Electrical Circuits: Power and Energy Calculations

[venom201]

Homework Statement

A student is asked to supply an electric motor with 1.0 mA of current at 3.0 V potential difference.
a. Determine the power to be supplied to the motor.
b. Determine the elctrical energy to be supplied to the motor in 60s.
c. Operating as designed, the motor can lift a .012 kg mass a distance of 1.0 m in 60s at a constant velocity.
d.To operate the motor, the student has available only a 9.0 v battery to use as the power source and 5 resistors: 1000 \Omega, 4000 \Omega, 4000\Omega, 5000\Omega, 10000\Omega.
How should the circuit be wired to power the motor.

Homework Equations

P= IV
efficiency = output/input

The Attempt at a Solution

a. (.001)(3)= .003 watts

b. Not too sure about this, I just multiplied the .003 by 60 to get .18 watts

c. mgh= (.012)(10)(1)= .12
So do I divide .12 by the answer from part b?, I feel like I'm missing a step in there.

d. For this I'm completely lost, I know the motor should be in parallel with another branch of the circuit but I'm not sure which resistors to use.

 

[Delphi51]

Welcome to PF!

a and b look good. There is no question in part c.
For d, I suggest a simple series circuit with the current from the battery flowing through a resistor on its way to the motor. Knowing the voltage drop across the resistor and the current, you can calculate the appropriate resistance. I'm sure you can see how to get that resistance from the collection of resistors you have available.

 

[gneill]

b should be in units of energy, not power. So joules, or volt-amp-hours, or watt-seconds, or some such.

 

[venom201]

Welcome to PF!

a and b look good. There is no question in part c.
For d, I suggest a simple series circuit with the current from the battery flowing through a resistor on its way to the motor. Knowing the voltage drop across the resistor and the current, you can calculate the appropriate resistance. I'm sure you can see how to get that resistance from the collection of resistors you have available.

Thanks for the welcome and sorry about part c not having a question, it asks for the efficiency of the motor if it can lift .012 kg 1 meter in 60 seconds. I came up with the change in potential energy of the mass which equals the work done by the motor and divided it by the energy supplied to get 66%@gneill, Thanks, it makes more sense that way now.

I'm still lost on part d though, if it is just a series circuit with one pathway for the current, the only way to get .001 Amperes of current would be with 9000 \Omega of resistance but that would also result in a total voltage drop of 12 V

 

[gneill]

I'm still lost on part d though, if it is just a series circuit with one pathway for the current, the only way to get .001 Amperes of current would be with 9000 \Omega of resistance but that would also result in a total voltage drop of 12 V

You'll want the motor to drop the same voltage as before (3.0V). So how much will the inserted resistor have to drop?

 

[venom201]

You'll want the motor to drop the same voltage as before (3.0V). So how much will the inserted resistor have to drop?

6 volts?

So do I use the 1000 ohm and 5000 ohm resistors in series with the motor and battery?

 

[gneill]

Sounds reasonable!

 

[venom201]

Wow, thanks for all the help guys!