Question

Let a series network be connected to a source voltage $\mathbf{V}$, drawing a current $\mathbf{l}$

(a) In terms of the load impedance $\mathbf{Z}\mathbf{=}\mathbf{Z}\mathbf{\angle }\mathbf{Z}$. Find expressions for $\mathbf{P}$and$\mathbf{Q}$,

(b) Express $\mathbf{p}\mathbf{(}\mathbf{t}\mathbf{)}$in terms of $\mathbf{P}$and $\mathbf{Q}$, by choosing$\mathbf{i}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\sqrt{\mathbf{2}\mathbf{l}}\mathbf{cos}\mathbf{\left(}\mathbf{\omega t}\mathbf{\right)}$.

(c) For the case of $\mathbf{Z}\mathbf{=}\mathbf{R}\mathbf{+}\mathbf{j\omega L}\mathbf{+}\frac{\mathbf{1}}{\mathbf{j\omega C}}$, interpret the result of part (b) in terms of$\mathbf{P}\mathbf{,}\mathbf{Q}\mathbf{}\mathbf{and}\mathbf{}{\mathbf{Q}}_{\mathbf{L}}$. In particular, if${\mathbf{\omega }}^{\mathbf{2}}\mathbf{LC}\mathbf{=}\mathbf{1}$, when the inductive and capacitive reactance cancel, comment on What happens.

Short answer

(a) The expression for the real power is ${\mathrm{Zl}}^2\cos \left(\theta \right)$and reactive power is $Z{l}^2\sin \left(\theta \right)$.

(b) The instantaneous power in term of$P$and$Q$is$P\left(1+\cos \left(2\omega t\right)\right)-Q\sin \left(2\omega t\right)$.
(c) The instantaneous power in term of$P,Q_{L}andQisP\left(1+\cos \left(2\omega t\right)\right)-Q\sin \left(2\omega t\right)$ and when the capacitive and inductive reactance cancel each other is$P\left(1+\cos \left(2\omega t\right)\right)$.

Step-by-step solution

determine the expression for Pand Q from the complex power consideration.

(a)

Write the expression for the apparent power.

$\mathbf{S}\mathbf{=}\mathbf{Re}\left({\mathrm{Vl}}^{*}\right)\mathbf{+}\mathbf{im}\left({\mathrm{Vl}}^{*}\right)$

Rewrite the expression as,

$$\begin{gathered} S=Re\left(Zl\times l^{*}\right)+Im\left(Zl\times l^{*}\right) \\ S=Re\left(Z{l}^2\right)+Im\left(Z{l}^2\right) \\ S=P+jQ \end{gathered}$$

Here, in the above equation the real power and the reactive power are,

$$\begin{gathered} P=Z{l}^2\cos \theta \\ Q=Z{l}^2\sin \theta \end{gathered}$$

Since the phase difference between the voltage and current is $\theta$.

Therefore, the expression for the real power is$Z{l}^2\cos \theta$ and reactive power is$Z{l}^2\sin \theta$.

Step 2:Express p(t) in terms of P and Q, by choosing i(t)=2lcos(ωt).

(b)

The instantaneous voltage can be calculated as,

$$\begin{gathered} v\left(t\right)=\sqrt{2}l\cos \left(\omega t+\theta \right)\left(Z\right) \\ =\sqrt{2}Zl\cos \left(\omega t+\theta \right)V \end{gathered}$$

The instantaneous power is the product of the instantaneous current and voltage.

$p\left(t\right)=v\left(t\right)\times i\left(t\right)$

Substitute the expressions and solve.

$$\begin{gathered} p\left(t\right)=\sqrt{2}l\cos \left(\omega t+\theta \right)\times \sqrt{2}l\cos \left(\omega t\right) \\ p\left(t\right)=2Z{l}^2\cos \left(\omega t+\theta \right)\times \cos \left(\omega t\right) \\ p\left(t\right)=Z{l}^{2}2\cos \left(\omega t+\theta \right)\times \cos \left(\omega t\right) \\ p\left(t\right)=Z{l}^2\left[\cos \left(2\omega t+\theta \right)+\cos \left(\theta \right)\right] \end{gathered}$$

Solve further as,

$$\begin{gathered} p\left(t\right)=Z{l}^2\left[\cos \left(2\omega t\right)\cos \left(\theta \right)-\sin \left(2\omega t\right)\sin \left(\theta \right)+\cos \left(\theta \right)\right] \\ =Z{l}^2\cos \left(2\omega t\right)\cos \left(\theta \right)-\sin \left(2\omega t\right)\sin \left(\theta \right)+\cos \left(\theta \right) \\ =\left[Z{l}^2\cos \left(\theta \right)\right]a\cos \left(2\omega t\right)-\left[Z{l}^2\sin \left(\theta \right)\right]\sin \left(2\omega t\right)+Z{l}^2\cos \left(\theta \right) \end{gathered}$$

Substitute the expression for instantaneous power and the reactive power and solve.

$$\begin{gathered} p\left(t\right)=P\cos \left(2\omega t\right)-Q\sin \left(2\omega t\right)+P \\ =P\left(1+\cos \left(2\omega t\right)\right)-Q\sin \left(2\omega t\right) \end{gathered}$$

Therefore, the instantaneous power in term of $P\mathrm{and}Q\mathrm{is}$

$P\left(1+\cos \left(2\omega t\right)\right)-Q\sin \left(2\omega t\right)$.

Interpret the instantaneous power in terms of P, Qc and QL.

(c)

The impedance of the $RLC$circuit is given by,

$Z=R+j\omega L+\frac{1}{j\omega C}$

The phase difference between voltage and current for the resistance is $0$.

$$\begin{gathered} P=R{l}^2\cos \left(\theta \right) \\ P=l^{2}R \end{gathered}$$

The phase difference between voltage and current for the inductance is$\theta =90°$.

$$\begin{gathered} Q_L=\omega L{l}^2\sin \left(90°\right) \\ {Q}_L=\omega L{l}^2 \end{gathered}$$

The phase difference between voltage and current for the capacitance is $\theta =-90°$.

$$\begin{gathered} Q_c=\omega L{l}^2\sin \left(-90°\right) \\ {Q}_c=\frac{-l^2}{\omega } \end{gathered}$$

The instantaneous power in term of $P,Q_L$and $Q_c$is $P\left(1+\cos \left(2\omega t\right)\right)-Q_L\sin \left(2\omega t\right)-Q_c\sin \left(2\omega t\right).$

When the capacitive and inductivereactance cancel each other. Then,

$$\begin{gathered} {\omega }^{2}LC=1 \\ {Q}_L{Q}_c=0 \end{gathered}$$

The instantaneous power can be calculated as,

$$\begin{gathered} p\left(t\right)=P\left(1+\cos \left(2\omega t\right)\right)-Q_L\sin \left(2\omega t\right)-Q_c\sin \left(2\omega t\right) \\ =P\left(1+\cos \left(2\omega t\right)\right)-\left(Q_L+Q_c\right)\sin \left(2\omega t\right) \\ =P\left(1+\cos \left(2\omega t\right)\right)-\left(0\right)\sin \left(2\omega t\right) \\ =P\left(1+\cos \left(2\omega t\right)\right) \end{gathered}$$

Therefore, the instantaneous power when the capacitive and inductive reactance cancel each other is$P\left(1+\cos \left(2\omega t\right)\right)$.