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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.

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.

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

Defining some variables:

α = R/2L for series circuits, or

α = 1/2RC for parallel circuits.

ω_{0}^{2} = 1/LC
(resonant frequency)

x_{0} = initial value

x_{f} = final value

where x represents v (voltage) or i (current).

**Overdamped** (α > ω_{o})

x(t) = x_{f} + A_{1}e^{s1t} +
A_{2}e^{s2t}

where

s_{1} = -α +
√α^{2} -
ω_{0}^{2}

s_{2} = -α -
√α^{2} -
ω_{0}^{2}

**Underdamped** (α < ω_{0})

x(t) = x_{f} + B_{1}e^{-αt}cos
ω_{d}t + B_{2}e^{-αt}sin
ω_{d}t *where* ω_{d} =
√ω_{0}^{2}
- α^{2}

**Critically damped** (α = ω_{0})

x(t) = x_{f} + D_{1}te^{-αt} +
D_{2}e^{-αt}

Click on this link: http://www.coilgun.info/mark2/rlcsim.htm for a simulation of a series RLC circuit. It was created by Barry Hansen.

Use the boundary value analysis from the lesson on first order
circuits to show that:

v_{0} = 220 V

v_{f} = 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,

i_{C} = C dv_{C}/dt

At t = 0, this becomes:

i_{C}(0) = Cv'(0)

You can use Kirchhoff's Current Law at t = 0^{+} to show
that,

i_{C}(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.

Now we calculate two constants:

α = 1/2RC = 1/(2 × 11 × 62.5×10^{-6})
= 727 radians/second

ω_{0}^{2} = 1/LC =
1/(.025 × 62.5×10^{-6}) = 64×10^{4}

ω_{0} = 800 radians/second
(Note that it's a radian frequency -- not in cycles/second.)

Since α < ω_{0},
the circuit is underdamped. We therefore use the following
equation:

x(t) = x_{f} + B_{1}e^{-αt}cos
ω_{d}t + B_{2}e^{-αt}sin
ω_{d}t *where* ω_{d} =
√ω_{0}^{2}
- α^{2}

Since we are looking for a voltage, this equation becomes:

v(t) = v_{f} + B_{1}e^{-αt}cos
ω_{d}t + B_{2}e^{-αt}sin ω_{d}t
*where* ω_{d} =
√ω_{0}^{2}
- α^{2}

Now let's evaluate some of the constants.

v_{f} = 0 V (previously calculated)

ω_{d} =
√ω_{0}^{2}
- α^{2} =
√800^{2}
- 727^{2} = 333 rad/sec

This leaves B_{1} and B_{2} as unknowns. These
must be determined from initial conditions. At t=0,

v(0) = 220 = 0 + B_{1}e^{0}cos(0) + B_{2}e^{0}sin(0)
= B_{1}

B_{1} = 220

To evaluate B_{2}, it will
now be useful to calculate v'(t), the first derivative of v(t):

v'(t) = d/dt[v_{f} + B_{1}e^{-αt}cos
ω_{d}t + B_{2}e^{-αt}sin ω_{d}t]

v'(t) = 0 + B_{1}e^{-αt}(-αcos ω_{d}t
- ω_{d}sin ω_{d}t)
+ B_{2}e^{-αt}(-αsin ω_{d}t
+ ω_{d}cos ω_{d}t)

We now evaluate the first derivative at t = 0, recalling that we
previously evaluated v'(0) to be 0:

v'(0) = 0 = B_{1}e^{0}[-αcos(0) - ω_{d}sin(0)]
+ B_{2}e^{0}[-αsin(0) +
ω_{d}cos(0)]

0 = B_{1}[-α - 0] + B_{2}[0 + ω_{d}] = 220[-727]
+ B_{2}[333]

B_{2} = 480

The final solution is therefore:

v(t) = 220e^{-727t}cos
333t + 480e^{-727t}sin 333t V

A plot of this voltage is shown below.

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