Exam Discussion!

[shroomer101]

I got 26V, may not be right tho.

yes i got 26V as well

okay this is making me feel better than how i did getting out of the exams

[KayKay]

Anyone pick up on the transformer having DC voltage? LOL completely blew right past me.

[Hiesenberg]

for the first electric power question, something like 10 V DC is supplied to a transformer whats the output?

I put 0 because an AC is needed to generate an output voltage. Right?

[KayKay]

for the first electric power question, something like 10 V DC is supplied to a transformer whats the output?

I put 0 because an AC is needed to generate an output voltage. Right?

Yep, that's correct.

[speedy]

3. 2 ms-1

To do this, do you take h and x as 0.2? Ie. halfway between the max's and min's of the oscillation?

[samroberts]

Yep got 0 V for the transformer question and 26 V for the light bulb question!
Sounds like that's the correct answer

[KayKay]

To do this, do you take h and x as 0.2? Ie. halfway between the max's and min's of the oscillation?

If I remember correctly, halfway between max and min of oscillation was an extension of 0.4 m. So 1.6=mgh + 1/2kx^2 + 1/2mv^2. h=x=0.4, sub and solve. But yeah, you do find the halfway point.

[Plitzer]

To do this, do you take h and x as 0.2? Ie. halfway between the max's and min's of the oscillation?

The halfway point is when the speed will be a maximum, yes, but the extension found from an earlier part is 0.8m. The value you found IS the extension as you are calculating from the unstretched length, so the 0.4m length of the unstretched spring doesn't apply.
Then the halfway point will be at x = 0.4m and you use conservation of energy principles

[Marrogi12]

I don't think time was a problem as most of the students in my class finished early , I finished 45 mins early ahah exactly at 11.00 ahah

[wanigara]

Can someone upload the paper pls...

[SeanC]

I think you guys are right with 26v but did something else. I assumed (probably incorrectly) that the power output of the power supply must be the same if you increased the voltage so you  would decrease the current. this led to something very complicated and i ended up using the quadratic formula to solve and find v to equal like 17.something i think. any chance of getting any marks for this? it was a 3 mark question

[chuck981996]

I don't think time was a problem as most of the students in my class finished early , I finished 45 mins early ahah exactly at 11.00 ahah

Yeah time was fine. I mean, I took the whole 2.5hours, but that's because I check as I go and go slowly hahaha

Did other people get 20Ω for the variable resistor when the voltage across the photodiode is 1V? Or maybe I'm thinking of the resistance in the coil… Everything is slipping away…

[Plitzer]

Yeah time was fine. I mean, I took the whole 2.5hours, but that's because I check as I go and go slowly hahaha

Did other people get 20Ω for the variable resistor when the voltage across the photodiode is 1V? Or maybe I'm thinking of the resistance in the coil…

Yeah I think so. One was 20ohm and another question had 60ohm
What did everyone get for the voltage divider circuit? I think i had 400ohm for Rv

[shroomer101]

I think you guys are right with 26v but did something else. I assumed (probably incorrectly) that the power output of the power supply must be the same if you increased the voltage so you  would decrease the current. this led to something very complicated and i ended up using the quadratic formula to solve and find v to equal like 17.something i think. any chance of getting any marks for this? it was a 3 mark question

Hmm I'm not too sure because when I calculated it I found that the power output at the generator was 104W which is different  from the original one :/

[KayKay]

Yeah I think so. One was 20ohm and another question had 60ohm
What did everyone get for the voltage divider circuit? I think i had 400ohm for Rv

For the 1V photodiode one I think I got a resistance of 187.5ohm... :/