Current and Resistance (with a flashback to work I think)

  • Thread starter Mjannell
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  • #1
Mjannell
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Hello,
Iam having difficulty with the following problem...

3. [GianPSE3 25.P.042.] A small immersion heater can be used in a car to heat a cup of water for coffee. Assume the manufacturer's claim of 60 percent efficiency. If the heater can heat 200 mL of water from 5°C to 95°C in 9.0 min, approximately how much current does it draw from the 12 V battery?
What is its resistance?

my approach is as follows...

I=P/V=(W/T)/V

Is this a reasonable approach? If so a gentle prod towards determineing the work would be most helpful, I'm having a case of rusty brain. :grumpy:
With current, the resistance isn't a problem.

Thank You!

Mike
 

Answers and Replies

  • #2
Doc Al
Mentor
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Start by figuring out how much power is needed to heat the water in the given time. Then, taking into account the efficiency, figure out how much electrical power the heater must use. Then you can use I = P/V to find the current. And V = IR to find the resistance.
 
  • #3
Mjannell
2
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Thanks, I got it sussed out.

sorry that it took so long for the gratitude, I'm only online once a week or so.

Mike
 

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