# ECE 252 Introduction to Electrical Engineering

## Lesson 8. First Order Circuits

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This lesson deals with the use of either capacitors or inductors (not both at the same time) in single-time-constant circuits. This usually means that the circuit consists of a single inductor or capacitor, several resistors, and one or more switches. Such a circuit is shown below. Assume that the switch has been open for a long time, and that at t = 0 the switch is closed. Intuitively, you should expect that before the switch is closed, the capacitor will charge up to 50 V. This is so because, if you wait long enough, the current in the 4000 Ω resistor will decay to zero, and there will be no voltage across the resistor. Then, by KVL, the capacitor and battery voltages must be equal.

After the switch is thrown, things get more complicated: The capacitor will begin to drain through the 1000 Ω resistor, but current will continue to be provided by the battery. You might guess that the capacitor voltage would decay, but wouldn't go all the way to zero, and you would be right.

Click on this link: http://micro.magnet.fsu.edu/electromag/java/timeconstant/ to see a demonstration of the RC time constant.

Below is a systematic approach to solving problems like this. Read it carefully, then we'll apply it to the above problem.

### Approach to Boundary Value Problems

This technique works for first order systems with constant sources.

For simplicity, assume ti = 0, where ti is the initial time, i.e. the time when the switch is closed (or opened).

1. Write down:

iC = C(dvC/dt)
vL = L(diL/dt)
i(t) = if + (i0 - if)e-t/τ
v(t) = vf + (v0 - vf)e-t/τ

2. For 0-

1. Recall that "steady state" has been reached.
2. Therefore, vC = constant, iL = constant
3. Therefore, iC = 0, vL = 0
4. Redraw the circuit, open circuiting capacitors (because iC = 0) and short circuiting inductors (because vL = 0)
5. Apply KVL and KCL to find unknown voltages and currents, particularly vC or iL .

3. For 0+

1. Recall that you can't change vC or iL instantaneously.
2. Therefore, vC(0+) = vC(0-) and iL(0+) = iL(0-)
3. Apply KCL and KVL to find v0 and i0.

4. For infinity

1. Repeat 2.
2. Calculate vf or if.

5. Finally

1. Redraw circuit, removing C or L, shorting voltage sources, and opening current sources.
2. Calculate RTH "looking into" the terminals where you took out C or L.
3. Calculate τ = RTHC or τ = L/RTH.
4. Plug values into i(t) or v(t) equation.
5. Simplify.

### Example Solution to RC Problem

1. Write down:

iC = C(dvC/dt)
v(t) = vf + (v0 - vf)e-t/τ

2. For 0-

1. Recall that "steady state" has been reached.
2. Therefore, vC = constant
3. Therefore, iC = 0
4. Redraw the circuit, open circuiting capacitors (because iC = 0) 5. Apply KVL to find the unknown voltage.
This is easy:
-50 + v4000 + v = 0
But v4000 = 0 because it caries no current, so
v = v(0-) = 50 V

3. For 0+

1. Recall that you can't change vC instantaneously.
2. Therefore, vC(0+) = vC(0-)
v(t = 0) = v0 = 50 V

4. For infinity

1. Repeat 2. By the voltage divider formula,
v = 50×1000/(4000 + 1000) = 10 V
2. Calculate vf .
vf = v = 10 V

5.

1. Redraw circuit, removing C, shorting voltage sources. 2. Calculate RTH "looking into" the terminals where you took out C.
RTH = 4000 × 1000/(4000 + 1000) = 800 Ω
3. Calculate τ = RTHC.
τ = 800 × 20×10-6 = 16×10-3 s = 16 ms = .016 s
4. Plug values into v(t) equation.
v(t) = vf + (v0 - vf)e-t/τ = 10 + (50 - 10)e-t/.016 V
5. Simplify.
v(t) = 10 + 40e-t/.016 V

Here's what a graph of the capacitor voltage looks like: ### Simulation

Check out this first order circuit explanation and example:  http://people.clarkson.edu/~svoboda/eta/plots/FOC.html (created by James A. Svoboda of Clarkson University).

At this time you should complete Tutorial 8 on first order circuits.