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AQA Physics PHYA1 - 24 May 2016 - RESIT [Exam Discussion Thread]

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Reply 320

Yavish13

7) Does resistance DECREASE as voltage increases??????????

Reply 321

HARRY108

6 marker - what did everyone put

Reply 322

Einstein1997

Original post by micycle

This is how I answered the heater question:

R = 12

V = 230

Resistor:

R = 0.015 (x2)

New current = 230/(12+2*0.015) = 19.11

V lost over cable (2 wires) = 19.11*0.015*2 = 0.574V

New V = 230 - 0.574 = 229.4V

It's 6.3 x resistance because they were 3.15m each.

Reply 323

HARRY108

Original post by Yavish13

7) Does resistance DECREASE as voltage increases??????????

I put decrease

Reply 324

Tayeb

Original post by micycle

Seems right. I STUPIDLY took away the resistance of the wires from 12 as a pose to adding them thinking 12 was the resistance of the whole circuit...And the length was 3.15m. You had to take that into consideration.

(edited 7 years ago)

Reply 325

TajwarC

Original post by Charlie543

Yeah but 229.5v to the same sig figs as question is 230, so mustve been series to get an acceptable different answer of 228v

I used a parallel arrangement and still got 229 volts? Found the lost volts in the wire and subtracted it from 230 - same method as someone above who used a series arrangement

Reply 326

JamieP123

Okay i'll try and explain that heater question.

Firstly you consider the wires to have NO resistance therefore the power of the heater is simply V^2/R

this is 230^2/12 = 4408 = 4400

After this it tells you that in reality the cables will have an internal resistance

You can consider the two wires as resistors (each with 3.15 * 0.0150 ohms of resistance = 0.04725) and place these in series with the 12 ohm resistor (heating element)

the terminal voltage was 230 volts meaning that the voltage would be shared (as a potential divider) between the three different resistors.

this means the voltage across the heating element would be (12/12+3.15*2*0.0150)*230. In other words, the resistance of the heating element over the total resistance of the circuit multiplied by the voltage. This gave 228.2029021 volts.

The energy wasted in the two wires would be as follows: 230-228.2029021 = 1.797...= voltage shared between two cables. One cables voltage =1.797/2 = 0.8985...V^2/R again = 0.8985^2/0.04725 * 2 (there are two cables) = 34.1 W

(edited 7 years ago)

Reply 327

ErwinJ

Original post by Rid123

For 5.
People whacking in a variable resistor in there to alter current?
Also to get readings before the spike in graph would you turn diode to reverse bias
And would resistance increase for increasing pd for last in 5?

Yes, variable resistor or source was needed. And correct on the reverse bias. However resistance would decrease as voltage increased

Reply 328

HARRY108

Original post by Einstein1997

It's 6.3 x resistance because they were 3.15m each.

what did you put for the 6 marker

Reply 329

Kmeister77

Original post by TheLifelessRobot

Why would the wires be in parallel? The circuit would be a cell and a resistor (filament) with each side of the resistor connected to the cell therefore it is in series.

Surely they are in parallel because you can't model it as a cell as there is no negative terminal, they are both connected to the positive terminal... I don't think it can make a full circuit because the voltage coming in would be equal to the voltage going out so no current current would flow... idk

Reply 330

Jonah1958

Original post by TajwarC

June 15 - Q7

There's a cable that's made from other cables, each with their own resistance. When calculating the resistance of the entire cable, you had to assume they were in parallel

That was the example i was trying to find!!!

Reply 331

Einstein1997

Original post by Kmeister77

That's what I thought.

Reply 332

Charlie543

Original post by TajwarC

I used a parallel arrangement and still got 229 volts? Found the lost volts in the wire and subtracted it from 230 - same method as someone above who used a series arrangement

Surely though anyway theyd had to have been in series othwerwise you'd only connevt to the heater by two wires in one end but no wire to connect to other ebd, so circuit wouldn't be complete

Reply 333

TheLifelessRobot

Original post by micycle

I did what you did the first time on that question, but the question says that the resistance of the wires were 0.0150 ohms per meter (length of wires were 3.15m). Therefore I did 0.0150*3.15*2 to get the total resistance of the wires and added that to 12 to get total overall resistance.

Reply 334

micycle

Original post by Einstein1997

It's 6.3 x resistance because they were 3.15m each.

So was the R of 0.015 for one meter of wire? I don't remember reading that!

Reply 335

MRuddyy

Personally I found that exam extremely easy and thrilling to read. I agree with most of the answers given in this post, however it must be said that some of the answers given here do not quite stand up to my par.
I was aiming to achieve 80UMS, but after sitting it I expect myself to get- at least- 98UMS. Question 6+7 were rather quite easy, and a lot of the people on this post seem to completely blow the questions out of proportion.

I'm praying to the Gods for you to all do well, however we all know that if you failed the exam then you yourself have no one to blame other than yourself- and no amount of attribution bias that is present will remove this fact.

At the end of the day I will be achieving an A*, whereas most of you commoners will, maybe, scrap a C overall.
See you all in the next post I grace you with my presence.

(edited 7 years ago)

Reply 336

TajwarC

Original post by micycle

I did it by calculating current (230/12)

Calculating lost volts - (3.15*0.0150*2)^-1 * 230

Then found the difference which was 229.5 or something

Reply 337

Yavish13

Original post by HARRY108

I put decrease

Thanks, Same

Reply 338

Lawliettt

Original post by TajwarC

June 15 - Q7

There's a cable that's made from other cables, each with their own resistance. When calculating the resistance of the entire cable, you had to assume they were in parallel

Think about it this way. It's a power source. 2 cables on either side of it. The other end of the cables connect to the heating system. How is it possible to make a parallel circuit from that set up?

Original post by micycle

It's 0.015 per meter. And there's 3.15 meters in each wire. I'm assuming factoring that in gets you 228V so your so your method is slightly incorrect. That being said, idk if 228V is right either