Rod's Physics 3/4 Questions Thread

[Rod]

And Lxnl, can you explain it to me. I don't want to simply plug in formulas I don't understand.

[lzxnl]

I just explained the formula to you in my previous post

[Rishi97]

Rishita, pleasssssseeeeeeee make your own thread

omg, oops.. sooo sorry. I tried to help you but somehow I was the one who ended up with help
Sorry again

[Rod]

Still confused with inverting and non-inverting amplifiers.

Gain >0 = non-inverting

Gain <0 = inverting

Inverting amplifiers = As input voltage increases output decreases and vice versa

Non-inverting - As input voltage increases, output voltage increases

Am I correcct? Can someone please add some stuff?

[Stevensmay]

Inverting amplifiers = As input voltage increases output decreases and vice versa

Non-inverting - As input voltage increases, output voltage increases

Inverting - Will output the opposite voltage to that of the input. Positive voltage in, negative voltage out. Negative voltage in, positive voltage out. It 'inverts' the polarity.

Non-inverting won't change the polarity.
Was a bit too general, see post below.

[PB]

Inverting - Will output the opposite voltage to that of the input. Positive voltage in, negative voltage out. Negative voltage in, positive voltage out. It 'inverts' the polarity.

Non-inverting won't change the polarity.

Hmm, not necessarily Stevensmay. Yes, your statement will hold true for the normal characteristic graph (aka. in this picture http://media.cheggcdn.com/media%2Fc20%2Fc205e5c5-b162-4ac5-b98e-961c8973fa09%2Fphp6yjl9G.png) However, it is possible to output the same polarity voltage as the input  (even though the amp is inverting).
Take a look at this characteristic graph of this amplifier.  http://6004.mit.edu/Fall12/tutprobs/digital07.gif

Its because of the base voltage, which shifts the characteristic graph up so that all values (input and output) within the linear range will always be positive!
  Rod, your statement, however, will always stand true : that an increase in input voltage will always result in a decrease in output voltage (this is because the gradient of the characteristic graph is negative).

[PB]

Still confused with inverting and non-inverting amplifiers.

Gain >0 = non-inverting

Gain <0 = inverting

Inverting amplifiers = As input voltage increases output decreases and vice versa

Non-inverting - As input voltage increases, output voltage increases

Am I correcct? Can someone please add some stuff?

There is nothing else to add really...the wording seems perfect Besides, the trickiest bit about amp questions is in drawing the output graph from the input graph so you might want to focus on that aspect.

[Rod]

There is nothing else to add really...the wording seems perfect Besides, the trickiest bit about amp questions is in drawing the output graph from the input graph so you might want to focus on that aspect.

Thanks PB!

Yeah I've been working on that as well. Could you please give me a trick/tip that would help me in doing so?

[PB]

Well, basically the things the examiners are looking for in these questions are whether you know when to clip the maximums/minimums of the graph if the input voltage goes out of the linear range and when to flip the graph (if the amp is inverting aka. the characteristic graph has a negative gradient)

So really, ALL the information you need should be in the characteristic graph that they should provide you with. It is actually quite straightforward and you will see that once you do the questions from Checkpoints etc.

However, many people do mistake the units and sketch the output graph as mV instead of V or something like that. So yeah, ALWAYS ALWAYS look at the units of the characteristic graph. Very important as VCAA likes to trip you up on that.

[Rod]

Need some help here;

So there is a voltage divider, R1 is a thermistor, which decreases in temperature as resistance increases. R2 is a variable resistor.

Now, to decrease the temperature, wouldn't the resistance in the variable resistor need to decrease? Because as the resistance in the variable resistor decreases, the output voltage of r2 would decrease.

This would result in an increase in output voltage across r1 (the thermistor), and if the output voltage across r1 increases, the resistance of the thermistor would need to increase as well. So then the temperature would decrease.

But the answer says the resistance of the variable resistor needs to increase, which changes my whole explanation

Can someone please point out where I am going wrong
thanks

[Rishi97]

A geosynchronous satellite is one with a period of 24 h positioned exactly above the equator. it appears motionless viewed from the surface of the Earth. Explain why it must be in an orbit above the equator?
I have no idea what the question is asking

Edit : OMG so sorry Rod... I thought I was posting in my own thread. Sorry. Pls forgive me

[Stevensmay]

A geosynchronous satellite is one with a period of 24 h positioned exactly above the equator. it appears motionless viewed from the surface of the Earth. Explain why it must be in an orbit above the equator?
I have no idea what the question is asking

Why must the satellite be orbiting above the equator? Why cannot it be just any orbit around the earth, say over the north pole?

[lzxnl]

It actually DOES have to be in an orbit above the equator. Note how orbits are elliptical (in general). Therefore, they lie in a plane. The plane of the orbit must intersect the centre of the Earth. After all, if we replace the Earth and the satellite with their centre of masses, the satellite's centre of mass must be orbiting around the Earth's centre of mass, or the Earth's centre.
Now, the only way for the satellite to appear motionless in the air is if the satellite is orbiting in the same direction as the Earth's rotation. If the satellite isn't in an orbit over the equator, you can see that it's rotating in a different direction to the Earth's rotation. Thus, a satellite with an orbit not over the equator will appear to be moving to someone standing on Earth.

[Rod]

Hey guys,

So a question with plotting the output vs time graph of an amplifier from looking at the input vs time graph + transfer chracteristic;

I've been getting all of them right, but I struggle on the 'clipping' questions. When do I know that I have to 'clip' but output vs time graph? And how do I do it?

Can someone also please provide an example

Thanks everyone

[Stevensmay]

Hey guys,

So a question with plotting the output vs time graph of an amplifier from looking at the input vs time graph + transfer chracteristic;

I've been getting all of them right, but I struggle on the 'clipping' questions. When do I know that I have to 'clip' but output vs time graph? And how do I do it?

Can someone also please provide an example

Thanks everyone

http://i.imgur.com/DMWBmRl.jpg
Isn't the best example due to time constraints sorry.
Clipping occurs when the output voltage of the amplifier is constant even though the input voltage is varying.
In the transfer characteristic graph, this would occur when the input voltage is below 3V or above 5V. We can see that even if we input 1V, we still get a 4V output, same with 2V and 3V.

Comparing the difference between the two left hand graphs, we can see that parts of the output graph have been 'clipped', with their tops and bottoms missing. Examining the top of the first peak, we can see that it goes above 5V all the way to 6V.
Back to the transfer graph, 5V will go to 10v but 6V will still only go to 10V.
Thus when we draw our output graph, we see a flat line.
Anything above the flat line has been 'clipped', in that we have lost the original signal, akin to distortion in speakers when they are too loud.

Essentially when doing clipping questions you do nothing different from the rest, simply take the input voltage, find the output voltage based on the amplifier and you are finished. There aren't any special steps etc.