Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

In an L-R-C series circuit, R = 300 Ω, XC = 300 Ω and XL = 500 Ω. The average electrical power consumed in the resistor is 60.0 W. (a) What is the power factor of the circuit? (b) What is the rms voltage of the source?

Short Answer

Expert verified

a) Power factor of the circuit is 0.832

b) The rms value of voltage is 161.17 V.

Step by step solution

01

Concept of impedance, power factor and average power

Impedance of a circuit is the total resistance offered by the circuit.Denoted by Z, the formula for impedance of a LCR circuit is Z=R2+XL-XC2, where, R is the resistance,XL is the inductive reactance,role="math" localid="1664194997760" XC is the capacitive reactance.

Power factor of a circuit is defined as the ratio of the power across the resistor and the power consumed by the circuit. For series combination, the power factor can also be defined as the ratio of the resistance to the impedance of the circuit.

Average power across the resistor in an A/C circuit is given by, Pavg=Irms2R

02

Calculation of impedance of the circuit and power factor

Given that

R=300ΩXL=500ΩXC=300Ω

Therefore the impedance can be calculated as-

Z=3002+500-3002=360.56Ω

Now, power factor can be calculated as-

RZ=300360.56=0.832

Hence, power factor of the circuit 0.832

03

Calculation of rms value of voltage

The average power across the resistor is given by, Pavg=Irms2R, therefore,

Irms=PavgR=60300=0.447A

Now,Vrms=IrmsZ , therefore,

Vrms=0.447×360.56=161.17V

Hence, the rms value of voltage is 161.7 V.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Questions: A conductor that carries a net charge has a hollow, empty cavity in its interior. Does the potential vary from point to point within the material of the conductor? What about within the cavity? How does the potential inside the cavity compare to the potential within the material of the conductor?

An emf source with E = 120 V, a resistor with R = 80.0 Ω, and a capacitor with C = 4.00 µF are connected in series. As the capacitor charges, when the current in the resistor is 0.900 A, what is the magnitude of the charge on each plate of the capacitor?

A 5.00-A current runs through a 12-gauge copper wire (diameter

2.05 mm) and through a light bulb. Copper has8.5×108free electrons per

cubic meter. (a) How many electrons pass through the light bulb each

second? (b) What is the current density in the wire? (c) At what speed does

a typical electron pass by any given point in the wire? (d) If you were to use

wire of twice the diameter, which of the above answers would change?

Would they increase or decrease?

In the circuit shown in Fig. E26.18,ε=36.V,R1=4.0Ω,R2=6.0Ω,R3=3.0Ω(a) What is the potential difference Vab between points a and b when the switch S is open and when S is closed? (b) For each resistor, calculate the current through the resistor with S open and with S closed. For each resistor, does the current increase or decrease when S is closed?

Electrons in an electric circuit pass through a resistor. The wire on either side of the resistor has the same diameter.(a) How does the drift speed of the electrons before entering the resistor compare to the speed after leaving the resistor? (b) How does the potential energy for an electron before entering the resistor compare to the potential energy after leaving the resistor? Explain your reasoning.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free