IB Physics: Potential Dividers

[@tomtucker4034]

My life saver, thank you. You helped me so much as my teacher did not teach this topic and the book explains it wrongly.

[@tumulkumar5626]

Mr Doner, you have been my go to person for all the physics topics. Thank you very much for all these great videos. As I am officially done with IB Physics, I will now be unsubscribing to this channel; but I will spread the word about this amazing channel to my juniors.

[@swasthikas7652]

Why unsubcribe?

[@aviationgio]

Why unsubscribe?

[@Piritiup]

you are a life saver

[@denzeltracy]

+1

[@jadkhoury6983]

At 13:18 , How did you know that the load resistor (2000 ohms) will be in parallel and not in series with the variable resistor (2 ohms)?

[@donerphysics]

We are not putting anything in series with the rheostat, we connect the load across.

[@jadkhoury6983]

@@donerphysics oh thank you!

[@piperbrooke2957]

Hello! Amazing video, but I just had one question. When calculating voltage across a load resistor with a resistance similar to that of the potentiometer, why would you need to find the equivalent resistance of the load and bottom? Isn't voltage the same across parallel branches, so I could just calculate the proportion of voltage across the bottom resistor?

[@donerphysics]

You are correct to say that the parallel resistors will have the same voltage across them, however, to calculate the proportion of the voltage, you need to determine the parallel resistance of the load and bottom of rheostat. If the load resistor is much larger than the equivalent resistance, this will simply equal the contribution from the rheostat.

[@ayanoaman3179]

Thank you for your efforts :) your explanation is very clear and helpful ❤

[@alaaosama9639]

such a lifesaver !!!

[@donerphysics]

Glad you think so.

[@pushpamahadevan7242]

A very helpful video. Thank you.

[@sondrewikberg3534]

Is output voltage the voltage across the load resistor?

[@donerphysics]

Essentially, yes. If you watch the video on internal resistance, you will see that not all the battery voltage will appear across the load because there is internal resistance within the battery.

[@nnivetha1986]

Thanks for the video! For the problem in 11:26, your method was amazing but is there a shortcut because I feel like I would spend too much time figuring out just one question! Could I use this logic: since X breaks, no current flows through Z (since Z and X are in series) so Z's brightness decreases. Since Z's brightness decreases, Y's brightness increases (since Y and Z are in series). Is this logic accurate?

[@donerphysics]

If X breaks current will still flow through Z.

[@thorchh]

thank you so much

[@donerphysics]

You're welcome!

[@muskaan4168]

thanks a lot

[@pubgplayer1720]

I live your videos, but I like them even better on 1.5 x speed :)

[@donerphysics]

Many do this .... makes me sound like a chipmunk though.

[@audrikachowdhury4316]

hi sir, what do you mean by load voltage or loads resistor

[@donerphysics]

The voltage across the load resistor. The load is really just the "thing" connected to the battery.