Sign Convention of \(V\) and \(I\)
Phasor Representation
Conversion of Time-domain to Phasor and Impedance Triangle
Powers: Instantaneous, Active, Reactive, and Complex
Relationship between the power quantities
Power Factor and its Significance
Advantage of Reactive Power Compensation
Convention for \(V\) and \(I\)
In linear circuit with sinusoidal \(V\) and \(I\) of frequency \(f\) applied for long so that steady state is reached, all \(V\) and \(I\) are at \(f=\omega/2\pi\)
Phasor converts \(v(t)\) and \(i(t)\) to complex variables \(\overrightarrow{V}\) and \(\overrightarrow{I}\)
\[\begin{aligned} v(t) & =\sqrt{2} V \cos \left(\omega t+\phi_{v}\right) \\ \Leftrightarrow \overrightarrow{V}& =V \angle \phi_{v} \\ i(t) & =\sqrt{2} I \cos \left(\omega t+\phi_{i}\right) \\ \Leftrightarrow \overrightarrow{I} & =I \angle \phi_{i} \end{aligned}\] Note: \(V\) and \(I\) are in RMS
To calculate \(i(t)\) use differential equation: \[R i(t)+L \frac{d i(t)}{d t}+\frac{1}{C} \int i(t) \cdot d t=\sqrt{2} V \cos (\omega t)\]
\[\begin{aligned} Z & =R+j X_{L}+j X_{c} ~~ =|Z|<\phi \\ X_{L} & =\omega L, ~~ \quad X_{c}=\left(\frac{1}{-\omega C}\right) \\ |Z| & =\sqrt{R^{2}+\left(\omega L-\frac{1}{\omega C}\right)^{2}} & \phi=\tan ^{-1}\left[\frac{\left(\omega L-\frac{1}{\omega C}\right)}{R}\right] \end{aligned}\] \(\checkmark\) Note: \(Z\) is complex but not a phasor, hence does not have time-domain expression
Instantaneous power \(p(t)=v(t)\cdot i(t)\) delivered by subcircuit-1 (Generator) is absorbed by subcircuit-2 (Load)
A negative value of \(p(t)\) reverses their role of delivering and absorbing power
Complex power (\(S\)), reactive power (\(Q\)), and power factor (pf) express how effectively active (average) power \(P\) transferred from one subcircuit to other
If \(v(t)\) and \(i(t)\) are in phase, \(p(t)\) pulsates at twice the steady-state \(f\)
\[\begin{aligned} p(t) & =\sqrt{2} V \cos \omega t \cdot \sqrt{2} I \cos \omega t \quad(i \text { in phase } \operatorname{with} v) \\ & = 2 V I \cos ^{2} \omega t=V I+V I \cos 2 \omega t \end{aligned}\] \(\ast\) \(\phi_{v}\) and \(\phi_{i}\) are assumed to be zero without loss of generality
In this case, for all times \(p(t) \geq 0\), and therefore power flows in only one direction from subcircuit 1 to 2
The average over one cycle of the second term of the RHS is zero, therefore average power is \(P= VI\)
Considering \(i(t)\) lags behind \(v(t)\) by \(\phi (t)\)
\(p(t)\) becomes -ve during a time-interval of \(\phi/\omega\) during each half-cycles as calculated by \[\begin{aligned} p(t)& =\sqrt{2} V \cos \omega t \cdot \sqrt{2} I \cos (\omega t-\phi) \\ & =V I \cos \phi+V I \cos (2 \omega t-\phi) \end{aligned}\]
Negative \(p(t)\) means power flow in opposite direction
This back and forth flow of power indicates real power is not transferred from one subcircuit to other
Avg. power \(P=VI\cos\phi\) is less than the previous case
\[\begin{aligned} S & =\overrightarrow{V}\overrightarrow{I}^{*} ~ *\rightarrow\text{conjugate}\\ & =V\angle\phi_{v}\cdot I\angle-\phi_{i}\\ & =VI\angle\left(\phi_{v}-\phi_{i}\right)\\ & =VI\angle\phi\\ & =P+jQ \end{aligned}\] where, \[\begin{array}{l} P=V I \cos \phi \\ Q=V I \sin \phi \end{array}\] \[\begin{array}{l} |S| =\sqrt{P^{2}+Q^{2}} \\ \phi =\tan^{-1}\left(\dfrac{Q}{P}\right) \end{array}\]
\(I\cos \phi\) is in phase with \(V\), and results in \(P\)
\(I\sin \phi\) is at \(90^\circ\) to \(V\), and results in \(Q\)
\[\begin{aligned}
P & =VI\cos\phi\\
\Rightarrow pf = \cos\phi & =\dfrac{P}{VI}
=\dfrac{\text{Real Power}}{\text{Apparaent Power}}\\
& =\dfrac{kW}{kVA} =\dfrac{R}{Z}
\end{aligned}\]
\[\begin{aligned} P & \rightarrow W(\text{Watts}); ~~ Q \rightarrow\text{VAR (Volt-Amperes Reactive)} \\ |S| & \rightarrow\text{VA (Volt-Amperes);} \\ \phi_{v},\phi_{i},\phi & \rightarrow\text{radians (+ve anticlockwise w.r.t real axis L to R)} \\ pf \rightarrow & \cos\phi = \text{ dimensionless, value between 0 to 1} \end{aligned}\]
Electrical usage cost is prop. to \(|S| = VI\)
Electrical insulation level and magnetic core size for a definite \(f\) depends on \(V\)
Conductor size depends on \(I\)
\(P\) represents (useful work \(+\) losses)
Desirable for \(Q = 0\) because it increases \(|S|\)
PF measure how effectively load draws \(P\)
Ideally \(P\) should be 1 (unity) i.e \(Q=0\)
Inductive load draws power at lagging pf (\(I\) lags \(V\))
Capacitive load draws power at leading pf (\(I\) leads \(V\))
\[\begin{aligned} \text{Total Real Power} & =\sum_{k}P_{k}=\sum_{k}I_{k}^{2}R_{k}\\ \text{Total Reactive Power} & =\sum_{k}Q_{k}=\sum_{k}I_{k}^{2}X_{k} \\ +Q & \rightarrow \text{lagging load (Inductive)} \\ -Q & \rightarrow \text{leading load (Capacitive)} \end{aligned}\]