# Circuit Theory/1Initially Excited

Circuits are used to charge capacitors and inductors. Then the
inductor/capacitor is switched out of the charging circuit and into the
discharging circuit. There are three concepts:

-   charging circuit \... assume on a long time if time on not specified
    \... assume steady state so capacitors are opens and inductors are
    shorts
-   switching circuit (different for capacitors and inductors)
-   discharging circuit \... assume initially capacitors are shorts and
    inductors are opens

There are two circuits to analyse: charging and discharging:

-   Charging circuit \... Assume initial conditions zero, find steady
    state (particular) solution \... these become the initial conditions
    to the discharging circuit.
-   Discharging circuit find the (homogeneous solution).

Solving the discharging circuit has these steps:

-   assume the solution is an exponential
-   find an exponential solution
-   use the initial conditions and final value conditions to find the
    constants.
-   If an exponential solution can be found (i.e. there is a real part),
    then the assumption was valid.

## Capacitors

<File:Ch7CapSwitchingCircuit.png%7CCapacitor> charging and discharging
circuit separated by a switch <File:Ch7capcharging.png%7CEquivalent>
charging circuit <File:Ch7capswitched.png%7CMost> switches are
BBM (break before make) so
the capacitor is an open while the single pole is being thrown between
the two poles. Super
capacitors store lots of
energy, charge up very fast, discharge very fast and could replace
batteries. <File:Ch7capDischarging.png%7CEquivalent> discharging circuit

### Charge Analysis

Charge Analysis is steady state analysis. Assume the circuit has been on
a long time. The capacitor has had a long time to charge up.

What is different is that instead of a sinusoidal source, there is a DC
voltage source. There is going to be a DC steady state voltage across
the capacitor.

A capacitor looks like an open when fully charged, so the full source
voltage is going to be across the capacitor. So in this case the initial
voltage across the capacitor is 1 volt.

### Discharge Analysis

!Looking at a capacitor discharging, marking up the current direction
and voltage polarity consistent with the normal terminal relations ..
for wikibook circuit
theory
A trivial mesh or loop analysis (both result in the same equation) would
be:

$$V_R + V_c = 0$$

$$R*i + V_c = 0$$ From the capacitor terminal relationship:

$$i=C{d Vc \over dt}$$ So substituting:

$$RC{d Vc \over dt} + V_c =0$$

Of course, after a long time, the capacitor discharges, the resistor
dissipates all the energy and both current and voltage everywhere will
be zero. But what is the equation in the time domain that describes how
the voltage and current go to zero?

Notice that the current is going down while charging, but switches to
going up when discharging. This can be instantaneous.

#### finding time constant

The general technique is to assume this form:

$$V_c = A*e^{-\frac{t}{\tau}}$$

Substituting into the above differential equation:

$$-\frac{RCA}{\tau}e^{-\frac{t}{\tau}} + A*e^{-\frac{t}{\tau}} = 0$$
Dividing through by A and the exponential, have:

$$-\frac{RC}{\tau} + 1 = 0$$ Solving for tau:

$$\tau = RC$$

#### finding the constants

So the formula for voltage across the capacitor is now:

$$V_c = A*e^{-\frac{t}{RC}} + C$$ This includes the steady state
particular solution of 0 and the constant C that comes from solving a
differential equation.

The initial voltage across the capacitor is +1 volt at t=0~+~. This
means that:

$$V_c(0_+) = 1 = A + C$$

After a long period of time, the V_c(t) = 0. This helps us find find C:

$$V_c(\infty) = A*0 + C = 0$$ So C = 0 and A = 1 and:

$$V_c(t) = e^{-\frac{t}{RC}}$$

#### finding the current

!Looking at a capacitor discharging, marking up the current direction
and voltage polarity consistent with the normal terminal relations ..
for wikibook circuit
theory
V~c~=-V~r~ or could plug into the capacitor terminal relation

$$I = \frac{V_r}{R} = \frac{-Vc}{R} = -\frac{e^{-\frac{x}{RC}}}{R}$$
Thus for this circuit:

$$V_C = e^{-\frac{t}{RC}}=e^{-\frac{t}{10\mu s}}$$

$$V_r = -e^{-\frac{t}{10\mu s}}$$ And:

$$I = - \frac{e^{-\frac{t}{10\mu s}}}{10}$$

#### Interpreting the results

V~c~ is in the polarity indicated. V~R~ is opposite the polarity drawn.
The current is going in the opposite direction drawn, which makes sense
if the capacitor is acting like a voltage source and dumping it\'s
energy into the resistor. Both are going to be essentially zero after 5
time constants which is 50 μs.

The homogeneous solution was easy to find because a DC source was used
to charge the circuit. The particular solution of the discharge circuit
was 0 because there was no forcing function.

