Calculating RMS and Peak Values of 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??

 

[OlderDan]

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

 

[xanthym]

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.


~~

 

[suf7]

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

 

[suf7]

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??