What is the maximum safe current and voltage for a 140 ohms, 1/2 watt resistor?

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SUMMARY

The maximum safe current for a 140 ohm, 1/2 watt resistor is 59.76 mA, calculated using the formula I = sqrt(P/R). The maximum electromotive force (e.m.f.) that can be applied to this resistor without overheating is 8.366V, derived from the equation P = V^2/R. It is essential to understand that e.m.f. is synonymous with voltage, and the watt rating indicates the power limit before failure. Always verify calculations to ensure accuracy in electrical applications.

PREREQUISITES
  • Understanding of Ohm's Law (V = IR)
  • Knowledge of power formulas (P = IV, P = I^2R, P = V^2/R)
  • Familiarity with resistor specifications (watt rating, resistance value)
  • Basic electrical concepts (current, voltage, power)
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  • Study the relationship between power, voltage, and current in electrical circuits
  • Learn about resistor ratings and their implications in circuit design
  • Explore advanced topics in electrical engineering, such as circuit analysis techniques
  • Research the historical context and evolution of electrical terminology, including e.m.f. vs. voltage
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Electrical engineering students, hobbyists working with circuits, and anyone involved in designing or analyzing electronic components will benefit from this discussion.

Bobzombie
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have looked thought all my notes and books and can not find the formal on how to do this question. I'm slowly going out of my mind any help would be great

Calculate the maximum current a 140 ohms, 1/2 watt resistor can have flowing through it safely.
Enter answer: mA

Then the 2nd part of the question is

What is the maximum e.m.f. which may be applied to the resistor above without causing overheating?

Enter Part 2 answer: V
 
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Do you know the relationship between ohms, amps, volts and watts? If so, using the formulas you can get it.
 
re

P = I^2 * R

I = sqrt(P/R) = sqrt(0.5/140) = 59.7 mA
 
emf? I have not encounter this question before :confused:
 
I have a question almost 100% identical to this one...
I know the first part, but is part two simply E=IR ??

I just have never seen e.m.f. being used to refer to voltage... and can't find any help anywhere on it
 
emf stands for electromotive force. Same thing as voltage.

In the first part it tells you what is the watt rating of the resistor. Power is determined by P = I^2 *R or P= V^2 / R or V=I*R.

If you have the value of the resistor and the watt rating you can easily figure out the answer.

PART 2: V or E is the variable to be found. Since we only have P and R let us use the relationship P = V^2 / R
0.5W = V^2/ 140 ohms

if the watt rating of a resistor is 0.5W that means the resistor will only be able to handle so much power until it burns out. So let us complete this problem.
sqroot(0.5W*140 ohms) = V or E
V or E = 8.366V

So if you apply 8.366V, you are on the border line of the max voltage the resistor can take.( in reality a resistor's watt rating is given a 10-20% margin I believe).
If you divide 8.366V by 140 ohms you get 59.76 mA which the maximum current the resistor can handle. How do you know this is the max current? well use P=V*I to check.
 
thanks a lot for you help ^_^

I only asked if I should use E = IR because the first part is calculating current, and once you have that voltage is easy

just wish my textbook/teacher had told me that electromotive force is the same thing as voltage, haha
 
True, you can use the current you calculated. But if you made a mistake in getting the current, the error would carry on.

Always recheck your values to see if they satisfy the equations.
 
Maeode said:
thanks a lot for you help ^_^

I only asked if I should use E = IR because the first part is calculating current, and once you have that voltage is easy

just wish my textbook/teacher had told me that electromotive force is the same thing as voltage, haha

They're trying to phase out the term. Voltage isn't a force so EMF is a misnomer.
 

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