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March 29, 2024, 12:42:15 am

Author Topic: VCE Physics Question Thread!  (Read 603230 times)  Share 

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Robert123

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Re: VCE Physics Question Thread!
« Reply #105 on: August 07, 2013, 02:07:09 pm »
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Having some difficulty understanding transformers:
So by the way I understand it is they change the current and voltage while (ideally) keeping the same power overall. A step up transformer increase voltage and decrease current while a step down transformer decrease voltage but increase current.
The difficulty I have is to relate this to Ohm's law (V=IR) which state that if you have an increase or decrease in voltage or current, you will get a proportional increase or decrease in the other. That is, you can't decrease one while increasing the other.

So, do transformers change the resistance? Could someone please explain to me how transformers work while taking into consideration ohms law.

Thank you

SocialRhubarb

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Re: VCE Physics Question Thread!
« Reply #106 on: August 07, 2013, 05:24:02 pm »
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A transformer works based on electromagnetic induction - the changing magnetic flux induced by one coil induces a voltage in the other coil. Furthermore, any resistors attached to the coils of the transformers are still subject to Ohm's laws. Changes in the secondary coil also affect the primary coil. This is probably best illustrated with an example.

Let's take a 5V RMS AC source, connected to a step-up transformer with a 1:10 ratio. A 100 Ohm resistor is connected to the secondary coil. Let's try to work out the current being drawn.

The voltage across the secondary coil is pretty straight forward to work out:



Applying Ohm's law:



Then we can find the current across the primary:



Notice that if we were to attach a 50 Ohm instead of a 100 Ohm resistor, we would double the current through the secondary coil and also double the current through the primary coil. This is probably one of the concepts students find hardest to grasp - changing the resistance across the secondary coil changes the current in the primary coil as well as the secondary coil. Why? It's a bit difficult to explain, but it's to do with the primary coil inducing a back EMF on itself. For this reason, you also can't treat the primary coil as a conventional 'circuit' - the coil induces a back EMF which almost totally opposes the voltage drawn, even if there are no physical resistors in its path. Also, the current produced by a power source is not fixed - it can vary and will vary depending on the external circuit, such as what kind of transformers and resistors it is attached to.

So Ohm's law still applies in circuits with transformers, but we also have to take into account the effect of electromagnetic induction, and remember that generally speaking the current generated by a power source is not constant, but varies depending on the circuit it's attached to.
« Last Edit: August 07, 2013, 05:27:38 pm by SocialRhubarb »
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Robert123

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Re: VCE Physics Question Thread!
« Reply #107 on: August 07, 2013, 05:56:35 pm »
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Ok, continuing on from your example, say we have a 10:1 to ratio (step down transformer). Now the voltage would 0.5V. Using ohms law on the resistance to find the current (V=IR)
I= 0.5/100= 5*10^-3A
P=VI
P= 0.5 * 5*10^-3= 2.5*10^-3 W
For the step up transformer in the example you provided, the power is 25W which is greatly differently to the result above (transformers are meant to have the same power). This is due to not taking into account the increase in current but that means ohms law is wrong. Correct?
So then, how can ohms law and transformers agree?

SocialRhubarb

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Re: VCE Physics Question Thread!
« Reply #108 on: August 07, 2013, 06:10:29 pm »
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The power on the coils of a transformer is the same. That is, the power you put into the primary of a transformer is equal to the power coming out of the secondary of a transformer. The rule is not that all transformers all have the same power, because we have different transformers in different circuits, but the power put into a transformer is equal to the power coming out of a transformer.
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Re: VCE Physics Question Thread!
« Reply #109 on: August 11, 2013, 02:18:28 pm »
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Just a diffraction query (wary clarification I guess) as the Heinemann textbook is slightly unclear...

Sometimes it talks about the extent of diffraction being most significant when the wavelength approaches the size of the gap, and other times it suggests/states the proportionality: extent of diffraction is proportional to wavelength/gap.

I presume the latter is most correct and - as such - wavelengths that are larger than the gap size will produce even more diffraction than wavelengths equal to the gap size?

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xenon2013

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Re: VCE Physics Question Thread!
« Reply #110 on: August 11, 2013, 03:00:09 pm »
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can't figure out how to do this:

A sprinter reaches his maximum speed vmax in 2.5 seconds from rest with constant
acceleration. He then maintains that speed and finishes the 100 meters in the overall
time of 10.40 seconds. Determine his maximum speed vmax.

