# A level physics Electricity question (5)

For this question the answer on the ms says Because the battery has to provide more energy/ power’ but I don’t understand how that answers the question
(edited 1 month ago)
Basically since the motor is doing work to lift smth up (ie doing smth that requires energy)
that energy has to come from somewhere so comes from the chemical store of the battery - ie "the battery has to provide more energy"
V = E/C, more energy -> more V -> V=IR -> greater I (resistance is constant)

tbh I would never just say "the battery has to provide more energy" I would always explain that in more detail, but that is I guess all of what the markscheme wants
hope that helps you
thank you thank you thank you ! that makes so much sense and your explanation was so easy to understand
Original post by mosaurlodon
Basically since the motor is doing work to lift smth up (ie doing smth that requires energy)
that energy has to come from somewhere so comes from the chemical store of the battery - ie "the battery has to provide more energy"
V = E/C, more energy -> more V -> V=IR -> greater I (resistance is constant)
tbh I would never just say "the battery has to provide more energy" I would always explain that in more detail, but that is I guess all of what the markscheme wants
hope that helps you

actually quick question, how do you know the resistance remains constant, does the internal resistance not change even when the battery has more current going through it ?
Original post by 1234kelly
For this question the answer on the ms says Because the battery has to provide more energy/ power’ but I don’t understand how that answers the question

Original post by 1234kelly
thank you thank you thank you ! that makes so much sense and your explanation was so easy to understand

A word of warning of the answer that you think you have understood.

If you see the circuit diagram, the motor “receives” the full emf of the battery (assume that the battery is ideal) and based on the “explanation in post number #2 or reply 2”, the emf of the battery should increase but this does not happen in the real world.

A way of thinking of “how” the battery provides more power is when we first connect a fixed resistance resistor say R1 to a battery and the battery would transfer energy to the resistor R1. If we connect another resistor R2 of the same resistance in parallel to R1, the current in the battery increases and more power is drawn or transferred to the resistors.
Note that this is not how the motor works

A detailed answer would involve electromagnetic induction.
Oh yeah that is completely my bad
I didnt even realise that meant that the emf of the battery would increase
Sorry @1234kelly
thank you for your help !

-but if we increase the current in the battery, the emf and I am assuming the internal resistance is constant wouldn't the lost volts increase causing the applied voltage supplied to the motor to decrease. then assuming the resistance of the motor is constant if P= V^2/ R it would cause the power supplied to reduce ? but in order to increase the current in the battery to begin with we reduced the resistance so I am guessing my logic is flawed somewhere
-Could we actually go over how electromagnetic induction relates to this question please because I would never have guess that was related to this question. Here my attempt of relating electromagnetic induction to answer the question but it's filled with plot holes
> Are we saying that we need to increase the current ( i.e larger than 0.82) by e.g adding more resistors in parallel to the cell to reduce the overall resistance to create the larger induced emf of the battery to increase power output and by doing so providing more power to the motor that is required for it to increase it work done when lifting the load. But in this case doesn't increasing the current increase the current flow in the motor as well so the motor itself would have it's own larger induced emf needed to increase it's work done so in this case how is the battery relevant . But also the induced emf would only be for a split second as the induced emf rely on a change in magnetic flux which I am guessing wouldn't be long enough to lift any load

so sorry in advance for this long follow up question, electricity is really a weak topic of mine
(edited 1 month ago)
Original post by mosaurlodon
Oh yeah that is completely my bad
I didnt even realise that meant that the emf of the battery would increase
Sorry @1234kelly

omg please don't apologise thank you for trying to help in the first place plus I still learnt a few things from your answer
Original post by 1234kelly
thank you for your help !

-but if we increase the current in the battery, the emf and I am assuming the internal resistance is constant wouldn't the lost volts increase causing the applied voltage supplied to the motor to decrease. then assuming the resistance of the motor is constant if P= V^2/ R it would cause the power supplied to reduce ? but in order to increase the current in the battery to begin with we reduced the resistance so I am guessing my logic is flawed somewhere
....

This is not how it works when we are dealing with electromagnetic induction.

Original post by 1234kelly
...
-Could we actually go over how electromagnetic induction relates to this question please because I would never have guess that was related to this question. Here my attempt of relating electromagnetic induction to answer the question but it's filled with plot holes
...

Have a look at this link. It has what you need at A level. If there is anything unclear, you can always ask. Just be careful about the word emf in the writing, it sometimes just means back emf NOT emf of the battery.

https://phys.libretexts.org/Bookshelves/University_Physics/Book%3A_Introductory_Physics_-_Building_Models_to_Describe_Our_World_(Martin_Neary_Rinaldo_and_Woodman)/23%3A_Electromagnetic_Induction/23.03%3A_Back_EMF_in_an_electric_motor

Original post by 1234kelly
....
> Are we saying that we need to increase the current ( i.e larger than 0.82) by e.g adding more resistors in parallel to the cell to reduce the overall resistance to create the larger induced emf of the battery to increase power output and by doing so providing more power to the motor that is required for it to increase it work done when lifting the load. But in this case doesn't increasing the current increase the current flow in the motor as well so the motor itself would have it's own larger induced emf needed to increase it's work done so in this case how is the battery relevant . But also the induced emf would only be for a split second as the induced emf rely on a change in magnetic flux which I am guessing wouldn't be long enough to lift any load

so sorry in advance for this long follow up question, electricity is really a weak topic of mine

Again, this is not how the motor works. You are over-reading the example that I provided. The example does not explain why the current increases in the motor but provides you a way of thinking about how we can increase current in the battery of a circuit without increasing the voltage or emf of the battery.