# Circuit Theory/Power Factor

Power factor is defined as:

$$p_f = \cos(\phi_v - \phi_i)$$

Facts:

-   The power factor is ideally 1.
-   Power factors less than 0.9 cause a visit by your utility company.
-   Building wire codes are designed to maximize the power factor.
-   Circuits are purely resistive (no capacitors or inductors) when they
    have a power factor of 1.
-   Capacitors can cancel out inductors (and vice versa) to create
    purely resistive circuits.
-   The Math of this is easiest in the phasor domain.
-   Exceptions to this are where building up oscillations are the design
    objective \... such as the cavity in a microwave oven or designing a
    musical instrument.
-   The turbulence that the power factor measures can only be seen in
    the phasor domain.

Let\'s consider two circuits: !series RLC
circuit !parallel RLC
circuit

## Series Circuit

The voltage across the inductor and capacitor have to cancel each other
out. This means that the power supply doesn\'t see either. This means
that the inductor and capacitor are passing energy back and forth, but
the resistor and power supply can not see them. This means that

$$\mathbb{V}_L + \mathbb{V}_C = 0$$

$$\mathbb{V}_L = - \mathbb{V}_C$$ The current is going to be the same:

$$\mathbb{I}$$ From the terminal relations:

$$\mathbb{V}_L = jwL\mathbb{I}$$

$$\mathbb{I} = jwC\mathbb{V}_C$$ Solving for $\mathbb{V}_C$:

$$\mathbb{V}_C = \frac{\mathbb{I}}{jwC}$$ Substituting into the voltage
equation:

$$jwL\mathbb{I} = - \frac{\mathbb{I}}{jwC}$$ Now solving for C:

$$C = \frac{1}{Lw^2}$$

## Parallel Circuit

The current through the inductor and capacitor have to cancel each other
out. This means that the power supply doesn\'t see either. This means
that the inductor and capacitor are passing energy back and forth, but
the resistor and power supply can not see them. This means that

$$\mathbb{I}_L + \mathbb{I}_C = 0$$

$$\mathbb{I}_L = - \mathbb{I}_C$$ The voltage is going to be the same:

$$\mathbb{V}$$ From the terminal relations:

$$\mathbb{V} = jwL\mathbb{I}_L$$

$$\mathbb{I}_C = jwC\mathbb{V}$$ Solving for $\mathbb{I}_L$:

$$\mathbb{I}_L = \frac{\mathbb{V}}{jwL}$$ Substituting into the current
equation:

$$jwC\mathbb{V} = - \frac{\mathbb{V}}{jwL}$$ Now solving for C:

$$C = \frac{1}{Lw^2}$$

So either way the formula for the capacitor to cancel out the inductor
is the same. This is strange because in the series circuit the L&C have
to combine to be a short to be invisible, and in the parallel circuit
they have to combine to be an open.

Obviously, there is never going to be a perfect match. If there were a
perfect match with ideal devices and a butterfly flapped its wings, the
circuit could blow up. But that is another story for later in this
course \...

## Dependence on Frequency

Eliminating the effects of inductance with capacitance is dependent upon
frequency. If the frequency deviates, then another imbalance occurs. In
power distribution systems, the 60 Hz or 50 Hz frequency doesn\'t
change. But in modems, radios and cable TV boxes, the frequency does
change \... when someone speaks, when data moves through a cable, when
changing channels. One of the objectives of this course is to understand
how to design/plan for these changes. 
