Calculating RMS and Peak Values of Electric Fire

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SUMMARY

The discussion focuses on calculating the RMS and peak values of current and voltage for a one bar electric fire with a power rating of 1000W and an RMS voltage of 240V. The RMS current is determined using the formula RMS Current = RMS Power / RMS Voltage, resulting in 4.167A. The peak voltage is calculated as Peak Voltage = √2 * RMS Voltage, yielding 339.4V, and the peak current is found using Peak Current = √2 * RMS Current, resulting in 5.893A. These calculations are based on the principles of AC circuits for purely resistive loads.

PREREQUISITES
  • Understanding of RMS (Root Mean Square) calculations
  • Basic knowledge of AC (Alternating Current) circuit theory
  • Familiarity with Ohm's Law (P = IV)
  • Knowledge of peak versus RMS values in electrical systems
NEXT STEPS
  • Study the concept of power factor in AC circuits
  • Learn about the implications of resistive versus inductive loads
  • Explore advanced AC circuit analysis techniques
  • Investigate the effects of varying voltage levels on power consumption
USEFUL FOR

Electrical engineers, students studying electrical engineering, and anyone involved in designing or analyzing AC electrical systems will benefit from this discussion.

suf7
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A one bar electric fire has a power rating of 1000W (1kW). If the RMS voltage of the mains is 240V, find the RMS value of the current flowing in the fire and the peak values of current and voltage?

Can some one please help me??..Its a past exam question and i don't even know how to start it??..What do i do??
 
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suf7 said:
A one bar electric fire has a power rating of 1000W (1kW). If the RMS voltage of the mains is 240V, find the RMS value of the current flowing in the fire and the peak values of current and voltage?

Can some one please help me??..Its a past exam question and i don't even know how to start it??..What do i do??

What is a "one bar electric fire"? The answer depends on the power factor of the circuit. If it is just a resistive heater, the voltage and the current are in phase, and the calculation of current is simple ohms law. The power would just be P = IV
 
suf7 said:
A one bar electric fire has a power rating of 1000W (1kW). If the RMS voltage of the mains is 240V, find the RMS value of the current flowing in the fire and the peak values of current and voltage?

Can some one please help me??..Its a past exam question and i don't even know how to start it??..What do i do??
SOLUTION HINTS:
This problem is designed to distinguish between RMS and Peak values of Voltage, Current, and Power for a purely resistive load. We are given:
{RMS Voltage} = (240 V)
{RMS Power Rating for Resistive Load} = (1000 W)

For standard AC systems, we know that:
{RMS Power} = {RMS Voltage}*{RMS Current}
Thus, we can solve for {RMS Current}:
{RMS Current} = {RMS Power}/{RMS Voltage} = (1000 W)/(240 V)

Also for standard AC systems:
{Peak Voltage} = √2{RMS Voltage} = (1.414)*{RMS Voltage}
{Peak Current} = √2{RMS Current} = (1.414)*{RMS Current}
{Peak Power} = {Peak Voltage}*{Peak Current} =
= √2{RMS Voltage}*√2{RMS Current} =
= (2)*{RMS Voltage}*{RMS Current}
::: ⇒ {Peak Power} = (2)*{RMS Power}

Use above basic information to determine required solutions to this problem.


~~
 
Last edited:
xanthym said:
SOLUTION HINTS:
This problem is designed to distinguish between RMS and Peak values of Voltage, Current, and Power for a purely resistive load. We are given:
{RMS Voltage} = (240 V)
{RMS Power Rating for Resistive Load} = (1000 W)

For standard AC systems, we know that:
{RMS Power} = {RMS Voltage}*{RMS Current}
Thus, we can solve for {RMS Current}:
{RMS Current} = {RMS Power}/{RMS Voltage} = (1000 W)/(240 V)

Also for standard AC systems:
{Peak Voltage} = √2{RMS Voltage} = (1.414)*{RMS Voltage}
{Peak Current} = √2{RMS Current} = (1.414)*{RMS Current}
{Peak Power} = {Peak Voltage}*{Peak Current} =
= √2{RMS Voltage}*√2{RMS Current} =
= (2)*{RMS Voltage}*{RMS Current}
::: ⇒ {Peak Power} = (2)*{RMS Power}

Use above basic information to determine required solutions to this problem.


~~

Thanks a lot!..Using your help i worked out the following:
{RMS Current} = 4.167A
{Peak Voltage} = 339.4V
{Peak Current} = 5.893A

Do these values seem ok?...If they are ok could i ask another question??..it carries on from the question I've already asked?

Thanks
 
OlderDan said:
What is a "one bar electric fire"? The answer depends on the power factor of the circuit. If it is just a resistive heater, the voltage and the current are in phase, and the calculation of current is simple ohms law. The power would just be P = IV

Sorry i don't quite understand what you mean??
 

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