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Question: An electric motor runs off a 12V d.v. supply and has an overall efficiency of 75%. Calculate how much electric charge will pass through the motor when it does 90J of work.

Can someone tell me where I went wrong?

What I did:

First of all, I don't know what "when it does 90J of work" means so I just assumed that because the efficiency is 75%, the "good" work it does will be 0.75*90=67.5J. Therefore, Q=W/V=67.5/12=5.625=5.6C to 2 s.f.

The answer at the back is given as 10C but I'm not sure what I did wrong. However, I did figure out that if you divide by 0.75 (i.e. 90/0.75) instead of multiplying by 0.75 like I did, you get the right answer.

Can someone run me through in detail how I should think about problems like this? I wouldn't mind some background regarding efficiency as well since I don't know what it is.

Thanks!

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    $\begingroup$ This is not a physics homework help site (including check my work problems there), it is a site for physics concepts. $\endgroup$
    – Kyle Kanos
    Commented May 11, 2015 at 14:56
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    $\begingroup$ @KyleKanos: This is not homework. I don't see the problem with asking for help on a concept I don't understand (electrical efficiency) $\endgroup$
    – user45220
    Commented May 11, 2015 at 14:57
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    $\begingroup$ Your question is asking for us to solve the problem for you. I fail to see how you can claim it is anything otherwise. If you had meant to ask about efficiency, go ahead and ask that, rather than ask us to do your homework. $\endgroup$
    – Kyle Kanos
    Commented May 11, 2015 at 15:03
  • $\begingroup$ @KyleKanos: I asked why my solution is wrong. I didn't just post the problem and ask you to solve it... $\endgroup$
    – user45220
    Commented May 11, 2015 at 15:05
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    $\begingroup$ Check the faq on meta.physics.se, it's stated there. We deal with concepts, not homework problems. If you want to understand electrical efficiency, just ask what's confusing you and forget about the numerics & problems. $\endgroup$
    – Kyle Kanos
    Commented May 11, 2015 at 15:11

1 Answer 1

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The "work" is the useful rotational pull you get from the motor, and excludes any wasted heat. If only 75% of the energy going in comes out as work, then you need to put in 90/0.75J of energy to get 90J of work out.

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  • $\begingroup$ Thanks so much! So in V=W/Q the "W" always stands for the useful work done by the motor? The book defines it as "The potential difference across a component is 1V when you convert 1J of energy moving 1C of charge through the component" and I'm failing to see the connection between the "W" as defined in this definition, and the "W" you defined (useful work done by motor) $\endgroup$
    – user45220
    Commented May 11, 2015 at 15:04
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    $\begingroup$ V=W/Q is measuring the amount of electrical energy going in to the motor. The phrase "how much electric charge will pass through the motor when it does 90J of work" is referring to 90J of work done by the motor. They are different measurements, but both measured in Joules. $\endgroup$
    – Simon B
    Commented May 11, 2015 at 15:43

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