ECE 311: Circuits II

AC Power

Sinusoidal Power

Voltage and Current:
$v(t) = \bold{V}{p} cos( \omega t + \theta )$
$i(t) = \bold{I}
{p} cos( \omega t + \phi )$

Electric power:
$p(t) = v(t) * i(t)$

$P = \frac{V_p * I_p}{2} cos( \theta - \phi )$

Larger Capacitance → Smaller Current Larger Inductance → Larger Current

If we substitute in the voltage and current equations into the power equations, we get:
$P = \frac{1}{T}\int^{t0 + T}{t_0} V_p I_p cos( \omega t + \theta) cos( \omega t + \phi ) dt$
$P = \frac{1}{T}\int^{t0 + T}{t_0} V_p I_p \frac{1}{2} [ cos( \theta + \phi) cos( 2 \omega t + \theta + \phi ) ] dt$

If you then integrate the expression, you get:
$P = \frac{V_p I_p}{2} cos( \theta - \phi)$

This even works with resistors:
$P_R = \frac{V_p I_p}{2} cos(0)$
$P_R = \frac{V_p I_p}{2}$

Power of a Resistor:
$P = \frac{1}{2} \frac{V_p^2}{R}$

Root Mean Square (rms) Voltage

Simplifies AC calculations of power. In this class, default is peak, rms is noted, unlike power systems.

$V_{rms} \equiv \sqrt{ \frac{1}{T} \int^{t0 + T}{t_0} v^2(t) dt }$

For harmonic voltage:
$V_{rms} = \sqrt{ \frac{1}{T} \int^{t0 + T}{t0} [ v p cos(\omega t + \theta ) ]^2(t) dt }$

$P = V{rms} * I{rms} cos( \theta - \phi )$

Example

$P = 100W$ Resistive Load: $\theta = \phi$, $V_{rms} = 120V$

$P = \frac{V{rms}^2}{R} => R = \frac{V{rms}^2}{P} 144 \Omega$

Note $V_{rms} = 120 V => Vp = \sqrt{2} * V{rms} \approx 170 V$

Power in an Inductive Load

$i(t) = I_p cos( \omega t + \phi)$

$v(t) = L \frac{di(t)}{dt}$
$= -L I_p sin( \omega t + \phi) * \omega$
$= -\omega L I_p cos( \omega t + \phi - \pi / 2)$
$= \omega L I_p cos( \omega t + \phi + \pi / 2)$
⇒
$V_p = \omega L I_p$
$\theta = \phi + \pi / 2$

Complex Power

Power is the voltage times the complex conjugate of the current, divided by two.

Complex Power: $\bold{S}$
Unit: VA (volt-ampere)
$\bold{S} = \bold{V}{rms} * \bold{I}{rms} \text{} = \frac{1}{2} V I \text{*}$

The symbol $\text{}$ means the complex conjugate.
Example 1:
$\bold{I} = 0.1 \angle 37 \degree$
$\bold{I} \text{
} = 0.1 \angle -37 \degree$
Example 2:
$\bold{I} = 0.35 + j0.28$
$\bold{I} \text{*} = 0.35 - j0.28$

Apparent Power: $S$
Unit: VA (volt-ampere)
$S = | \bold{S} | = | \bold{V}{rms} | * | \bold{I}{rms} \text{} | = V_{rms} I_{rms}$

Average Power: $P$
Unit: W (watt)
$P = \text{Re}{ \bold{V}{rms} * \bold{I}{rms} \text{} } = \frac{1}{2} \text{Re}{ \bold{V} \bold{I} \text{*} }$

Reactive Power: $Q$
Unit: VAR (volt-ampere reactive)
$Q = \text{Im}{ \bold{V}{rms} * \bold{I}{rms} \text{} } = \frac{1}{2} \text{Im} { \bold{V} \bold{I} \text{*} }$

Therefore: $\bold{S} = P + jQ$