# 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 vise 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.

Lets 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 it's 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.