A \(1500-\) W heater runs on \(120 \mathrm{V}\) ms. What is the peak current through the heater? (Wtorial: power in ac circuits)

Short Answer

Expert verified
Answer: The peak current through the heater is approximately 17.68 A.

Step by step solution

01

Determine the rms current

To find the rms current through the heater, use the power formula for AC circuits: $$ P = V_{rms} \cdot I_{rms} $$ Rearrange the formula to solve for rms current: $$ I_{rms} = \frac{P}{V_{rms}} $$ Substitute the given values for power and voltage: $$ I_{rms} = \frac{1500 \mathrm{W}}{120 \mathrm{V}} $$
02

Calculate the rms current

Divide the power by the voltage to find the rms current: $$ I_{rms} = \frac{1500 \mathrm{W}}{120 \mathrm{V}} = 12.5 \ \mathrm{A} $$ So, the rms current through the heater is 12.5 A.
03

Determine the peak current

Since we have the rms current, we can now find the peak current using the relationship between the rms and peak currents: $$ I_{peak} = \sqrt{2} \cdot I_{rms} $$ Substitute the given rms current value: $$ I_{peak} = \sqrt{2} \cdot 12.5 \ \mathrm{A} $$
04

Calculate the peak current

Multiply the rms current by the square root of 2 to find the peak current: $$ I_{peak} = \sqrt{2} \cdot 12.5 \ \mathrm{A} \approx 17.68 \ \mathrm{A} $$ The peak current through the heater is approximately 17.68 A.

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

A series \(R L C\) circuit has \(R=500.0 \Omega, L=35.0 \mathrm{mH},\) and $C=87.0 \mathrm{pF} .$ What is the impedance of the circuit at resonance? Explain.
A certain circuit has a \(25-\Omega\) resistor and one other component in series with a \(12-\mathrm{V}\) (rms) sinusoidal ac source. The rms current in the circuit is 0.317 A when the frequency is \(150 \mathrm{Hz}\) and increases by \(25.0 \%\) when the frequency increases to \(250 \mathrm{Hz}\). (a) What is the second component in the circuit? (b) What is the current at $250 \mathrm{Hz} ?$ (c) What is the numerical value of the second component?
A 40.0 -mH inductor, with internal resistance of \(30.0 \Omega\) is connected to an ac source $$\varepsilon(t)=(286 \mathrm{V}) \sin [(390 \mathrm{rad} / \mathrm{s}) t]$$ (a) What is the impedance of the inductor in the circuit? (b) What are the peak and rms voltages across the inductor (including the internal resistance)? (c) What is the peak current in the circuit? (d) What is the average power dissipated in the circuit? (e) Write an expression for the current through the inductor as a function of time.
(a) What is the reactance of a \(10.0-\mathrm{mH}\) inductor at the frequency \(f=250.0 \mathrm{Hz} ?\) (b) What is the impedance of a series combination of the \(10.0-\mathrm{mH}\) inductor and a \(10.0-\Omega\) resistor at $250.0 \mathrm{Hz} ?$ (c) What is the maximum current through the same circuit when the ac voltage source has a peak value of \(1.00 \mathrm{V} ?\) (d) By what angle does the current lag the voltage in the circuit?
A large coil used as an electromagnet has a resistance of \(R=450 \Omega\) and an inductance of \(L=2.47 \mathrm{H} .\) The coil is connected to an ac source with a voltage amplitude of \(2.0 \mathrm{kV}\) and a frequency of $9.55 \mathrm{Hz}$. (a) What is the power factor? (b) What is the impedance of the circuit? (c) What is the peak current in the circuit? (d) What is the average power delivered to the electromagnet by the source?
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