Question

Consider a load of impedance Z = jωL connected to a voltage and V let the current drawn be I .

(a) Develop an expression for the reactive power Q in terms of ωt, L and I , from complex power considerations.

(b) Let the instantaneous current be I (t)=$\sqrt[]{\mathbf{2}}\mathbf{Icos}$ (ωt + θ). Obtain an
expression for the instantaneous power p (t) into and then express it in terms of Q .

(c) Comment on the average real power P supplied to the inductor and the instantaneous power supplied.

Short answer

(a) The expression for the reactive power in term of and is ω,L and I is ωLI² .

(b) The power in term of Q is -Q sin2 (ωt+θ) .

(c) The real power supplied to the inductor is 0 .

Step-by-step solution

Step 1:Develop an expression for the reactive power  Q in terms of  ω, L and I  , from complex power considerations.

(a)

The apparent power is given by,

$\mathit{S}\mathbf{}\mathbf{=}\mathbf{}\mathbf{VI}\mathbf{*}$

Solve for the reactive power Q .

$\begin{array}{l}\mathrm{Q}=\mathrm{lm}\left({\mathrm{VI}}^{*}\right)\\ \mathrm{Q}=\mathrm{lm}\left({\mathrm{j\omega Lll}}^{*}\right)\\ \mathrm{Q}=\mathrm{im}\left({\mathrm{j\omega Ll}}^2\right)\\ \mathrm{Q}={\mathrm{\omega Ll}}^2\end{array}$

Hence, the expression for the reactive power in term of ω, L and I is ωLI² .

Determine expression for the instantaneous power p (t)  .

(b)

The instantaneous current is given by,

i(t) = $\sqrt[]{2}$I cos (ωt +θ)

The instantaneous voltage can be calculated as,

$$\begin{gathered} \mathrm{V}=\mathrm{L}\frac{\mathrm{di}\left(\mathrm{t}\right)}{\mathrm{dt}} \\ \mathrm{V}=\mathrm{L}\frac{\mathrm{d}}{\mathrm{dt}}\left(\sqrt{2}\mathrm{lcos}\left(\mathrm{\omega t}+\mathrm{\theta }\right)\right) \\ \mathrm{V}=\mathrm{L}-\sqrt{2}\mathrm{\omega Llsin}\left(\mathrm{\omega t}+\mathrm{\theta }\right) \end{gathered}$$

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

p (t) = v (t) x i (t)

Substitute the expression and solve.

$$\begin{gathered} p\left(t\right)=-\sqrt{2}\omega Ll\sin \left(\omega t+\theta \right)\times \sqrt{2}l\cos \left(\omega t+\theta \right) \\ p\left(t\right)=-2\omega L{l}^2\sin \left(\omega t+\theta \right)\cos \left(\omega t+\theta \right) \\ p\left(t\right)=-\omega L{l}^2\times 2\sin \left(\omega t+\theta \right)\cos \left(\omega t+\theta \right) \\ p\left(t\right)=-\omega L{l}^2\times \sin 2\left(\omega t+\theta \right) \end{gathered}$$

Substitute the reactive power expression from part (a) and solve as,

p (t) = -Q x sin2 (ωt+θ)

Hence, the power in term of Q is -Q x sin2 (ωt+θ)

Calculate the average real power  P supplied to the inductor and the instantaneous power supplied.

(c)

The average real power is given by

$$\begin{gathered} P=\mathrm{Re}\left({\mathrm{VI}}^{*}\right) \\ P=\mathrm{Re}\left(\mathrm{j\omega L}\times \mathrm{l}\times {\mathrm{l}}^{*}\right) \\ P=\mathrm{Re}\left(\mathrm{j\omega L}\times {\mathrm{l}}^2\right) \\ P=0 \end{gathered}$$

Here, the real power is zero and reactive power Q would be maximum.

Hence, the real power supplied to the inductor is 0.