Calculating Maximum Permissible Current

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Discussion Overview

The discussion revolves around calculating the maximum permissible current and voltage for a 1 watt resistor with specific color coding. Participants explore the implications of resistor tolerance on current calculations and the relationship between power, current, and resistance.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant calculates the maximum current using the formula I = sqrt(P/R) and suggests a maximum current of 0.0105 mA, incorporating a 5% tolerance factor.
  • Another participant challenges the initial calculation, stating that the tolerance should not be used to calculate maximum current and provides alternative current values based on varying resistance (9500 ohms to 10500 ohms).
  • There is confusion regarding the calculation of square roots and unit conversions, with one participant incorrectly calculating sqrt(0.00001) as 3.16 and questioning if the current would then be 3.16 amps.
  • A later reply confirms the need to recalculate maximum current using the maximum resistance and provides a new current value of 9.76 mA based on the maximum resistance of 10500 ohms.
  • Another participant proposes a method to calculate maximum voltage using the lowest resistance value (9500 ohms) and arrives at a voltage of 92.72 volts.

Areas of Agreement / Disagreement

Participants express differing views on how to incorporate tolerance into the calculations, leading to multiple competing approaches for determining maximum current and voltage. The discussion remains unresolved regarding the correct application of tolerance in these calculations.

Contextual Notes

There are unresolved issues regarding the accuracy of square root calculations and unit conversions, as well as the proper application of tolerance in resistor specifications.

licklecee
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Homework Statement



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 ) ?


Homework Equations



P = I^2 * R



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!
 
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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.
 
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..
 
licklecee said:
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..

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

2. No.
 
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
 

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