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# finding the power

## 5 posts in this topic

A.The battery has 7.5 V output voltage and 19 Ohm internal resistance. The battery is used to boot a small electric motor with 155 Ohm resistance. Find the power delivered to the motor.

so my first attempt was 7.5/19 =.394737A

then .394737^2*155 = 24.1517W

that was wrong so i I'm thinking this but I have only one try left:

7.5/(19+155) = .043103A

.043103^2*155 = .287976W

B.Two batteries with 2.8 voltage and 9.5 Ohm internal resistance each are connected in series. This battery pack is used to boot a small electric motor with 270 Ohm resistance. Find the power delivered to the motor.

This second question i did the same as the first but only adding the voltage = 2.8 +2.8 = 5.6 and adding the internal resistance R = 19

the 5.6/19 = .294737

.294737^2*270 = 23.4548W

And I also have only one try left with this guess:

5.6/(19+270) = .019377

.01937^2*270 = .101378W

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[ 7.5V/(19+155) ]^2 * 155 is right for A. Alternatively you could have used a voltage divider, but this works the same way in simple logic.

For B, [ (2.8+2.8)/(9.5+9.5+270) ]^2 * 270 is right. Too lazy to work out the math. Again, you could have used a voltage divider, but using I^2 R also works.

(Man, I miss doing basic circuits.)

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Thanks.

I'm having trouble with this one too:

The battery is made by connecting 6 unit cells in series and then by connecting 8 of these series cells in parallel. Each unit cell has the voltage of 1.25 V and the internal resistance if 2.2 Ohm. What is the maximum power that can be delivered to the load by this battery?

My first try is this:

V(series) = 6*V

R(series) = 6*R

R(total) = 1/[8/(6R)]

so then P = V^2/R(total) = 48*V^2*R = 165W

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(Man, I miss doing basic circuits.)

Looking at this brings back painful memories of higher level electrical classes. :cry:

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Man, I love electrostatics, current electricity, and magnetism!

BTW, OP I have replied to the new thread you created for the second question.

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