switch moves to b. Find the expressions for the vc(t), ic(t) andvo(t).

*SolutionAt t < 0, the switch was at a. the capacitor behaves like an opencircuit as it is being supplied by a constant source.

At t > 0, the instant when the switch is at b. vc(0+) = vc(0-) = 60Vthe voltage across capacitor remains the same at this particular instant

*

RT = (18 k + 12 k) // 60 k = 20 ktime constant, = RTC = 20k x 0.1 F = 2msVc(t) = 60e-500t VUsing voltage divider rule, Hence, Vo(t) = 24e-500t V

*Example 4The switch in the circuit below has been closed for a long time, and it is opened at t = 0. Find v(t) for t 0. Calculate the initial energy stored in the capacitor.

*SolutionFor t0, the switch is open and the RC circuit is

*vc(0) = vc(0+) = vc(0-) = 15 VReq = 1+9 = 10Time constant, = ReqC = 0.2sBecause;So ;Because;Voltage across the capacitor;v(t) = 15e-5t VInitial energy stored in the capacitor is;wc(0) = 0.5Cvc2 =2.25J

*Summary

No RL circuitRC circuit12Inductor behaves like a short circuit when being supplied by dc source for a long timeCapacitor behaves like an open circuit when being supplied by dc source for a long time3Inductor current is continuousiL(0+) = iL(0-) Voltage across capacitor is continuousvC(0+) = vC(0-)

*5-3 Step Response of RL CircuitThe step response is the response of the circuit due to a sudden application of a dc voltage or current source.Consider the RL circuit below and the switch is closed at time t = 0.

After switch is closed, using KVL

(1)

*Cont.Rearrange the equation;(4)(3)(2)(5)(6)Therefore:

*Cont.Hence, the current is;

Or may be written as;

Where i(0) and i() are the initial and final values of i, respectively.

The voltage across the inductor is;

Or;

*Example 5The switch is closed for a long time at t = 0, the switch opens.Find the expressions for iL(t) and vL(t).i

*Solution When t < 0, the 3 resistor is short circuit, and the inductor acts like short circuit.So; i(0) = i(0+) = i(0-) = 5ABecause inductor current cannot change instantaneouslyWhen t< 0, the switch is open and the both resister are in series.iL(0+)

*Time constant;When t = , the inductor acts as short circuit again.Thus: iL(t) = i() +[i(0) i()]e-t/ = 2 + 3e-20t A

And the voltage is:= -15e-20t V

*5-4 Step Response of RC CircuitConsider the RC circuit below. The switch is closed at time t = 0

From the circuit;

Division of Equation (1) by C gives;

(1)(2)

*Cont.

Same mathematical techniques with RL, the voltage is:

Or can be written as:

And the current is:

Or can be written as:

v(t) = v() + [v(0) v()]e-t/

*Example 6 The switch has been in position a for a long time. At t = 0, theswitch moves to b. Find Vc(t) for t > 0 and calculate its value att=1s and t=4s

- *Solution When t
*At t = , the capacitor again behaves like an open circuit.Hence;Since, v(t) = v() + [v(0) v()]e-t/So; vc(t) = 30-15e-0.5t VAt , t = 1s, Vc(t) = 20.9VAt , t = 4s, Vc(t) = 28 VAnd;