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lzxnl

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Re: VCE Physics Question Thread!
« Reply #111 on: August 11, 2013, 04:21:37 pm »
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So...his maximum speed is 2.5a
Maintains this for a time of 10.4 - 2.5 = 7.9 seconds
Distance travelled = 7.9 * 2.5a + 1/2*at^2 = 1/2*a*6.25 + 7.9*2.5a
22.875a = 100
a = 4.37 m/s^2
Max speed = 2.5a = 10.9 m/s
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Re: VCE Physics Question Thread!
« Reply #112 on: August 11, 2013, 04:50:19 pm »
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an alternative:









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xenon2013

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Re: VCE Physics Question Thread!
« Reply #113 on: August 11, 2013, 04:57:17 pm »
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So...his maximum speed is 2.5a
Maintains this for a time of 10.4 - 2.5 = 7.9 seconds
Distance travelled = 7.9 * 2.5a + 1/2*at^2 = 1/2*a*6.25 + 7.9*2.5a
22.875a = 100
a = 4.37 m/s^2
Max speed = 2.5a = 10.9 m/s

im a bit confused because, if he's maintaing the vmax, shouldnt the acceleration be zero instead of 4.37? and within that 7.9 seconds, he travelled a distance less than 100m because he took 10.4s to finish, so i dont understand why t=7.9 and dist=100 were subbed together into one eqn, to me this implies that he ran 100m in the 7.9s

sorry if this is confusing but im having trouble understanding it all

lzxnl

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Re: VCE Physics Question Thread!
« Reply #114 on: August 11, 2013, 05:15:05 pm »
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im a bit confused because, if he's maintaing the vmax, shouldnt the acceleration be zero instead of 4.37? and within that 7.9 seconds, he travelled a distance less than 100m because he took 10.4s to finish, so i dont understand why t=7.9 and dist=100 were subbed together into one eqn, to me this implies that he ran 100m in the 7.9s

sorry if this is confusing but im having trouble understanding it all

OK. I'll clarify my working.
a is the acceleration in the period where he is accelerating. The acceleration is only equal to a in that time. When he is travelling at max speed, the acceleration is zero.

He accelerated for 2.5 seconds. He ran for 10.4 seconds in total. Therefore the time in which he maintained the max speed is 7.9 seconds. What I subbed in was 1/2*at^2, the distance travelled during the acceleration, and 7.9*vmax, or 7.9*2.5a, which is the distance travelled at constant speed.
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lolipopper

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Re: VCE Physics Question Thread!
« Reply #115 on: August 15, 2013, 06:47:40 pm »
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i gotta question:

Why is the EMF induced in a DC generator a maximum when the coil is parallel to the magnetic field, and a minimum when perpendicular to the field?
 
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lzxnl

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Re: VCE Physics Question Thread!
« Reply #116 on: August 15, 2013, 07:11:02 pm »
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i gotta question:

Why is the EMF induced in a DC generator a maximum when the coil is parallel to the magnetic field, and a minimum when perpendicular to the field?

Assuming that the coil is rotating at a constant rate:

The magnetic flux is actually given by BA*sin(theta) where theta is the angle between the magnetic field lines and the coil. As the rotation is at a constant rate, theta = wt where w is the angular velocity and t is time. I am assuming that at t = 0 there is no flux for simplicity.

Then, flux = BA*sin(wt)
By Faraday's law, emf = -rate of change of flux = -d(flux)/dt = -wBA*cos(wt)
So, for the magnitude to be a maximum, we want wt = some integer multiple of pi. This corresponds to the field being parallel to the coil.
When the field and coil are perpendicular, cos(wt) = 0, i.e. minimum magnitude of the emf.
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sydneyboy

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Re: VCE Physics Question Thread!
« Reply #117 on: August 18, 2013, 07:32:20 pm »
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We are doing a prac at school with a circular motion kit, I've stumbled on a task that i have no clue how to go about:

"Given Fnet = mv^2/r

Determine a relationship between F and m so everything else must be kept constant. Think about how best to keep v^2/r constant while collecting your f vs m data.

Clue: what is another way of expressing v^2/r and how can this be kept constant even when radius is changing. "

Not sure how to star, any pointers would be appreciated.

lolipopper

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Re: VCE Physics Question Thread!
« Reply #118 on: August 18, 2013, 08:54:00 pm »
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well v^2/r = acceleration. and if you keep the magnitude of the velocity constant, than the acceleration should be constant.

also when radius increases, increase the velocity so the acceleration remains same. and vice versa when radius decreases.

hope it helps.
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lzxnl

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Re: VCE Physics Question Thread!
« Reply #119 on: August 18, 2013, 11:10:19 pm »
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We are doing a prac at school with a circular motion kit, I've stumbled on a task that i have no clue how to go about:

"Given Fnet = mv^2/r

Determine a relationship between F and m so everything else must be kept constant. Think about how best to keep v^2/r constant while collecting your f vs m data.

Clue: what is another way of expressing v^2/r and how can this be kept constant even when radius is changing. "

Not sure how to star, any pointers would be appreciated.

v = radius * angular velocity
= rw

v^2/r = (rw)^2/r = w^2*r
Or v^2/r = (2pi*r/t)^2/r = 4pi*r/t^2

The second form is probably more useful, so if you keep the same distance and period you'll be fine. I don't know about the practicalities of that though.
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