## Inductors

<File:Ch7IndustorSwitchingCircuit.png%7CThis> circuit shows an inductor
sitting between a charging circuit and discharging circuit. Because most
switches are BBM, a push button was added to short the inductor before
the SPDT switch is thrown. Both switches could be replaced by a single
SPDT MBB switch. However
simulation software typically just has BBM switches.
<File:Ch7InductorCharging.png%7CThe> equivalent inductor charging
circuit <File:Ch7IndustorSwitchingCircuitEffective.png%7CThe> equivalent
circuit after the push button switch is pressed, but before the SPDT BBM
switch is thrown. <File:Ch7InductorSwitched.png%7CThe> equivalent
circuit after the SPDT BBM switch is thrown while the push button switch
maintains contact. Superconductors can be energy
storage devices just like super capacitors. Current moves at the speed
of light through a circular wire that acts like a continuous inductor.
<File:Ch7InductorDischarging.png%7CThe> equivalent inductor discharging
circuit after the push button switch is released.

### Charge Analysis

!When inductor has same initial current as the current source, then no
current is going to flow through the short. Superposition says the
currents are going to cancel in the
short.
Charge Analysis is steady state analysis. Assume the circuit has been on
a long time. The inductor has had a long time to build up its magnetic
fields.

What is different is that instead of a sinusoidal source, there is a DC
current source. There is going to be a DC steady state current through
the inductor.

An inductor looks like a short when fully charged, so the full source
current is going through the inductor. So in this case the initial
current is 1 amp.

### Shorted source and inductor Analysis

!An inductor stores it\'s energy in a magnetic field, current has to
continue to move to keep the magnetic field from collapsing \... this is
the basis of a
superconductor
!Inductor with initial current discharging through resistor, for
wikibook circuit
theory
Shorting an inductor and current source by pushing the button of SW1 is
safe. (It is dangerous to open wires to an inductor or current source).
With ideal components, there will be no current through the short.
However current through the inductor remains the same and the inductor
remains charged up.

### Shorted inductor energy storage

The instant the SPDT switch cuts out the current source, the inductor\'s
current appears in the short. The inductor and shorting wire do not
store energy very long (unless frozen because the
wire acts like a resistor).

### Discharge Analysis

Immediately after SW2 has completed moving the throw over to the second
pole and the push button switch is released, the current of 1amp still
flows through the inductor. A voltage instantly appears across the
inductor and resistor. A trivial mesh or loop analysis (both result in
the same equation) would be:

$$V_R - V_L = 0$$ Looking at the circuit markup, the current and voltage
don\'t have the positive sign convention relationship so the inductor\'s
terminal relationship has a negative sign:

$$V_L= - L\frac{di}{dt}$$ So:

$$R*i + L\frac{di}{dt} = 0$$ Of course, after a long time, the inductor
discharges, the resistor dissipates all the energy and both current and
voltage everywhere will be zero. But what is the equation in the time
domain that describes how the voltage and current go to zero?

Notice that the voltage is positive on top of the inductor but switches
polarity when discharging. This can be instantaneous.

#### finding time constant

The general technique is to assume this form:

$$i = A*e^{-\frac{t}{\tau}}$$

Substituting into the above differential equation:

$$R*A*e^{-\frac{t}{\tau}} + L*\frac{d (A*e^{-\frac{t}{\tau}})}{dt} = 0$$
Dividing through by A and then evaluating the derivative, have:

$$R*e^{-\frac{t}{\tau}} - \frac{L*e^{-\frac{t}{\tau}}}{\tau} = 0$$
Dividing through by the exponential:

$$R - \frac{L}{\tau}=0$$ Solving for tau:

$$\tau = \frac{L}{R}$$

#### finding the constants

So the formula for current now is:

$$i = A*e^{-\frac{t}{\frac{L}{R}}} + C$$ The steady state particular
solution is 0, and the constant C comes from solving the differential
equation.

The initial current is 1 amp at t=0~+~. This means that:

$$1 = A + C$$ After a long time, there is nothing going on in the
circuit so:

$$i(\infty) = A*0 + C$$ So A=1 and C=0 thus:

$$i(t) = e^{-\frac{t}{\frac{L}{R}}}$$

#### finding the voltage across the Inductor/Resistor

V~L~=V~r~ or could plug into the inductor terminal relation

$$V_R = R*i = R * e^{-\frac{t}{\frac{L}{R}}}$$ of plugging into the
terminal relation:

$$V_L = - L*\frac{d (e^{-\frac{t}{\frac{L}{R}}})}{dt} = -L*(\frac{-R}{L})e^{-\frac{t}{\frac{L}{R}}} = Re^{-\frac{t}{\frac{L}{R}}}$$
So in summary:

$$V_R = V_L = 10*e^{-\frac{t}{0.1\mu s}}$$

$$i = e^{-\frac{t}{0.1\mu s}}$$

#### Interpreting the results

Maintaining the current and magnetic field energy of an inductor
requires shorting it during the switching between circuits. This would
be just as simple as switching a capacitor, but a different type of
switch is required: \... a make before break switch .. so that two
circuits are connected simultaneously.

The discharge circuit was marked up with the final directions and
polarity of everything. But the terminal relationship for the inductor
had to have a negative sign in it because the voltage and current
directions did not follow the positive sign convention.

Again, the homogeneous solution was easy to find because a DC source was
used to charge the circuit. The particular solution of the discharge
circuit was 0 because there was no forcing function.

