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College Physics (Urone)
Found in: Page 864
College Physics (Urone)

College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

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Short Answer

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 by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

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 \(0.0011781{P_{\max }}\).

Most popular questions for Physics Textbooks

The motor in a toy car is powered by four batteries in series, which produce a total emf of \(6.00V\). The motor draws \(3.00A\) and develops a \(4.50V\) back emf at normal speed. Each battery has a \(0.100\Omega \) internal resistance. What is the resistance of the motor?

A plug-in transformer, like that in Figure \(23.29\), supplies \(9.00{\rm{ }}V\) to a video game system.

(a) How many turns are in its secondary coil, if its input voltage is \(120{\rm{ }}V\) and the primary coil has \(400\) turns?

(b) What is its input current when its output is \(1.30{\rm{ }}A\)?

Suppose you have a supply of inductors ranging from 1.00 nH to 10.0 H , and capacitors ranging from 0.100 pF to 0.100 F . What is the range of resonant frequencies that can be achieved from combinations of a single inductor and a single capacitor?

The capacitor in Figure \(23.55\) (a) is designed to filter low-frequency signals, impeding their transmission between circuits. (a) What capacitance is needed to produce a \(100{\rm{ }}k\Omega \) reactance at a frequency of \(120{\rm{ }}Hz\)? (b) What would its reactance be at \(1.00{\rm{ }}mHz\)? (c) Discuss the implications of your answers to (a) and (b).

(a) An inductor designed to filter high-frequency noise from power supplied to a personal computer is placed in series with the computer. What minimum inductance should it have to produce a \(2.00{\rm{ }}k\Omega \) reactance for \(15.0{\rm{ }}kHz\) noise? (b) What is its reactance at \(60.0{\rm{ }}Hz\)?
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