Power puzzled

At a rough guess - your resistance reading is in error.

resistance is relative. You need to include impedance of the total cct in parallel.

p=ir

4500=970*I I=4500/970 4.6amps......

1/970 etc

total resistance is 323 ohms

more confusion?

Single phase or three phase star/y connected ?

Most home water heaters have 2 elements - and they do not operate simultaneously. Each one is 4.5 kW by itself. -- 240^2/4500 = 12.8 ohms

RE

To get 1.5 kW out of an element at 240 volts requires 6.25 amps. To draw 6.25 amps at 240 volts requires 38.4 ohms. You have a number wrong, somewhere, likely the resistance measurement. Ed

I bet he dropped a decimal point and these are 9.xx ohms. A water heater usually just uses one elememt at a time so they are all rated at the heater rating.

You have that backward or a lamp filiment would explode. Nichrome wire increases resistance as it heats up. I imagine he is really reading

9.7 ohms cold and that goes to 12.8 hot. Do the math, that works out to 4.5kw @ 240v. Water heaters usually only use one element at a time.

No I measured all 3 on the 200ohms scale of the meter. All came back with 97 ohms.

I'll go with the idea that the resistance will drop as the elements get warm.

Can't be 9.7 ohms as the current would be much too high.

snipped-for-privacy@aol.com wrote:

This is a spa bath heater. I think all the elements are on at once. I have traced the wiring back the to contactor so they all come on at once.

snipped-for-privacy@aol.com wrote:

probably your measurement - the heating element changes (increases) resistance as it heats up.

It may be time to replace the battery in your meter. In fact, get two new batteries - one for the meter and the other for the paragraph below. Proceed with the paragraph, after replacing the battery in your meter. A weak battery can cause erroneous resistance readings.

Would you like to test your theory? It's a simple test. First, be safe and kill the power to your heater. Install a 100 ohm resistor in series with 1 element and a new 9 volt battery. Measure the voltage across the 100 ohm resistor. If the elment is 97 ohms, you should get about 4.56 volts across the 100 ohm resistor. If it is 9.7, you should get about 8.2 volts across the 100 ohm resistor.

Finally, re-check the resistance of the element with your meter.

Ed

------------------- Since when is p=ir?? Also the apparent resistance of an element is 97 ohms, not 970. It appears that the ohmmeter reading is in error but his arithmetic is correct.

---------- An ohmmeter is often unreliable in making measurements on power circuits. At

240V, the effect of some surface film is often negligable (it breaks down and conducts well ) but at the voltage provided by a meter battery, it is an important factor adding to the apparent resistance as seen by the ohmmeter. You may be able to get around this by scratching the terminals of the element (likely copper) to get bare metal and then using your meter. If the results still seem off, then measure the current, as indicated below, at 240V. In addition, the hot resistance will be higher than that of a cold element so the chances are that the cold element resistance will be less than 38 ohms and rise to 38 ohms after it heats up.

If you do not have an ammeter I suggest that you put a 0.1 ohm, 5 watt ohm resistor in series with the element and apply 240V to the combination (measure it), and measure the voltage drop across the resistor. Multiply this by 10 to get the current.