I think it's an interpretation of Kirchhoff's Second Law:

0 = -0.2(R1) - (-0.3)(R2)

I can't remember what R1 and R2 was but there were values such that the equation above equals 0.

(I think people just remember that in parallel the pds. are the same, they just remember what they're taught)
:smile:

kirchhoffs second law is about emfs=p.ds not about p.d in a parallel circuit

Reply 1343

geditor

3

Original post by Theafricanlegend

kirchhoffs second law is about emfs=p.ds not about p.d in a parallel circuit

Actually, although we learn the definite of Kirchhoff's second law as something along the lines of "The sum of the emf's around a closed loop is equal to the sum of the potential differences around the same closed loop", Kirchhoff's law can actually be written "The vector sum of the electrical potential differences/voltage around any closed loop is zero".

[Thus, EMFs are interpreted as positive p.d.s. in the definition we learn]

Therefore, if you go around the closed loop containing the two parallel resistors, the current in the first resistor will be positive and as you go around the closed loop in this case, the current in the second resistor is interpreted as being negative within this loop only. This means if you use V=IR on both resistors within Kirchhoff's second law, the voltages cancel each other out to 0, meaning my earlier post is correct.

So yes, Kirchhoff's second law can be about emfs=pd.s. but it also applies to a larger variety of applications such as parallel circuits.

BAAM! That's Electronics A-Level for you :smile:

Reply 1344

Theafricanlegend

15

Original post by geditor

Actually, although we learn the definite of Kirchhoff's second law as something along the lines of "The sum of the emf's around a closed loop is equal to the sum of the potential differences around the same closed loop", Kirchhoff's law can actually be written "The vector sum of the electrical potential differences/voltage around any closed loop is zero".

[Thus, EMFs are interpreted as positive p.d.s. in the definition we learn]

Therefore, if you go around the closed loop containing the two parallel resistors, the current in the first resistor will be positive and as you go around the closed loop in this case, the current in the second resistor is interpreted as being negative within this loop only. This means if you use V=IR on both resistors within Kirchhoff's second law, the voltages cancel each other out to 0, meaning my earlier post is correct.

So yes, Kirchhoff's second law can be about emfs=pd.s. but it also applies to a larger variety of applications such as parallel circuits.

BAAM! That's Electronics A-Level for you :smile:

yeah this physics and we never got taught that and kirchhoffs first law makes sense

(edited 11 years ago)

Reply 1345

Sarabande

15

Original post by Theafricanlegend

yeah this physics and we never got taught that and kirchhoffs first law makes sense

How does it make sense in regards to the question that was asked. It was asking which law did you take into account in your proof of the 0.2A resistor. You didn't account for the 1st law ONCE if you proved it correctly whether through use of ratios between resistance and current (V=IR) or flat out using V=IR

Reply 1346

tooty_fruit

I think full ums will be around 87/88. They are usually rather picky on marking with the written sections so alot of people might not get marks they expected too there

Reply 1347

username916760

Original post by tooty_fruit

I think full ums will be around 87/88. They are usually rather picky on marking with the written sections so alot of people might not get marks they expected too there

but although it is gonna be high (somewhere like 95 for full ums), there were quite a few questions that had never come up (e.g.the pulse). i think that might be a good reason for the boundaries not to go up supersonicly, in spite of being fairly high. :ssaw:

Reply 1348

username916760

Original post by geditor

Actually, although we learn the definite of Kirchhoff's second law as something along the lines of "The sum of the emf's around a closed loop is equal to the sum of the potential differences around the same closed loop", Kirchhoff's law can actually be written "The vector sum of the electrical potential differences/voltage around any closed loop is zero".

[Thus, EMFs are interpreted as positive p.d.s. in the definition we learn]

Therefore, if you go around the closed loop containing the two parallel resistors, the current in the first resistor will be positive and as you go around the closed loop in this case, the current in the second resistor is interpreted as being negative within this loop only. This means if you use V=IR on both resistors within Kirchhoff's second law, the voltages cancel each other out to 0, meaning my earlier post is correct.

So yes, Kirchhoff's second law can be about emfs=pd.s. but it also applies to a larger variety of applications such as parallel circuits.

