# Resistive heating formula?

1. Dec 4, 2008

### chitrageetam

Hi
Heat generated by the conductor because of DC is i*i*R
where i is the current and R is the resistance.

Does anybody know the heat generated by the same conductor for an AC?

For example frequency 5kHz, DC resistance-20 ohms, RMS current is 3A..
How do we calculate this?
I do not think that the DC formula apply here.

Thanks

2. Dec 5, 2008

### f95toli

The RMS value of an alternating current is by definition the value that gives the same amount of heating in a resistor as a dc current of the same value.
Hence, the same formula applies.

If the current is sinnusoidal ($$i(t)=i_0 \sin \omega t$$, (which I assume is the case here) the RMS value is simply given by $$i_0/\sqrt{2}$$ where $$i_0$$ is the amplitude of the current.

Note that multimeters always show the RMS values of current and voltage.

3. Dec 5, 2008

### chitrageetam

Hi Toli

Thanks but.. this formula does not account for eddy current / induction
heating due to the high frequency.
If we have to use RMS value, then heating for 1kHz will be same as
heating for 5 kHz..which is not true in reality..higher frequency gets it more hot.
I think the formula must be frequency dependent.

Another thing is..if the conductor is wound like an inductor or solenoid..
what is the formula to calculate the heating --is there any different
formula for induction heating?

Thanks

4. Dec 5, 2008

### Redbelly98

Staff Emeritus
5 kHz is not really considered "high frequency". You'd probably have to get at least into the MHz range before you'd have to worry about stray inductances and capacitances.

For a definitive answer, you'd have to know just what the inductance and resistances are in the circuit. As long as

$$2 \pi f L \ << \ R$$

then inductive effects can be ignored.

5. Dec 7, 2008

### Phrak

What kind of resistor are you using? Is it a ceramic nichrome type? As Redbelly says, the inductance is not that great in value at 5Khz, but if it's wire wound, it's easily calculated. But inductance doesn't lead to much inductive heating through eddy currents at that frequency. If anything it reduces the current a small amount through inductive reactance. But you claim a given current rather than voltage...

6. Dec 13, 2008

### Gnosis

Hello chitrageetam,

Ultimately, Ohm’s Law applies to every closed electrical circuit without exception, whether it’s a DC or an AC circuit. DC circuits are typically the most straightforward in which to apply Ohm’s Law, as they possess no frequency component however, the operating characteristics of the particular components remain a consideration.

AC circuits tend to be more complicated due to operating characteristics of specific components, the manner in which they are configured, and commonly require that the circuit’s “impedance” be resolved in order to apply Ohm’s Law to derive power consumption or derive current based upon applied AC voltages at a given frequency. The “impedance” of a circuit (“impedance” indicating that it’s an AC circuit) essentially refers to the equivalent of a DC circuit’s resistance loading.

If the 20 ohm resistor you’re using is a typical carbon resistor and there are no additional components in the circuit to consider, then it makes no difference whether you’ve applied a 60 Hz or a 5 KHz AC voltage. All you need to do is calculate the RMS (Root Means Square) voltage of the applied AC voltage (no need to consider its frequency). Once you have the RMS voltage, simply apply that RMS voltage as though it were a DC voltage per Ohm’s Law.

Typically, multiplying the peak of the applied AC voltage by .707 is close enough to derive the RMS voltage. This RMS voltage performs the equivalent work of a Direct Current of the same voltage meaning, 120 Volts RMS = 120 Volts DC in the potential work that they can perform.

NOTE: Anytime an RMS voltage is mentioned, it is referring to an AC voltage, which is equivalent to the same voltage in DC.

In U.S. wall outlets, the 120 VAC has a peak voltage of 170 volts (with its peak-to-peak voltage being 340 volts) therefore, .707 x 170 volts = 120 Volts RMS.

Once you’ve calculated the RMS voltage, you could simply calculate the power dissipation via Ohm’s Law:

P = (E)^2 / R = (60 Volts RMS)^2 / (20 ohms) = 180 watts

You could easily derive the current:

I = E / R = (60 Volts RMS) / 20 ohms = 3 amps

I hope you found this somewhat helpful.

7. Jun 24, 2010

### huma ozair

8. Jun 24, 2010

### Integral

Staff Emeritus
Probably not. Getting the amount of heat generated is easy, arriving at a actual temperature is very hard. There are many factors to be considered, like size of the wire, insulation on the wire (surprisingly more insulation can mean lower temperatures), environmental factors are huge, what is the ambient temperature, is there air flow. That is just an example of the factors to be considered.

The best way to find the temperature is to measure it.