# What is the minimum wattage rating the resistor should have?

1. Oct 13, 2014

### zug

1. The problem statement, all variables and given/known data

The question "What is the minimum wattage rating the resistor should have?" comes after a previous problem I did to determine the maximum and minimum current when given a resistors tolerance. Here are all the values from that problem, some aren't really important but want to show all that I have. In the problem I used the Minimum Current as my value for I:

$$V = 7.0V$$
$$R = 82 \Omega +/- 5%$$
Max. Resistance = 86 ohms
Min. Resistance = 78 ohms
Minimum Current = 81.4 A
Maximum Current = 89.7 A

2. Relevant equations
1.$$P = IV$$
2. $$P = \frac{V^{2}}{R}$$
3. $$P = {I^{2}}{R}$$

3. The attempt at a solution

What I'm not sure about if should I use the 1st, 2nd or 3rd equation?
When I used the 1st one I got $$(81.4 A) \times (7.0 V) = 569 W$$
If I use the 2nd one I get $$\frac{7.0^{2}}{86} = 569 mW$$
For the third I get: $$(81.4^2) \times (86) = 569 kW$$

Last edited: Oct 13, 2014
2. Oct 13, 2014

### phinds

The point of determining a minimum wattage is to assure that a resitor won't burn out. Given that, do you really think it is the right thing to do to use the MINIMUM current?

3. Oct 13, 2014

### zug

So maybe use the maximum current value 89.7 A and the minimum resistance value 78 ohms? Once I did this I get 628W, 628 kW and 628 mW. I'm still unsure of what formula to use.

4. Oct 13, 2014

### Matterwave

You should check your current values...those looks WAY too high considering the voltage and resistance...

5. Oct 13, 2014

### zug

OMG, the current values are actually suppose to be in mA!!!! So then, once I used the value (89.7 mA) in each equation it simply gives me the SAME value, not matter which equation I use. Thanks for pointing out the current values, I kind of feel dumb now.

Last edited: Oct 13, 2014
6. Oct 13, 2014

### Matterwave

Yeah, all 3 equations should give you the same number if you use the correct current, voltage, and resistance, but Phinds is right, you probably shouldn't be using the minimum current to do this calculation.