# Calculating RMS and Peak Values of Electric Fire

• suf7
In summary, this conversation is about finding the RMS and peak values of current and voltage for a one bar electric fire with a 1000W power rating and a 240V RMS voltage. The solution involves using basic information about standard AC systems to calculate the values, and the final values are 4.167A for RMS current, 339.4V for peak voltage, and 5.893A for peak current. The conversation also mentions the importance of considering the power factor of the circuit when determining the answer for a "one bar electric fire."
suf7
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??

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 ALOT!..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??

## 1. What is the difference between RMS and peak values in electric fire?

The RMS (Root Mean Square) value of an electric fire is the effective or average value of its alternating current. It is calculated by squaring the instantaneous values of the current, taking the mean, and then finding the square root. On the other hand, the peak value is the highest value that the current reaches in one cycle of its AC waveform.

## 2. How do you calculate the RMS value of an electric fire?

To calculate the RMS value of an electric fire, you need to take multiple instantaneous readings of the current over one cycle of its AC waveform. Square each of these readings, find the mean, and then take the square root of that value. The result will be the RMS value of the current.

## 3. Why is it important to know the RMS value of an electric fire?

Knowing the RMS value of an electric fire is important because it is the value that determines the power output of the device. It is also used to calculate the heat produced by the fire and to ensure the safety of the electrical system.

## 4. Can the peak value of an electric fire be higher than the RMS value?

Yes, the peak value of an electric fire can be higher than the RMS value. This occurs because the peak value represents the maximum value of the current during one cycle of its AC waveform, while the RMS value represents the average value.

## 5. How does the RMS value of an electric fire affect its efficiency?

The RMS value of an electric fire is directly related to its efficiency. A higher RMS value means a higher power output, which results in a more efficient electric fire. However, it is important to note that the efficiency of an electric fire also depends on other factors such as the design and materials used.

• Introductory Physics Homework Help
Replies
5
Views
1K
• Engineering and Comp Sci Homework Help
Replies
4
Views
6K
• Electrical Engineering
Replies
4
Views
694
• Engineering and Comp Sci Homework Help
Replies
1
Views
997
• Introductory Physics Homework Help
Replies
13
Views
2K
• Introductory Physics Homework Help
Replies
4
Views
763
• Engineering and Comp Sci Homework Help
Replies
8
Views
4K
• Introductory Physics Homework Help
Replies
5
Views
2K
• Introductory Physics Homework Help
Replies
16
Views
2K
• Introductory Physics Homework Help
Replies
5
Views
2K