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Q103PE

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Found in: Page 864

College Physics (Urone)

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 $$0.1498{P_{\max }}$$.

See the step by step solution

Step 1: Concept Introduction

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).

Step 2: Given Information

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

Step 3: Calculate the power factor

(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 $$0.1498$$.

Step 4: Find the phase angle

(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 $$81.4^\circ$$.

Step 5: Find the average power

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 $$0.1498{P_{\max }}$$.

Step 6: Find the average power at the circuit's resonant frequency.

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 $$0.1498{P_{\max }}$$.