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#### LazyZombie

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##### Re: VCE Physics Question Thread!
« Reply #15 on: April 17, 2013, 08:46:40 pm »
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General questions about voltage;
Why can you have negative voltage (AC) ? Doesn't that imply we can have "negative" energy? Wouldn't it be best if we have negative and positive current for saying which way the current flow?
Also;
I lost a mark on a SAC with the question being about explainging how seatbelt and crumpling reduce the severity of the drivers injury, it was a 3 mark question. What type of exam "style" answers should I have. The teacher noted on it "how does the seat belt increase the time of collision?", how would I undergo answering that for an 'exam' quality answer?
Thanks
I'm not sure about the voltage thing but since voltage is relative, I think it is possible. But I'm prettyy sure its not necessary in the course. (not in the study design) I might get back to this later

As for the crumple zone and seatbelt
$F=\frac{\Delta p}{t}$ therefore the crumble zone increases the time of the collision, thus decreasing the force.
Inertia causes the body to continue moving in the original motion of the car in a collision and cause injury or fatality, and seatbelts help to prevent this from happening.

Seatbelts don't increase the time of collision. I think that's what your teacher means. (?)
« Last Edit: April 17, 2013, 08:58:11 pm by LazyZombie »
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#### availn

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##### Re: VCE Physics Question Thread!
« Reply #16 on: April 17, 2013, 08:53:02 pm »
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Put simply, if voltage is negative, then the current is also negative. As P = VI, the negative cancels out.

And yeah, seat belts aren't crumple zones, they don't increase the time of collision, they just make sure you don't shoot out of your car.
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##### Re: VCE Physics Question Thread!
« Reply #17 on: April 17, 2013, 10:45:43 pm »
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Ok, thanks for that. I will ask my teacher what her opinion of what the answer "should" be (even though I find people on this forum more knowledgable and reliable with answer).
Also, another question (this is more a general education question).
Does the only thing that matters with SACs is your ranking? Please only answer if you are 100% sure due to reading a report from VCAA or talking to an examiner?
If so, then does that means there is no different between averaging  90% and 100% on SACs if you go to a low scoring school?
Split this post and moved it to the Technical Score Discussion boards

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##### Re: VCE Physics Question Thread!
« Reply #18 on: April 25, 2013, 04:41:12 pm »
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Jasper has a transistor radio with ﬂat batteries. Suggest a reason to explain why the sound is distorted when he tunes in to a strong radio signal.

^help anyone? thanks in advance

#### ~T

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##### Re: VCE Physics Question Thread!
« Reply #19 on: April 26, 2013, 05:16:06 pm »
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Jasper has a transistor radio with ﬂat batteries. Suggest a reason to explain why the sound is distorted when he tunes in to a strong radio signal.

^help anyone? thanks in advance

I'm not entirely sure on this one but I'll give you my thoughts...

Basically, any amplification system works of a small varying AC signal (the radio signal here) and a larger DC input (the battery here) that amplifies the input signal to create a much larger output (the sound here).

So, if the batteries are running flat, then they will not supply enough power to the transistor(s) in order to amplify the larger voltages, and the linear amplification region may be smaller. The input voltages will still be the same, but the extreme values will no longer be amplified linearly, distorting the signal.

This only half makes sense in my head. But pretty much, there is no reason that the input signal would be larger (resulting in clipping outside of the region) so it must be that the linear region itself becomes smaller.
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#### Alwin

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##### Re: VCE Physics Question Thread!
« Reply #20 on: April 26, 2013, 09:11:33 pm »
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Jasper has a transistor radio with ﬂat batteries. Suggest a reason to explain why the sound is distorted when he tunes in to a strong radio signal.

^help anyone? thanks in advance

For this one, I decided to add an image! just because.. I can haha

Quote
The input signal may be too large, causing the amplifier to be limited by the supply voltage.

This is true in the case of Jasper and his poor flat batteries. The DC power supply is the two parallel lines, where the voltage supplied is +Vcc and grounded at 0V. I won't go into the specifics of the transistor, since it is no longer on the study design, but simply the "in" signal is the current that enters the middle of the transistor and allows the larger current to flow through (top to bottom) from the DC power source.
Hence, if the DC source is not large enough then it cannont properly amplify the "in" signal because it cannot amplify a strong signal, ie it cannot amplify the strong +ve and strong -ve signals, ie smaller clipping range

Hope that makes sense!
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#### Robert123

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##### Re: VCE Physics Question Thread!
« Reply #21 on: April 27, 2013, 03:04:38 pm »
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What does a negative gain amplifier actually achieve?
I know it inverts the Vout graph but how does that have a meaningful purpose? What would be the difference in a microphone that is hooked up to an amplifier with a negative gain rather than a positive one?
Thanks

#### Alwin

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##### Re: VCE Physics Question Thread!
« Reply #22 on: April 27, 2013, 03:19:04 pm »
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What does a negative gain amplifier actually achieve?
I know it inverts the Vout graph but how does that have a meaningful purpose? What would be the difference in a microphone that is hooked up to an amplifier with a negative gain rather than a positive one?
Thanks

There actually is no "purpose" for wanting negative gain. It is just a side effect of using a simple single transistor amplifier, the npn transistor inverts the input signal. If you want a non-inverting amplifier, you need a two-stage amplifier (two transistors). That's the only reason why a lot of amplifiers in physics questions have negative gain since we deal with simple amplifiers.

