Rod's Physics 3/4 Questions Thread

[Nato]

Sorry for not being clear! I don't use the total current for the individual resistor, but the resistors broken down into one! So for example, when I do a circuit question I break down the circuit until I have one resistor. With that I can find the total current. Then I back track until the last stage in which the resistors are in series, so I use the total current to figure out the voltage drop of each 'broken down' resistor. Then, after finding the voltage drops, I back track again to the normal circuit I was given by the question, the voltage drop values do not change because resistors that were previously in parallel would have the same voltage because of Kirchoff's laws. And the answers I have been getting are correct so I think my method is working. I'm just thinking a bit more and confusing myself, and want to know more about why this is happening.

Rod

If you are still confused, I recommend you post a question, your solution and how when you trace back you get different values.

[Rod]

If you are still confused, I recommend you post a question, your solution and how when you trace back you get different values.

Yeah thank you . I'm planning to do it tomorrow because I have to figure out a way of drawing/posting the circuits I've been drawing to find the values.

Thanks for all the help Nato!

[Rod]

Hey everyone,

Just having trouble with the 'skin effect'. Is it in the course?

It's the effect that occurs in copper wires that causes attenuation (loss of strength of a signal of a communication channel). I usually give a lot more info about something I have trouble with but am unable to provide any now, I don't understand the skin effect at all.

If it's not in the course please let me know!

ALSO, I was going through Lenz law. Apparently it is also in unit 4 chem, what is the difference in knowledge required between the two in both subjects?

Rod

[hongkyho]

Just having trouble with the 'skin effect'. Is it in the course?

No.

ALSO, I was going through Lenz law. Apparently it is also in unit 4 chem, what is the difference in knowledge required between the two in both subjects?

Don't remember much (if any) about Lenz's Law in Chemistry, maybe someone else can help you there.

But, for physics, there are two main things to note about Lenz's Law:
- Being able to differentiate it from Faraday's Law (i.e. Faraday's Law is the formula while Lenz's Law is the statement saying that the induced current in a loop will be in the direction so that the flux it creates will oppose the change in the flux that produced it.)
- Being able to apply it to find the direction of the current induced due to the change in flux.

PS: Hope Daisy is BOG everytime he plays Collingwood

[Rod]

No.
Don't remember much (if any) about Lenz's Law in Chemistry, maybe someone else can help you there.

But, for physics, there are two main things to note about Lenz's Law:
- Being able to differentiate it from Faraday's Law (i.e. Faraday's Law is the formula while Lenz's Law is the statement saying that the induced current in a loop will be in the direction so that the flux it creates will oppose the change in the flux that produced it.)
- Being able to apply it to find the direction of the current induced due to the change in flux.

PS: Hope Daisy is BOG everytime he plays Collingwood

Cool! Understood, thank you so much!

Haha hope he breaks his ankle xD. Absolute dud, he wore a navy tie at Alan Didak's wedding :\/

[Rod]

After doing some exam style questions I've confused myself with a topic I learnt really well back in electricity.

It's diodes and the current over voltage graph (or characteristic graph). Does this graph tell you about the TOTAL amount of voltage the diode eats up or at a specific amount of current? Or does it only eat some.
Example;


As the characteristic curve shows above, after a voltage of 0.6 volts, the gradient increases abruptly. Now, I always thought that the lower, constant line is just the total amount of voltage the diode eats up before passing on the current (so in this case it's 0.6 V). But after doing a question, which had a circuit with a total voltage of 8 v, but a resistor with 6 v meant the diode takes up 2 volts despite the characteristic curve showing that the gradient starts increasing after 0.8 V.

So yeah, I've confused myself after doing this question. Any help to clarify if what I'm saying is right, or I've just confused myself.

Rod

Mod Edit: Resizing image - Phy124

[hongkyho]

Does this graph tell you about the TOTAL amount of voltage the diode eats up or at a specific amount of current? Or does it only eat some.

There is basically three sections to this graph:
1. The horizontal part of the graph: in this part, the diode does not conduct so there is no current going through the circuit.
2. The vertical part of this graph: this part shows the total voltage drop across the diode, which is only true if the current going through the circuit is above a certain value (i.e above the value which marks the point where the graph starts becoming vertical). This is where you get your 0.6/0.7V from.
3. The in-between the above sections: Not really looked at in VCE, but if the current is lower than the value discussed above, then you have to use the graph to match up the current in the circuit to the voltage drop of the diode. Won't be likely to be discussed in a VCAA exam.

Any help to clarify if what I'm saying is right, or I've just confused myself.

