Question

Referring to Example 23.14, find the average power at \({\rm{10}}{\rm{.0}}\;{\rm{kHz}}\).

Short answer

The average power at\({\rm{10}}{\rm{.0}}\;{\rm{kHz}}\)is\({\rm{16}}{\rm{.0\;W}}\).

Step-by-step solution

Definition of capacitor

A capacitor is a two-terminal electrical component that may store energy in the form of an electric charge. It is made up of two electrical wires separated by a certain distance. The space between the conductors can be filled with the vacuum or a dielectric, which is an insulating substance.

Find the value of power

We have a circuit with a resistor of\({\rm{40}}\;\Omega \), an inductor of\({\rm{3}}\;{\rm{mH}}\), and a capacitor of\({\rm{5}}\;{\rm{\mu F}}\)that is fed at \({\rm{120\;V,10}}\;{\rm{kHz}}\).

The authority will be granted by

\(\begin{aligned} {\rm{P}} &= {\rm{UIcos}}\phi \\ &= \left( {\frac{{{{\rm{U}}^{\rm{2}}}}}{{\rm{Z}}}} \right){\rm{cos}}\phi \end{aligned}\)

That is, we must determine the impedance and the power factor.

The capacitive and inductive components are found as follows to determine the impedance:

\(\begin{aligned} {{\rm{X}}_{\rm{C}}} &= \frac{{\rm{1}}}{{{\rm{2\pi fC}}}}\\ &= \frac{{\rm{1}}}{{{\rm{2\pi }} \times {\rm{10}}\;{\rm{kHz}}\left( {\frac{{{{10}^3}\;Hz}}{{{\rm{1}}\;{\rm{kHz}}}}} \right) \times {\rm{5}}\;\mu {\rm{F}}\left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 6}}}}\;F}}{{1\;\mu {\rm{F}}}}} \right)}}\\ &= {\rm{3}}{\rm{.183}}\;\Omega \\{{\rm{X}}_{\rm{L}}} &= {\rm{2\pi fL}}\\ &= {\rm{2\pi }} \times {\rm{10}}\;{\rm{kHz}}\left( {\frac{{{{10}^3}\;Hz}}{{{\rm{1}}\;{\rm{kHz}}}}} \right) \times {\rm{3}}\;{\rm{mH}}\left( {\frac{{{{10}^{ - 3}}\;H}}{{{\rm{1}}\;{\rm{mH}}}}} \right)\\ &= {\rm{188}}{\rm{.5}}\;\Omega \end{aligned}\)

As a result, the impedance will be

\(\begin{aligned} {\rm{Z}} &= \sqrt {{{\rm{R}}^{\rm{2}}} + {{\left( {{{\rm{X}}_{\rm{L}}} - {{\rm{X}}_{\rm{C}}}} \right)}^{\rm{2}}}} \\ &= \sqrt {{{\left( {{\rm{40}}\;\Omega } \right)}^{\rm{2}}} - {{\left( {{\rm{188}}{\rm{.5}}\;\Omega - {\rm{3}}{\rm{.183}}\;\Omega } \right)}^{\rm{2}}}} \\ &= 189.6\;\Omega \end{aligned}\)

As a result, the power factor will be

\(\begin{aligned} {\rm{cos}}\phi &= \frac{{\rm{R}}}{{\rm{Z}}}\\ &= \frac{{{\rm{40}}\;\Omega }}{{{\rm{189}}{\rm{.6}}\;\Omega }}\\ &= 0.21\end{aligned}\)

Thus, the power is.

\(\begin{aligned} {\rm{P}} &= \frac{{{{\left( {{\rm{120}}\;{\rm{V}}} \right)}^{\rm{2}}}}}{{{\rm{189}}{\rm{.6}}\;\Omega }} \times 0.21\\ &= {\rm{16\;W}}\end{aligned}\)

Therefore, the average power at\({\rm{10}}{\rm{.0}}\;{\rm{kHz}}\)is\({\rm{16}}{\rm{.0\;W}}\).