Exam revision Hard questions

[m@tty]

Yea it's 6V

It works like this. Technically there is a current running through the circuit, it's just that this current is extremely small so we can give it an effective value of 0A. The PD across the ohmic resistor is given by the formula V = IR, so if I = 0, then V = 0. Hence 0V across ohmic resistor. However the supply voltage is 6V. Hence voltage across LED (non-ohmic resistor) is 6-0 = 6V.

If the LED is in reverse bias no current reaches the resistor therefore Voltage drop must be 6 across LED.

If the LED is in reverse bias, what is the potential difference across it? Solutions says 6V, but I thought there would be none because the circuit is incomplete?  Probably wrong lol..

My one chance to help you M@tty...

Because the current flowing through the resistor is essentially 0
When using V=IR and I is 0Amperes the voltage drop across the resistor is 0 therefore the voltage drop must be across the diode.

Also correct me if i am wrong but i believe that when a diode is placed reverse biased it has an incredibly high resistance there the ratio between the variable resistor is tiny compared the resistance of the diode placed in reverse bias hence the voltage drop across the diode

PS: Sorry for the random stuff a kid posted on my account, tell me where more are so i can edit them out...

I forgot that there is still an infinitesimal current running, Thanks


Just so that I know I have the foundations right. If there was no current flowing at all, would the pd be 0V. I guess my question is this: If the circuit is truly in complete(no current flow at all) is there any voltage across it? I think no, right?

From Insight 2010

For this question, friction on the container is 1000N.

Would the equation I set up look like this:

10000-7000-1000=2400a

I think it is almost correct, just change the 2400 to 1400.





up the incline.

I think this is it

But since we're doing F=ma on the whole system, don't we have to consider both masses?

You're not considering the whole system though. You want the acceleration of the thing on the slope, and you have found the net force on it. So you need to divide by that mass. Have you done VCAA 08? I remember screwing the first question up for exactly this reason.

[qwertyda2nd]

The TSSM '09 paper substituted in the normal force of an inclined plane, N=mgsin(theta). But this assumes there is downward motion, or a friction force acting to stop the car, this may be why they substituted it in, as they state that there is no friction acting downwards, and in the previous stages of the question you were asked to draw a friction force acting up the hill.
This wasn't stated very clearly in the question, and i think that in the exam they will ask questions clearer than that.

[kyzoo]

Hmmmm...I think it's possible to have a potential difference between two points without a current running between the two points. It's just that that the potential difference doesn't do anything until you make a conduction path between the two points.

[cipherpol]

Ah, yeh, I remember that question. It just seems weird that the thing on the slope is accelerating at a different rate compared to the weight when you do it like this.

Thanks m@tty

[kyzoo]

Hmm is relative velocity on the exam? Or is that old study design stuff?

And CSE 2007 and VCAA 2006 have been the hardest exams I have encountered. Any others?

[m@tty]

Like if you took the LED completely out of the circuit, would there still be a difference across the gap? Again, I guess the question is: If you apply a voltage to one end of a wire, while the other end is not connected, does the entire wire carry the voltage?

EDIT: Wait I think I realised 0 current = 0 voltage because of ohms law..


Lol, I thought VCAA 2006 wasn't that bad...

[lachymm]

nothing compares to itute..

[kyzoo]

oh, i forgot about iTute ~~, but I haven't had time to do those unlike last year for Methods =(

[Blakhitman]

oooooh I want CSE 07, you have it virtual Kyzoo?

Relative velocity is old stuff yes you're right, lol that question in VCAA 06 got me...then I realised it was relative stuff! But I liked VCAA 06, wasn't that bad.

[kyzoo]

=( Sorry I got it from my Physics teacher >.<

[Blakhitman]

All good, about to do QAT apparently it's wack?

[TrueTears]

Hmm is relative velocity on the exam? Or is that old study design stuff?

And CSE 2007 and VCAA 2006 have been the hardest exams I have encountered. Any others?

teacher said it wasn't but doesn't hurt to know, kinda abstract to visualise but v_{a relative to b} = v_a - v_b, they are all vectors.

[kenhung123]

Can someone show me how the current flows in this circuit?

[/0]

Two resistors on top are in parallel with the bottom one. Each of the parallel routes receive the same voltage. From this, and the resistance of the resistors, you can work out the current in each path.

[kyzoo]

Meh I just get two objects A and B moving towards each other on a line. I know that the magnitude of the velocity of A relative to B = |V(A)| + |V(B)|. And the direction of the velocity of A relative to B has to be in the direction of A.

Then from this understanding I derive the formula.

Since
~ V(B) is in the opposite direction of V(A)
~ V(A relative to B) has the same direction of V(A)
~ |V(A relative to B)| = |V(A)| + |V(B)|

Therefore

V (A relative to B) = V(A) - V(B)