A2 Physics | Post your doubts here

[smzimran]

Assalamoalaikum wr wb!
For application part, in inverting amplifier...it says input impedance is infinite so current in R(in) is equal to the current in R(f)
I dont get why? can someone plz explain this? :s

AoA.
Input impedance is infinite means no current enters the input,
so current has just one path to follow (i.e. through feedback resistor), therefore all of the current goes to feedback resistor.
Thus, current in R(in) is equal to the current in R(f)

 

[smzimran]

Assalamu alaikum,
Calculation is done, but I don't completely understand the explanation given in the marking scheme. It got calculated the other way, I guess. Method is different but the answer is same. Do you want me to post the solution??

this question is from the cie coursebook..please explain b (ii)

A strain gauge contains 15 cm of wire of resistivity 5.0 × 10−7 Ω m. The resistance of the
strain gauge is 150 Ω.
i Calculate the cross-sectional area of the wire in the strain gauge. [2]
ii Calculate the increase in resistance when the wire extends by 0.1 cm, assuming that
the cross-sectional area and resistivity remain constant.


the answer to b (ii) is Extends by 1/150 so resistance increases by a factor of 1/150 , which is 1 Ω.

AoA,
(i)
R = ρl / A
A = ρl / R
A = (5.0 * 10^-7) (15 * 10^-2) / (150)
A = 5.0 * 10^-10 m^2

(ii)
∆R = ρ∆l / A
∆R = (5.0 * 10^-7) (0.1 * 10^-2) / (5.0 * 10^-10)
∆R = 1 Ω

 

[smzimran]

in Q10..for V-.. why do we divide 2000 by 4100.. why not the resistance of the thermistor .ie 2100 divded by 4100?
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w09_qp_41.pdf

AoA,
Because the input to the inverting input of the amplifier is the p.d across P so we have to take its resistance for comparison (coloured red below)
So using the potential divider method:
Inverting input = (Resistance of P / total resistance of thermistor and P) * 2V
Inverting input = (2000 /2000 + 2100) * 2
Inverting input = 0.97 V
This has to be compared with the constant 1V input to the non-inverting input!

 

[smartangel]

AoA,
Because the input to the inverting input of the amplifier is the p.d across P so we have to take its resistance for comparison (coloured red below)
So using the potential divider method:
Inverting input = (Resistance of P / total resistance of thermistor and P) * 2V
Inverting input = (2000 /2000 + 2100) * 2
Inverting input = 0.97 V
This has to be compared with the constant 1V input to the non-inverting input!
View attachment 6990

umm so basically we take resistance of P(red one) to compare it with the other 2 resistors? had there been a thermistor in place of one of the two resistors we could have taken the pd across thermistor?

 

[smartangel]

in Q 11 b(ii)..how do we calculate the frequency of the information signal?
http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w09_qp_42.pdf

 

[smzimran]

umm so basically we take resistance of P(red one) to compare it with the other 2 resistors? had there been a thermistor in place of one of the two resistors we could have taken the pd across thermistor?

yes!

 

[unique840]

in Q 11 b(ii)..how do we calculate the frequency of the information signal?
http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w09_qp_42.pdf

the frequency of info wave is the freq of the modulated wave to change from one max to the other or from one min to the other. here it takes 100 microseconds to move from one max to other. so freq of info will be 1/100μ

 

[hassam]

also this one​

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w07_qp_4.pdf​

Q1.B(ii)​

u cn solve it urself b4....bt here;s what is confusing me......we got an equation for t as T=MG + Fc...........Now the second equation that we use i.e T=Kx.....yar ab iss mein why we are just usiong T=Kx......neeche ki taraf weight hay ....ooper kki taraf tension....and tension greater than weight bt phir bhi extension.......i m just cnfused here.......just tell me why only T=Kx...why we are neglecting weight here...​

 

[leadingguy]

A little confusing fr me
when calculatiing attenuation or gain in signals........

