Hey!

I built a circuit with a 5V power supply and a 300 ohms resistor. With a multimeter I did the measurements of resistance, voltage and current across the resistor. The values was:

R = 298 ohms
E = 4.94 V
I = 16 mA

So I made the calcs to obtain the power dissipated by the resistor:

P = EI = 4.94 * .016 = 79 mW
P = E²/R = 4.94² / 298 = 81.9 mW
P = I²*R = .016² * 298 = 76.3 mW

My question is: Why there are differences between the power results? It's because of the lack of precision in measurements, the increase of resistor's resistance when powered or a combination of both?

 

Hello,

The precision of the current is to low.
The current will be 4.94 Volts / 298 Ohms = 16.5771 mA.
When you do the calculations with this value the power will be 81.89 mWatt.

Bertus

 

Precision problems. I calculate 16.577 ma, but most meters don't go that far down in the digits. Neither do they measure closer than a tenth of an ohm.

 

So the problem is my HM-1000 multimeter.

 

You can spend big bucks for something that has 5 or 6 digits, but it usually doesn't matter unless you're servicing precision analog meters.

Most people (including me) get along with a $10 POS except for rare occasions.

 

Hello,

High precision multimeters do not have a hobby price:
http://www.newark.com/keithley/2000...2/dp/58T8718?MER=PPSO_N_P_EverywhereElse_None

We will have to do with that what is in our budget.

Bertus

 

You can spend big bucks for something that has 5 or 6 digits, but it usually doesn't matter unless you're servicing precision analog meters.

Most people (including me) get along with a $10 POS except for rare occasions.

Hello,

High precision multimeters do not have a hobby price:
http://www.newark.com/keithley/2000...2/dp/58T8718?MER=PPSO_N_P_EverywhereElse_None

We will have to do with that what is in our budget.

Bertus

Yes! I know. I was just justifying the lack of precision in my measurements. The HM-1000 is good enough for me.

 

Did you actually measure the 16mA current, meaning your meter was part of the circuit? That would introduce some series resistance and lower the current. If you have a second meter, you could measure the ∆V with the first meter in place.

 

Also, high precision meters take longer to get accurate measurements.