Single-Phase AC Power Analysis Quiz

$P = \frac{1}{T} \int_{0}^{T} p(t) dt$ or $P = \frac{1}{T} \int_{0}^{T} v(t)i(t) dt$

$P = VI cos\Phi = I^2R

$P = VI cos\Phi = 0

A resistive load (R)

A reactive load (L or C)

The average power is the average of the instantaneous power over one period and can be calculated using the formula $P = \frac{1}{T} \int_{0}^{T} p(t) dt$. It is more convenient to measure than instantaneous power, and the wattmeter, the instrument for measuring power, responds to average power.

The expression for average power in electrical circuits using phasor notation is given by $P = \frac{V_{rms}I_{rms}}{2} \cos(\theta_v - \theta_i)$, where $V_{rms}$ and $I_{rms}$ are the rms voltage and current, and $\theta_v$ and $\theta_i$ are the phase angles of the voltage and current, respectively.

In a resistive circuit, the average power is given by $P = V_{rms}I_{rms}\cos(\Phi)$, where $\Phi = 0$. In a pure inductive or capacitive circuit, the average power is zero, as $\Phi = \pm 90$ and $\cos(\Phi) = 0.

The complex power $S$ absorbed by the AC load is a combination of real power (average power) and reactive power. It is defined as $S = P + jQ$, where $P$ is the average power and $Q$ is the reactive power.

A resistive load (R) always absorbs power, while a reactive load (L or C) absorbs zero average power. This is because the average power in a reactive load is dependent on the phase angle, leading to a power factor of zero.