ECE 252 Introduction to Electrical Engineering

Lesson 9. Second Order Circuits

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In this lesson, you will study circuits that contain one capacitor, one inductor, resistors, sources, and switches. The significant difference from the lesson on first order circuits is the inclusion of both a capacitor and an inductor in the circuit. This dramatically changes the character of the resulting solutions. Although the boundary value analysis you have previously studied will still work, the solutions will typically no longer be simple exponentials. The capacitor and the inductor will exchange energy in ways that produce solutions with such components as damped sinusoids.

Consider the circuit below.

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It is desired to find the voltage v across the capacitor after the switch is closed. The direct approach to solving this problem would be to write differential equations for the currents and the voltages in the circuit, then solve them. Instead, we will solve it using a systematic approach assuming that the solution falls within one of three general forms.

Generic Solution for RLC problems

These solutions work if the circuit can be reduced to look like either of the two figures below.

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Defining some variables:
α = R/2L for series circuits, or
α = 1/2RC for parallel circuits.
ω02 = 1/LC (resonant frequency)
x0 = initial value
xf = final value
where x represents v (voltage) or i (current).

Overdamped (α > ωo)
x(t) = xf + A1es1t + A2es2t
s1 = -α + √α2 - ω02
s2 = -α - √α2 - ω02

Underdamped (α < ω0)
x(t) = xf + B1e-αtcos ωdt + B2e-αtsin ωdt where ωd = √ω02 - α2

Critically damped (α = ω0)
x(t) = xf + D1te-αt + D2e-αt


Click on this link: for a simulation of a series RLC circuit.  It was created by Barry Hansen.

Solution of Sample Problem

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Use the boundary value analysis from the lesson on first order circuits to show that:
v0 = 220 V
vf = 0 V

Also use boundary value analysis to find v'(0) -- that's the first derivative of v(t) evaluated at t = 0. This can be done by noting that,
iC = C dvC/dt
At t = 0, this becomes:
iC(0) = Cv'(0)
You can use Kirchhoff's Current Law at t = 0+ to show that,
iC(0) = 0, and therefore
v'(0) = 0

Next, we must reduce the circuit to the form of the parallel RLC circuit. This can be done with a source conversion on the voltage source and resistor.

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Now we calculate two constants:
α = 1/2RC = 1/(2 × 11 × 62.5×10-6) = 727 radians/second
ω02 = 1/LC = 1/(.025 × 62.5×10-6) = 64×104
ω0 = 800 radians/second (Note that it's a radian frequency -- not in cycles/second.)

Since α < ω0, the circuit is underdamped. We therefore use the following equation:
x(t) = xf + B1e-αtcos ωdt + B2e-αtsin ωdt where ωd = √ω02 - α2
Since we are looking for a voltage, this equation becomes:
v(t) = vf + B1e-αtcos ωdt + B2e-αtsin ωdt where ωd = √ω02 - α2

Now let's evaluate some of the constants.
vf = 0 V (previously calculated)
ωd = √ω02 - α2 = √8002 - 7272 = 333 rad/sec

This leaves B1 and B2 as unknowns. These must be determined from initial conditions. At t=0,
v(0) = 220 = 0 + B1e0cos(0) + B2e0sin(0) = B1
B1 = 220
To evaluate B2, it will now be useful to calculate v'(t), the first derivative of v(t):
v'(t) = d/dt[vf + B1e-αtcos ωdt + B2e-αtsin ωdt]
v'(t) = 0 + B1e-αt(-αcos ωdt - ωdsin ωdt) + B2e-αt(-αsin ωdt + ωdcos ωdt)
We now evaluate the first derivative at t = 0, recalling that we previously evaluated v'(0) to be 0:
v'(0) = 0 = B1e0[-αcos(0) - ωdsin(0)] + B2e0[-αsin(0) + ωdcos(0)]
0 = B1[-α - 0] + B2[0 + ωd] = 220[-727] + B2[333]
B2 = 480

The final solution is therefore:
v(t) = 220e-727tcos 333t + 480e-727tsin 333t V

A plot of this voltage is shown below.

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At this time you should complete Tutorial 9 on second order circuits.