by the formula = log(p1/p2) which value should be p1 and which value should be p2 the lost power or the input, out put?????? try to understand mY question and ans. it plz

 

[unique840]

A little confusing fr me
when calculatiing attenuation or gain in signals........

by the formula = log(p1/p2) which value should be p1 and which value should be p2 the lost power or the input, out put?????? try to understand mY question and ans. it plz

for gain: p1 will be output power and p2 will be input
for attenuation: p1 will be input nd p2 will be output

 

[hassam]

just remeber that larger value shud in the numerator

 

[hassam]

http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_w03_qp_4.pdf
q2.c

 

[XPFMember]

Assalamoalaikum wr wb!

got a doubt in the application part...in the reproduction of the analogue signal, like when converting digital to analogue, they ask to state ways in which it could be improved. One reason is to increase the samlpling frequency. That's ok to me. I don't understand the oother reason => increase the no. of bits. Can someone plz explain me this?

 

[badrobot14]

Assalamoalaikum wr wb!

got a doubt in the application part...in the reproduction of the analogue signal, like when converting digital to analogue, they ask to state ways in which it could be improved. One reason is to increase the samlpling frequency. That's ok to me. I don't understand the oother reason => increase the no. of bits. Can someone plz explain me this?

no. of bit = bit depth/resolution..
you understand the concept of sampling rate.. a greater sampling rate means on horizontal scale there are more points(data) available so better signal wd be recoverd. But what about on the vertical scale.?? lets say you are converting analog wave to a 4bit digital signal, means on vertical scale you have 2^4 = 16 differernt levels and any signal sample falling b/w these levels wd be rounded off to the nearest of these levels (to make digital)... if bit depth increased to 2^8 = 256 levels means round off won't be too much and wd result in more accurate recovery of signal...

 

[XPFMember]

no. of bit = bit depth/resolution..
you understand the concept of sampling rate.. a greater sampling rate means on horizontal scale there are more points(data) available so better signal wd be recoverd. But what about on the vertical scale.?? lets say you are converting analog wave to a 4bit digital signal, means on vertical scale you have 2^4 = 16 differernt levels and any signal sample falling b/w these levels wd be rounded off to the nearest of these levels (to make digital)... if bit depth increased to 2^8 = 256 levels means round off won't be too much and wd result in more accurate recovery of signal...

is it possible, to show with an example(like some diagroam or something??) :s i know this is an easy part but i dont seem to get it :$

 

[badrobot14]

is it possible, to show with an example(like some diagroam or something??) :s i know this is an easy part but i dont seem to get it :$

We measure resolution in the terms of the number of bits of resolution. For example, a 1-bit resolution would only allow two (two to the power of one) values – zero and one. A 2-bit resolution would allow four (two to the power of two) values – zero, one, two and three. If we tried to measure a five volt range with a two-bit resolution, and the measured voltage was four volts, our ADC would return a value of 3 – as four volts falls between 3.75 and 5V.

(Notice I how a 1-bit resolution approx may look vs a 2 bit approx.)

 

[XPFMember]

We measure resolution in the terms of the number of bits of resolution. For example, a 1-bit resolution would only allow two (two to the power of one) values – zero and one. A 2-bit resolution would allow four (two to the power of two) values – zero, one, two and three. If we tried to measure a five volt range with a two-bit resolution, and the measured voltage was four volts, our ADC would return a value of 3 – as four volts falls between 3.75 and 5V.

(Notice I how a 1-bit resolution approx may look vs a 2 bit approx.)

jazak Allahu khairen..
i get it now

 

[badrobot14]

jazak Allahu khairen..
i get it now

Wa iyyakum....

 

[XPFMember]

Assalamoalaikum wr wb!

9702_w09_qp_41.pdf
Q:7 c ii

Can someone show me how exactly will the diagram (graph) look like?

 

[Mustehssun Iqbal]

What happened to this thread:
http://www.xtremepapers.com/community/threads/chemistry-biology-physics-applications.11156/ ??