A 1.0 kΩ resistor is rated ½ W. (It will bedestroyed if the rate of heat dissipation is greater than ½ W.)?

A 1.0 kΩ resistor is rated ½ W. (It will be destroyed if the rate of heat dissipation is greater than ½ W.) What is the maximum voltage you can apply to this resistor without risking damage to it?

This is the only one I couldn't get off the worksheet, of course its the last one. Tried a bunch of stuff manipulating the P = W/t and the P = IV, V = IR but still couldn't get it and don't wanna lose any more sleep from it.

3 Answers

  • 10 years ago
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    Yeah, it's a little tricky. You need to combine the P=VI and V=IR formula to create the one you need.

    You have a power P (0.5 W), and a resistance, R (1kΩ) and you want to find a voltage, V.

    That means you want to get rid of the current, I, term from the equations.

    Rearrange V = IR to give you I=V/R

    Now substitute V/R for I in the P= VI formula, giving you P = V^2/R

    You want V as the subject, though, so multiply both sides by R to give PR=V^2

    Then take the square roots of both sides V = sqrt(PR) = sqrt (0.5*1000) = 22.36 volts

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  • 4 years ago

    I'm gonna say Willie Stargell in '79, sure I was only 5 months old at the time of the World Series, but First Base has been pretty weak for the Buccos since then. I mean with the likes of John Milner, Jason Thompson, Sid Bream, Gary Redus, Orlando Merced, Kevin Young, Brian Hunter, Mark Johnson, Kevin Young part 2, Randall Simon, Daryle Ward, Sean Casey, Adam LaRoche, Garrett Jones and now Lyle Overbay following up Stargell, 1st Base has been garbage in Pittsburgh. Willie was the BEST! So Stargell was a 10, Overbay, not so much, maybe a 3 or so. That may be generous with the numbers he's put up.

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  • 10 years ago

    P = V^2/R

    V^2 = (0.5 watts)(1,000 ohms)

    V = 22.4 volts

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