In any circuit question relating to diodes, the voltage drop of the diode is your starting point to work out the rest of the question, i.e. in this case, you know the voltage drop across the diode is 0.6V, you should then work out the voltage of the resistor and then the resistor current by using the voltage and resistance and Ohm's Law.

[Rod]

There is basically three sections to this graph:
1. The horizontal part of the graph: in this part, the diode does not conduct so there is no current going through the circuit.
2. The vertical part of this graph: this part shows the total voltage drop across the diode, which is only true if the current going through the circuit is above a certain value (i.e above the value which marks the point where the graph starts becoming vertical). This is where you get your 0.6/0.7V from.
3. The in-between the above sections: Not really looked at in VCE, but if the current is lower than the value discussed above, then you have to use the graph to match up the current in the circuit to the voltage drop of the diode. Won't be likely to be discussed in a VCAA exam.
In any circuit question relating to diodes, the voltage drop of the diode is your starting point to work out the rest of the question, i.e. in this case, you know the voltage drop across the diode is 0.6V, you should then work out the voltage of the resistor and then the resistor current by using the voltage and resistance and Ohm's Law.

Understood, understood.

Thankyou

[Orb]

Hey guys, need some help with this question: (kinematics)

At time = 0 seconds a blue van passes a red stationary car. The blue van travels at a constant speed of 60 km h-1. The red car accelerates at a rate of 4.6 m s-2 for 5.6 seconds, and then travels at a constant speed.

How long does it take for the red car to catch the blue van?

Bit lost!

[rhinwarr]

Since you're trying to find the time, you need to make equations for the distance travel with the variable t for time.

The blue car is moving at constant speed so for the whole time it's equation for distance is:
d = 50/3 t

The red car accelerates for 5.6 seconds so we need to find out the distance it travels in that time.
To do this we can use the equation x = ut + 0.5at^2
u=0 t=5.6 a=4.6 x=?
so putting these in the equation,
X = 72.128m

Now to check that the red car has not taken over the blue car yet, we find how far the blue car has travelled.
d = st
d = 50/3 x 5.6
d = 93.33m
The distance is greater which means the red car has not taken over yet.

Now we need to find the equation for the red car now that it is travelling at constant speed. This can be done with the equation
v = u + at
V = 4.6 x 5.6
v = 25.76 m/s

The equation for the red car would be
d = 25.76(t-5.6)+72.128
It is t-5.6 because the car only starts travelling at 25.76m/s after 5.6s. We also need to add the distance travelled while it was accelerating which was the 72.128m.

Now that we have both equations, we can equate them to find when the cars have travelled he same distance, aka when they take over.

50/3 t = 25.76(t-5.6)+72.128
t = 7.93s

[Rod]

Hey everyone,

Thinking about what detail study I should do. Throughout the holidays I went over materials and construction and I was just like hell no!
Einstein's relativity looks interesting but HAARD.

My school is doing sound, but I'm really, really interested in the electricity + further photonics stuff.

Should I do both sound and electricity? What are your thoughts?

Thanks guys, been a long time since I posted a question xD.

[lzxnl]

Hey everyone,

Thinking about what detail study I should do. Throughout the holidays I went over materials and construction and I was just like hell no!
Einstein's relativity looks interesting but HAARD.

My school is doing sound, but I'm really, really interested in the electricity + further photonics stuff.

Should I do both sound and electricity? What are your thoughts?

Thanks guys, been a long time since I posted a question xD.

Only one of them counts. AND sound is the easiest to do well on. Except for that one funny trial paper in which I swear the sound questions were rigged and the relativity questions were a joke.

[Rod]

Only one of them counts. AND sound is the easiest to do well on. Except for that one funny trial paper in which I swear the sound questions were rigged and the relativity questions were a joke.

Thanks Lxznl!

What year was this 'rigged paper' done?

[Phy124]

Thinking about what detail study I should do. Throughout the holidays I went over materials and construction and I was just like hell no!

Interesting, I found Materials and their use in structures to be the easiest of the detailed studies. Then again, I am a Civil Engineering student so I suppose that makes sense 

I'd agree with lzxnl and say that sound is also an easy detailed study. If I had to choose between sound and further electronics I would definitely opt for the former. Fortunately, if you do choose the latter you shouldn't be as bad off as the students who studied it in 2011 (I think there was 8(?) questions on the 2011 paper from that section where less than 55% students chose the correct answer)

[lzxnl]

Thanks Lxznl!

What year was this 'rigged paper' done?

Hahaha that was a trial paper from last year. In which I swear I could have taught any student how to do the relativity part in an hour.

The thing about sound is that it does sort of build from existing knowledge of waves and interference, so it's not ALL new stuff.