Question

For the circuit element of Problem 2.3, calculate (a) the instantaneous power absorbed, (b) the real power (state whether it is delivered or absorbed), (c) the reactive power (state whether delivered or absorbed), (d) the power factor (state whether lagging or leading).

[Note: By convention the power factor $\mathbf{cos}\mathbf{(}\mathbf{\delta }\mathbf{-}\mathbf{\beta }\mathbf{)}$ is positive. If$\left|\delta -\beta \right|$ is greater than 90°, then the reference direction for current may be reversed, resulting in a positive value of $\mathbf{cos}\mathbf{(}\mathbf{\delta }\mathbf{-}\mathbf{\beta }\mathbf{)}$.

Short answer

(a) Therefore, the expression for the instantaneous power is

$\mathrm{p}\left(\mathrm{t}\right)=20000\left[\cos \left(2\mathrm{\omega t}-{50}^0\right)+0.342\right]$.

(b) Therefore, the value of the real power is $6840.4\mathrm{W}$ .

(c) Therefore, the value of the reactive power is$-18793.85\mathrm{VAR}$ .

(d) Therefore, the power factor is$0.342$ leading.

Step-by-step solution

Define formula for the instantaneous power, real power reactive power and the active power.

Consider the formula for the average power absorbed by the element.

$\mathbf{p}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{v}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{}\mathbf{i}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$ …… (1)

Here, v (t) is the instantaneous voltage and i (t) is the instantaneous current.

Consider the formula for the real power.

role="math" localid="1655118856768" $\mathbf{P}\mathbf{=}\mathbf{VIcos}\mathbf{(}\mathbf{\delta }\mathbf{-}\mathbf{\beta }\mathbf{)}$ …… (2)

Here,$(\delta -\beta )$ is the phase angle, V and I are the rms voltage and current respectively.

Consider the formula for the reactive power.

$\mathbf{P}\mathbf{=}\mathbf{VIsin}\mathbf{(}\mathbf{\delta }\mathbf{-}\mathbf{\beta }\mathbf{)}$ …… (3)

Consider the formula for the power factor.

$\mathbf{PF}\mathbf{=}\mathbf{cos}\mathbf{(}\mathbf{\delta }\mathbf{-}\mathbf{\beta }\mathbf{)}$ …… (4)

Determine the value of the instantaneous power absorbed.

Write the equation of the instantaneous current and the voltage.

$$\begin{gathered} \mathrm{v}\left(\mathrm{t}\right)=400\cos \left(\mathrm{\omega t}-{60}^0\right) \\ \mathrm{i}\left(\mathrm{t}\right)=100\cos \left(\mathrm{\omega t}+{10}^0\right) \end{gathered}$$

Substitute the values in equation (1) and solve as,

$$\begin{gathered} \mathrm{p}\left(\mathrm{t}\right)=400\cos \left(\mathrm{\omega t}-{60}^0\right)100\cos \left(\mathrm{\omega t}+{10}^0\right) \\ =\frac{1}{2}\left(400\right)\left(100\right)\left[2\cos \left(\mathrm{\omega t}-{60}^0\right)\cos \left(\mathrm{\omega t}+{10}^0\right)\right] \\ =\frac{1}{2}\left(400\right)\left(100\right)\left[2\cos \left(\mathrm{\omega t}-{60}^0+\mathrm{\omega t}+{10}^0\right)+\cos \left(\mathrm{\omega t}-{60}^0-\mathrm{\omega t}+{10}^0\right)\right] \\ =20000\left[\cos \left(2\mathrm{\omega t}-{50}^0\right)+\cos \left(-{70}^0\right)\right] \end{gathered}$$

Solve further as,

$\mathrm{p}\left(\mathrm{t}\right)=20000\left[\cos \left(2\mathrm{\omega t}-{50}^0\right)+0.342\right]$

Therefore, the expression for the instantaneous power is

$\mathrm{p}\left(\mathrm{t}\right)=20000\left[\cos \left(2\mathrm{\omega t}-{50}^0\right)+0.342\right]$.

Determine the value of the real power.

The maximum voltage and current are,

$$\begin{gathered} {\mathrm{V}}_{\mathrm{m}}=400\mathrm{V} \\ {\mathrm{I}}_{\mathrm{m}}=100\mathrm{A} \end{gathered}$$

The rms voltage and current are,

$$\begin{gathered} \mathrm{V}=\frac{400\mathrm{V}}{\sqrt{2}} \\ \mathrm{I}=\frac{100\mathrm{A}}{\sqrt{2}} \end{gathered}$$

Substitute the values in the equation (2) and solve as,

role="math" localid="1655120195415" $$\begin{gathered} \mathrm{P}=\left(\frac{400\mathrm{V}}{\sqrt{2}}\right)\left(\frac{100\mathrm{A}}{\sqrt{2}}\right)\cos \left(-{60}^0-{10}^0\right) \\ =\left(\frac{400}{\sqrt{2}}\right)\left(\frac{100}{\sqrt{2}}\right)\cos \left(-{60}^0-{10}^0\right) \\ =6840.4\mathrm{W} \end{gathered}$$

Therefore, the value of the real power is$6840.4\mathrm{W}$ .

Determine the value of the reactive power.

Substitute the values in the equation (3) and solve as,

$$\begin{gathered} \mathrm{P}=\left(\frac{400\mathrm{V}}{\sqrt{2}}\right)\left(\frac{100\mathrm{A}}{\sqrt{2}}\right)\cos \left(-{60}^0-{10}^0\right) \\ =\left(\frac{400}{\sqrt{2}}\right)\left(\frac{100}{\sqrt{2}}\right)\sin \left(-{60}^0-{10}^0\right) \\ =-18793.85\mathrm{VAR} \end{gathered}$$

Therefore, the value of the reactive power is $-18793.85VAR$.

Determine the value of the power factor.

Substitute the values in the equation (4) and solve as,

$$\begin{gathered} \mathrm{PF}=\cos \left(-{60}^0-{10}^0\right) \\ =\cos \left(-{70}^0\right) \\ =0.342 \end{gathered}$$

Since, current leads the voltage so the power factor is $0.342$ leading.

Therefore, the power factor is $0.342$ leading.