Capicitor Discharge Curve (URGENT!)

What values do you get?
Initial current at start of graph = V/R (V is initial pd on capacitor=12V)
Current after 35ms is V/R where V is pd after 35ms. This V is 0.37 times the start value.

I recognise this question. I spoke to my physics teacher about it, and it turns out the markscheme is wrong.

V = IR, I = V/R which means 12V/16 = 0.75 A

The initial value is 750mA and the current at the time constant is around 275mA. you know the graph shape

How do you know that the V is 0.37 times the start value?

Because the graph is a reverse exponential graph, and according to the equation It = Io x e^(- time/time constant), at the time where t = time constant, it becomes It = Io x e^-1, which is the same as multiplying by 1/e, which is approximately 0.368.

Io is the initial current, which we now know to be 750mA,

So current at the time constant = 750 x 1/e = 750 x 0.368 = 276mA

I'll see if i can upload a graph

just in case :P as you can see my MS Paint skills are exquisite

attachment below (i hope)

Are you ready for the exam tomorrow?

I guess I'm pretty ready, although I still get a bit confused about electromagnetic induction sometimes As it turns out, the explanation for electromagnetic induction is just a model! So we might actually be learning wrong stuff hehe

good luck!

for this type of question, would you estimate the value of I when time is 35ms using the graph or do an actual working and multiply 750ms by 1/e?

That would make so much sense!
Thank you!

Although I don't get what you mean by the time constant being around 275mA? don't you mean that the current at 35ms is around 275mA?

Graph I vs t , where I is calibrated as 0, 20, 40, 60 .. should be calibrated as 0, 200,400,600 ..
and then all is well.