As for the difference, well there is none. Remember when a microphone or speaker works, it is a diaphragm that vibrates back and forth creating a sound wave. If the initial current is positive, then the diaphragm moves forward, then back, then forward etc. But, if the signal had been inverted, initially the current flows in the other direction and so the diaphragm moves back first then forwards then back again if you get what I mean. So, the frequency of sound produced / recorded it the exact same.

Hope that clears things up for you!
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#### Robert123

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##### Re: VCE Physics Question Thread!
« Reply #23 on: April 27, 2013, 03:42:51 pm »
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Thank you for that,  clears things up a lot

#### Homer

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##### Re: VCE Physics Question Thread!
« Reply #24 on: May 02, 2013, 07:08:52 pm »
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Hey guys how would you calculate the value of R?

Thanks

ANS: 309.1ohms
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#### sin0001

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##### Re: VCE Physics Question Thread!
« Reply #25 on: May 02, 2013, 07:48:45 pm »
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For part b of the attached question, I get that the resistor, parallel to the LED, will have the same potential difference as the LED-2.5 V, so doesn't that mean the Resistor should have the same current flowing through it as the LED-11 mA; so can't we then use these values for current & voltage, to work out the resistance of Resistor-'R'?
But in the worked solutions they haven't assumed that current across 'R' will be same as I(LED), instead this is what how they solved resistance:

Vtherm. = 10 – 2.5 = 7.5 V, Itherm. = 7.5/500 = 0.015 A
since ILED = 0.011, IR = 0.004 A and RR = 2.5/0.004 = 625 Ω
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#### Alwin

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##### Re: VCE Physics Question Thread!
« Reply #26 on: May 02, 2013, 08:04:22 pm »
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Hey guys how would you calculate the value of R?

Thanks

ANS: 309.1ohms

Well, first you look at the diodes:
Diode 1:
Since the current is 20 mA, V = 1. Hence, R = 1/0.020 = 50 ohm. Thus, you have a 50ohm 'resistor' and a 500 ohm resistor in parallel
$\text{effective resistance of vertical branches} = 45.45 \Omega$

Diode 2:
This diode is in reverse bias, hence no current flows. There are only the 2 100ohm resistors in parallel
$\text{effective resistance of horizontal branches} = 50 \Omega$

Total resistance:
$R_T=45.45+50+50+R$

Total current:
$I_T = 0.022 \text{A} \text{ (through parallel circuits ratios)}$

From Ohm's law, R = V/I
$45.45+50+50+R = \frac{10}{0.022}$
$R=309.1 \Omega$

Hope this makes sense!
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#### Alwin

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##### Re: VCE Physics Question Thread!
« Reply #27 on: May 02, 2013, 08:08:37 pm »
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For part b of the attached question, I get that the resistor, parallel to the LED, will have the same potential difference as the LED-2.5 V, so doesn't that mean the Resistor should have the same current flowing through it as the LED-11 mA; so can't we then use these values for current & voltage, to work out the resistance of Resistor-'R'?
But in the worked solutions they haven't assumed that current across 'R' will be same as I(LED), instead this is what how they solved resistance:

Vtherm. = 10 – 2.5 = 7.5 V, Itherm. = 7.5/500 = 0.015 A
since ILED = 0.011, IR = 0.004 A and RR = 2.5/0.004 = 625 Ω

I think you got a bit confused between parallel and series circuits. In parallel circuits, the potential difference across both branches is the same, BUT the current in each branches is different. The sum of the currents in each of the branches is equal to the total current, in this case 0.015A. This is different from series circuits.

Hopefully you can see how the solutions got the answer from here!
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#### sin0001

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##### Re: VCE Physics Question Thread!
« Reply #28 on: May 02, 2013, 08:16:53 pm »
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I think you got a bit confused between parallel and series circuits. In parallel circuits, the potential difference across both branches is the same, BUT the current in each branches is different. The sum of the currents in each of the branches is equal to the total current, in this case 0.015A. This is different from series circuits.

Hopefully you can see how the solutions got the answer from here!
Ahhh yeah I had a mental blank lol
Also, how would you work out the resistance of these diodes, would you just plug it into the formula: R = V/I?
Thanks!
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#### Alwin

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##### Re: VCE Physics Question Thread!
« Reply #29 on: May 02, 2013, 08:53:31 pm »
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Ahhh yeah I had a mental blank lol
Also, how would you work out the resistance of these diodes, would you just plug it into the formula: R = V/I?
Thanks!

It's okay! just try not to have a mental blank in the exam

Yeah, just R = V/I. I have an example in a previous post:
Re: Physics [3/4] Question Thread!
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A pessimist says a glass is half empty, an optimist says a glass is half full.
An engineer says the glass has a safety factor of 2.0