Question: How do I calculate the power delivered to the motor using Ohm's Law?

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SUMMARY

The discussion focuses on calculating the power delivered to a motor using Ohm's Law with a battery output of 1.35 V and an internal resistance of 0.15 Ohm. The motor has a resistance of 285 Ohm. The correct approach involves determining the total resistance in series (0.15 Ohm + 285 Ohm) to find the current using Ohm's Law, followed by calculating power with the formula P = I²R. The second part of the problem involves repeating the calculation with two batteries in series, effectively doubling the voltage to 2.7 V.

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1. Homework Statement

The battery has 1.35 V output voltage and 0.15 Ohm internal resistance. The battery is used to boot a small electric motor with 285 Ohm resistance. Find the power delivered to the motor. What the power will be if we use two batteries of this type in series?



3. The Attempt at a Solution

I thought the formula to use was

PL=VB2-(RL/(Ri+RL)2)

Ri= internal resistance
VB2=output voltage
RL=resistance

When I use this formula my answer is .00639W which is incorrect.
And I'm not sure how to do the second part of the problem

Any help on what I am messing up on would be appreciated.

Thanks
 
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Don't look for complex formulae. This is pure Ohm's Law stuff.

The total resistance across the 1.35 Volts of the single battery is what?

(The internal resistance plus the motor resistance. They are in series.)

So, what current flows?

The motor has this current flowing in it and a known resistance, so what is the power delivered to the motor?

Now put another battery in series with the first one and do it again.
 

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