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Expert-verified 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 $$1.00\;k\Omega$$ resistor, a $$150\;\mu H$$ inductor, and a $$25.0\;nF$$ capacitor. (a) Find the power factor at $$f = 7.50\;Hz$$. (b) What is the phase angle at this frequency? (c) What is the average power at this frequency? (d) Find the average power at the circuit's resonant frequency.

(a.)The power factor is $$0.0011781$$.

(b.) The phase angle at this frequency is $${89.9325^\circ }$$.

(c.) The average power at this frequency is $$0.0011781{P_{\max }}$$.

(d.) The average power at the circuit's resonant frequency is $$0.0011781{P_{\max }}$$.

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.

The electrical energy is absorbed by the resistor in the process of acting as a barrier to the flow of electricity by lowering the voltage, and it is released as heat.

## Step 2: Given Information

• The resistance value: $$1.00\;k\Omega$$
• The inductance value: $$150\;\mu H$$
• The capacitance value: $$25.0\;nF$$
• The frequency value: $$f = 7.5Hz$$

## Step 3: Find 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 $$f = 7.50\;Hz$$ which means we will have,

\begin{aligned} {X_C} &= \frac{1}{{2\pi \times 7.5{\rm{ }}Hz \times 80.0\;nF \times {{10}^{ - 6}}\;F}}\\ &= \frac{1}{{2\pi \times 120{\rm{ }}Hz \times 8.00 \times {{10}^{ - 6}}\;F}}\\ &= 848826\;\Omega \end{aligned}

\begin{aligned} {X_L} &= 2\pi \times 7.5\;Hz \times 100\;\mu H\left( {\frac{{{{10}^{ - 6}}\;H}}{{1\;\mu H}}} \right)\\ &= \;0.00471\;\Omega \end{aligned}

As a result, the reactance will be

\begin{aligned} Z &= \sqrt {{{1000}^2} + {{\left( {0.00471\;\Omega - 848826\;\Omega } \right)}^2}} \\ &= 848827\;\Omega \end{aligned}

As a result, the power factor will be

\begin{aligned} \cos \phi &= \frac{R}{Z}\\ &= \frac{{1000}}{{848827}}\\ &= 0.0011781\end{aligned}

Hence, the result is obtained as $$0.0011781$$.

## Step 4: Find the phase angle at this frequency

(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 \\ \Rightarrow \phi &= \arccos {\rm{ }}k\end{aligned}

As a result, our angle will be in degrees.

\begin{aligned} \phi &= \arccos 0.0011781\\ &= {89.9325^\circ }.\end{aligned}

Therefore, the phase angle at this frequency is $${89.9325^\circ }$$.

## Step 5: Find the average power at this frequency and the average power at the circuit's resonant frequency

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 \\ &= 0.0011781{P_{\max }}\end{aligned}

Therefore, the average power is $$0.0011781{P_{\max }}$$.

## 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 \\ &= 0.0011781{P_{\max }}\end{aligned}

Hence, the average power is $$0.0011781{P_{\max }}$$. ### Want to see more solutions like these? 