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

  • Thread starter Thread starter truettct
  • Start date Start date
  • Tags Tags
    Load Power
AI Thread Summary
To calculate the power delivered to the motor using Ohm's Law, first determine the total resistance by adding the internal resistance of the battery (0.15 Ohm) to the motor's resistance (285 Ohm), resulting in 285.15 Ohm. Next, calculate the current flowing through the circuit using the formula I = V/R, where V is the battery voltage (1.35 V). The power delivered to the motor can then be found using P = I^2 * RL. For two batteries in series, the output voltage doubles, allowing for a similar calculation with the new voltage.
truettct
Messages
8
Reaction score
0
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
 
Physics news on Phys.org
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.
 
Back
Top