BAAM! That's Electronics A-Level for you :smile:

thats what we learnt in gcse in Iran.

Reply 1349

username916760

Original post by geditor

Actually, although we learn the definite of Kirchhoff's second law as something along the lines of "The sum of the emf's around a closed loop is equal to the sum of the potential differences around the same closed loop", Kirchhoff's law can actually be written "The vector sum of the electrical potential differences/voltage around any closed loop is zero".

[Thus, EMFs are interpreted as positive p.d.s. in the definition we learn]

Therefore, if you go around the closed loop containing the two parallel resistors, the current in the first resistor will be positive and as you go around the closed loop in this case, the current in the second resistor is interpreted as being negative within this loop only. This means if you use V=IR on both resistors within Kirchhoff's second law, the voltages cancel each other out to 0, meaning my earlier post is correct.

So yes, Kirchhoff's second law can be about emfs=pd.s. but it also applies to a larger variety of applications such as parallel circuits.

BAAM! That's Electronics A-Level for you :smile:

but its not about correct physics. you (tbh we all) need to get the marks. :juggle:

Reply 1350

Nick_

The definition of kirchhoff's second law as "The sum of the emf's around a closed loop is equal to the sum of the potential differences around the same closed loop" clearly implies that the p.d across all parallel branches must be the same.

Reply 1351

ahct_952

does anyone remember what they got for the second to last question about the wavelength of the microwave?

Reply 1352

Sarabande

15

Original post by ahct_952

does anyone remember what they got for the second to last question about the wavelength of the microwave?

I got 3.8mm. Probably wrong, I just guessed.

Reply 1353

tooty_fruit

Original post by ahct_952

does anyone remember what they got for the second to last question about the wavelength of the microwave?

I think it was 30mm, as the node-antinode separation is wavelength/4

Reply 1354

Z REFAN Z

Original post by tooty_fruit

I think it was 30mm, as the node-antinode separation is wavelength/4

yeah that's what i got, because it was antinode to node therefore 1/4 wavelength then multiple 7.5mm to get 30mm

Reply 1355

ahct_952

Original post by Z REFAN Z

yeah that's what i got, because it was antinode to node therefore 1/4 wavelength then multiple 7.5mm to get 30mm

yeah, that's what i put too, but some people at my school said they put 15mm because they thought the question meant the distance between two nodes/two anti-nodes was 7.5mm, so now i'm not sure... i think the wording of the question was 'the distance between adjacent maxima and minima is 7.5mm', which in my opinion does not make clear what they mean

Reply 1356

tp1995

Original post by mashmammad

but although it is gonna be high (somewhere like 95 for full ums), there were quite a few questions that had never come up (e.g.the pulse). i think that might be a good reason for the boundaries not to go up supersonicly, in spite of being fairly high. :ssaw:

Yeah good point, it seems like most people ether found it awful, or easy with the odd mistake? I think that will level out the statistics?

Also for the question about explaining why the current through the resistors in parallel was 0.2 and 0.3, I said you could work out the current by: using V=IR, then I1*R1=I2*R2, so you can work out 0.2 (call it X, as if we were working out an unknown value) by 4*0.3=6*X? i might have left out the rearangement of V=IR and just subbed in the values and said the the current is split with respect too R1/(1/6)=R2/(1/4), I know its a bit of a backward way, but will i still get marks do you think?

Reply 1357

geditor

3

Original post by Nick_

The definition of kirchhoff's second law as "The sum of the emf's around a closed loop is equal to the sum of the potential differences around the same closed loop" clearly implies that the p.d across all parallel branches must be the same.

Thank You. :smile:

People these days, it's not rocket science, just physics, oh wait...

Reply 1358

NerdyMcNerdNerd

Seriously has no one done an unofficial markscheme??

also what did everyone get for the very last question? I think i did 0.8^2 / 0.6^2 *30mm = 53.3

Reply 1359

Sarabande

15

I got maximum points of intensity are where there is maximum constructive interference and minimum points of intensity are where there is maximum destructive interfence.

So:

\frac{(0.8+0.6)^2}{(0.8-0.6)^2} = 49

OCR Physics A G482, Electrons, Waves and Photons, 25th May 2012

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