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Q103PE

Expert-verifiedFound in: Page 864

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**An RLC series circuit has a \(2.50\;\Omega \) resistor, a \(100\;\mu H\)** **inductor, and an \(80.0\;\mu F\) capacitor. (a) Find the power factor at** \(f = 120\;Hz\)**. (b) What is the phase angle at** \(120\;Hz\)**? (c) What is the average power at** \(120\;Hz\)**? (d) Find the average power at the circuit's resonant frequency.**

(a.)The power factor at \(f = 120\;Hz\) is \(0.1498\).

(b.) The phase angle at \(120\;Hz\)is \(81.4^\circ \).

(c.) The average power at \(120\;Hz\)is \(0.1498{P_{\max }}\).

(d.) The average power at the circuit's resonant frequency is \(0.1498{P_{\max }}\).

**A passive electrical component called an inductor stores energy in the form of a magnetic field. An inductor is a wire loop or coil in its most basic form. The coil's inductance is proportional to the number of turns it has.**

**A capacitor (also called a condenser) is a two-terminal passive electrical component that stores energy electrostatically in an electric field. Practical capacitors come in a variety of shapes and sizes, but they all have at least two electrical conductors (plates) separated by a dielectric (i.e., insulator).**

- The resistance value: \(2.50\;\Omega \)
- The inductance value: \(100\;\mu H\)
- The capacitance value: \(80.0\;\mu F\)
- The frequency value: \(f = 120\;Hz\)

(a)

We are aware that the power factor is expressed as:

\(\cos \phi = \frac{R}{Z}\)

This means we'll have to figure out the circuit's reactance, \(Z\). We already know that the reactance is expressed as,

\(Z = \sqrt {{R^2} + {{\left( {{X_L} - {X_C}} \right)}^2}} \)

This means we'll need to figure out the capacitor and inductor's active resistances, which are provided as

\(\begin{aligned} {X_C} &= \frac{1}{{2\pi fC}}\\{X_L} &= 2\pi fL\end{aligned}\)

The frequency in our case is \(120{\rm{ }}Hz\) which means we will have,

\(\begin{aligned} {X_C} &= \frac{1}{{2\pi \times 120{\rm{ }}Hz \times 80.0\;\mu F\left( {\frac{{{{10}^{ - 6}}\;F}}{{1\;\mu F}}} \right)}}\\ &= \frac{1}{{2\pi \times 120{\rm{ }}Hz \times 8.00 \times {{10}^{ - 5}}\;F}}\\ &= 16.57\;\Omega \end{aligned}\)

\(\begin{aligned} {X_L} &= 2\pi \times 120\;Hz \times 100\;\mu H\left( {\frac{{{{10}^{ - 6}}\;H}}{{1\;\mu H}}} \right)\\ &= \;0.075\;\Omega \end{aligned}\)

As a result, the reactance will be

\(\begin{aligned} Z &= \sqrt {{{\left( {2.50\;\Omega } \right)}^2} + {{\left( {0.075\;\Omega - 16.57\;\Omega } \right)}^2}} \\ &= 16.68\;\Omega \end{aligned}\)

As a result, the power factor will be

\(\begin{aligned} \cos \phi &= \frac{R}{Z}\\ &= \frac{{2.5\;\Omega }}{{16.68\;\Omega }}\\ &= 0.1498\end{aligned}\)

Therefore, the power factor is** \(0.1498\)**.

(b) We know the value of the power factor, let's call it \(k\), and that it is the cosine of the desired angle.

As a result, we can identify the latter as

\(\begin{aligned} k &= \cos \phi \\\phi &= \arccos \left( k \right)\end{aligned}\)

As a result, our angle will be in degrees.

\(\begin{aligned} \phi &= \arccos \left( {0.1498} \right)\\ &= 81.4^\circ \end{aligned}\)

Therefore, the phase angle is \(81.4^\circ \).

c)

For this, we can expect the power to be

\(P = {U_{rms}}{I_{rms}}\)

However, we have no idea what the current or voltage values are. To conclude, one of these would be required. We can only predict that we will succeed.

\(\begin{aligned} {P_{rms{\rm{ }}}} &= {P_{\max }}\cos \phi \\ &= 0.1498{P_{\max }}\end{aligned}\)

Therefore, the average power is \(0.1498{P_{\max }}\).

d)

For this, we can expect the power to be

\(P = {U_{rms}}{I_{rms}}\)

However, we have no idea what the current or voltage values are. To conclude, one of these would be required. We can only predict that we will succeed.

\(\begin{aligned} {P_{rms{\rm{ }}}} &= {P_{\max }}\cos \phi \\ &= 0.1498{P_{\max }}\end{aligned}\)

Therefore, the average power is \(0.1498{P_{\max }}\).

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