Calculating Maximum Permissible Current

1. Aug 13, 2009

licklecee

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

B) What is the maximum permissible current which can flow through a 1 watt resistor colour coded Brown, Black Orange, Gold ?

C ) what is the highest voltage that can be applied to the resistor in part ( B ) ?

2. Relevant equations

P = I^2 * R

3. The attempt at a solution

P = 1w
R = 10,000

I = sqrt(P/R)
I = sqrt(1/10,000)
I = sqrt 0.00001
I = 0.01 Ma

Variance Factor :

Gold = 5%

0.01 * 1.05 = 0.0105Ma

Maximum Current = 0.0105Ma

Can someone please double check this for me ? i'm struggling abit with this problem. Thank you!!

2. Aug 13, 2009

skeptic2

1. Recheck sqrt(1/10000) = sqrt(0.00001).

2. Why are you expressing current in mA?

3. Yes the tolerance is 5% but that should not be used to calculate maximum current. (The 5% refers to the resistance, so the resistance can vary from 9500 ohms to 10500 ohms. This means that in order to dissipate 1 W, the 9500 ohm resistor would require sqrt(1/9500) amps or 10.26 mA and the 10500 ohm resistor would require sqrt(1/10500) amps or 9.76 mA.) Note that a 5% change in resistance results in approximately 2.5% change in current.

3. Aug 13, 2009

licklecee

oh okay,

1.so sqrt 0.00001 = 3.16

2. Would it be 3.16 amps then ?

3. Okay so is it correct to say that to calculate maximum current, i need to calculate the maximum resistance, using the tolerance of in this case 5 percent,. Then repeat the previous formulae I = sqrt(P/R) with the new resistance.

I = sqrt 1/10500
I = sqrt 9.523
I = 9.76 Ma

Thank you for your help. I often get mixed up with the units and make very basic mistakes, something i need to work on thanks..

4. Aug 13, 2009

skeptic2

1. recalculate sqrt(1/10000) and watch your decimal places.

2. No.

5. Aug 13, 2009

licklecee

Also to calculate the maximum voltage, would it be correct to say ...

V=IR
V = 9.76*9500 ( 9500 because that is the lowest resistance point )
V = 0.00976*9500
V = 92.72 volts

Thank you