Exam Discussion!

[the_analyser]

I got 187.5 as well, to find voltage across it was 1.3-1 or something similar

[Plitzer]

I got 187.5 as well, to find voltage across it was 1.3-1 or something similar

The photodiode is acting as the battery in the circuit, isn't it? So if it was producing how ever many amps that corresponded to 1V, then the resistor would also have 1V across it?

[SeanC]

My working was
Vtotal=vloss+6
vtotal=iline*rline+6
vtotal=5i+6       if power output must be the same then p=26 so i=26/v
v=130/v +6
v^2-6v-130=0 and solve with quadratic formula

[Chalkhous]

2b was really odd cause if you used f=kx you got 0.4m as the extension. if you use mgx=1/2kx^2 you got 0.8m
But looking at the replies it looks like 0.8m was right which is great cause i guessed that way.

[patel.aayush]

with that photodiode question part b when the switch is open is the voltage 0? cause the voltmeter is in series and cause it has infinite resistance (or so did the q say) the voltage across photodiode is 0? or did i overthink and u were just supposed to read off the graph

[RKTR]

I got 187.5 as well, to find voltage across it was 1.3-1 or something similar

got 187.5 as well

[Ya Habibi]

2b was really odd cause if you used f=kx you got 0.4m as the extension. if you use mgx=1/2kx^2 you got 0.8m
But looking at the replies it looks like 0.8m was right which is great cause i guessed that way.

Was it a three mark question? I got an answer of 0.8 but wasn't sure if the process was worth the three marks.

[KayKay]

got 187.5 as well

I put 187.5ohm as well but I think it might not be right. If you take the photodiode to be the power source, if it produces 1V, then the resister would also have 1V. The current is 1.6x10^-3A, which means the resister is 625ohm :/ I have no idea if this is right or not, I don't even understand how the current was changing when the light intensity on the photodiode was kept constant. It was a really odd question.

[SeanC]

I put 187.5ohm as well but I think it might not be right. If you take the photodiode to be the power source, if it produces 1V, then the resister would also have 1V. The current is 1.6x10^-3A, which means the resister is 625ohm :/ I have no idea if this is right or not, I don't even understand how the current was changing when the light intensity on the photodiode was kept constant. It was a really odd question.

I think i put 625ohm

[Rishi97]

got 187.5 as well

same wooo!!!

[evogamerz]

mc answers for materials and structures

a,b,c,d,a,c,b,b,c,d,a

[d4m0]

I think i put 625ohm

I got 625 ohms as well.

[theshunpo]

What did you guys get for the '' eV question?

[S61778]

Could you determine the mass of the planet?

[bpearson]

I couldn't