Great video. I've spent a while looking around the internet to help with an assignment I have and after watching yours I was finally able to figure it out. You have a good way of explaining things.
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Hi. Pot’s are usually used in a device to vary the voltage to different parts of the device. See example near the end of the video (volume control with a preamplifier/power amplifier). In audio-visual systems, rotary or linear potentiometers are often used to control volume, tone, balance, brightness, etc. (though this is less true in digital circuits). In servomechanisms, potentiometers can be used to monitor the position of an object (by attaching the object to the sliding contact).
The output voltage (across R1 say) is given by: Output voltage = (R1/(R1+R2) x Input voltage. (this is the potential divider formula as explained in the video). Providing R1 and R2 are fixed, then R1/(R1+R2) is constant. This means the output voltage is proportional to the input voltage. E.g. if you double the input voltage, you will double the output voltage.
Hi. The video is correct. If the voltmeter were connected between the TOP of the red wire and the slider, the voltmeter would give the voltage across the upper (30cm) section of wire. This would be 2.0V x (30cm/100cm) = 0.6V. But in the video, he voltmeter is connected between the slider and the BOTTOM of the red wire, i.e. across the lower (70cm) section of wire. The reading is therefore 2.0V x (70cm/100cm) = 1.4V (Note that the 2 voltages add up to the supply voltage: 0.6V + 1.4V = 2.0V.)
Hi. Good question. A 'good' voltmeter has a very high impedance (resistance), so only a negligible curent flows through it. The same applies to a 'good' amplifier. If the input imepedance (of voltmeter or amplfiier) are small, this is no longer true but this doesn't prevent the operation of the potentiometer, but makes it non-linear in that the voltage from it is not proportional to the slider's distance from the end. This is not usually important. My video on potential dividers might help.
The battery's voltage is exactly divided in 2 because the resistors (each half of the wire) are equal. The resistors could be 100Ω or 999Ω each for example - it makes no difference. This is because the resistors have the same current (as they are in series). I and R are the same for both halves of the of the wire. V=IR, so V is the same for both halves. Since the 2 voltages must add up to 2volts, each part has 1volt. Increasing R would reduce I, so IxR remains 1volt. Hope that makes sense.
A low resistance potentiometer wire (PW) would produce a large current, which causes overeating of the power supply and the PW. It is basically a short-circuit, so it is avoided. There are also technical reasons for avoiding a low resistance PW (to minimise the effect of slider contact resistance and resistance-change due to heating). You CAN have a low resistance, but usually a small current is adequate. Potentiometers vary from a few ohms to millions of ohms depending on the application.
Q1 (Volume control) .If: - the volume control is a simple pot’ between a mic’ and preamp, and; - the input impedance of the preamp is much larger than the mic’s, and; - by ‘ground’ you mean the common connection (usually grounded for safety and noise reduction); then, in answer to your question, yes. But the current is tiny. It is generally better to think in terms of voltages at the input stage.
The syllabus is easily found by doing a search on “OCR Physics A Specification”. RUclips blocks links in comments, so I can’t send you it. In section 2.3.2, potentiometers are mentioned in passing. But potential dividers ARE expected. Since a potentiometer is simply a type of potential divider, I would advise being familiar with it, if only as a way of helping your understanding of potential dividers. It is well worth looking through your syllabus by the way!
To TheMichaeljohnny. The cell (or battery) can be either way around. It would only matter if the polarity (which way around + and -are) was important, e.g. if the circuit included a diode.
To ZerkosXD. Yes - GCE physics isn't as hard as people make out. But students with weak maths, tend to find it hard going. When choosing GCE subjects, you need to think ahead to what you want to do after the GCEs are over. But physics is a great choice if you’re planning to do anything technical/scientific.
Q4 (Impedance matching for guitar.) Maybe this will help.. Referring to high and low impedances is not clear - actual figures are needed, A guitar pickup’s impedance is quite high - typically around 50kΩ. It should really be connected to a much higher input impedance amplifier (say 500kΩ) to minimise voltage-loss (see answer to Q3). Many amplifiers do not have a high enough input impedance to suit a guitar pickup, so you then use a DI unit.
Hi. I think you have asked 4 different questions. I’m no expert on audio systems but I’ve tried to answer. I’ve spread the answers over several messages due to message-size limits.
Hi. Unfortunately I have a large backlog of videos I need to make, but lack of time is preventing me from working on them. However, if you search RUclips for ‘emf potentiometer’ you will find a couple of videos that explain it.
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Hi, do I have your permission to use some of the concept described in this video for a video on controller aiming? I will put citations in the description and a link to this video if you would like!
Since we assume the current flows form positive to negative, the in you example circuit, the current will only flow through the 30cm part and will not flow through the 70cm part. So shouldn't the voltage be: 2*0.3 instead of 2*0.7?
+Mohammed Ali You asked about the question at about 5 minutes in. A voltmeter has a very high (ideally infinite) resistance. That means negligible current (basically zero current) flows through the voltmeter. Therefore all the current flowing through the 30cm section of wire also flows through the 70cm section. The current flows from the cell's positive terminal, through the 30cm length, then through the 70cm length, then back to the cell's negative terminal. No current flows through the voltmeter. The 30cm and 70cm wire sections are simply 2 resistors in series carrying the same current. Pretend the 2 resistors are 3Ω and 7Ω for example. Total resistance = 3+7 = 10Ω. Current I = emf/R = 2/10 = 0.2A Voltage across 3Ω = IR = 0.2x3 = 0.6V Voltage across 7Ω = IR = 0.2x7 = 1.4V It gets more complicated if the voltmeter (or anything else in the same place) doesn't have a very large resistance. In this case the current splits. Some goes through the voltmeter and the rest goes through the 70cm wire section. But these two currents then combine and flow back to the cell's negative terminal. Hope that helps.
@@alexandrucoca9770 Well I'm doing OK for myself I guess haha. I studied Electrical engineering at uni. I'm now an engineer now for the railways. If you want to get rich do computer science
Q2 (Poweramp) Yes, But it’s more usual to think about poweramps in terms of power-gain (output power /input power) than in terms of current-gain. Q3 (Voltage transfers from lo-Z to hi-Z) Yes. Low-Z to high Z minimises voltage loss, because the current is kept small by the high total impedance. By keeping the current small, voltage losses are minimised (V=IR or V=IZ; small I means small V).
There are different types of DI unit but the simplest is a transformer. If you want to understand how it works, you will have to search on ‘impedance matching transformer’ as the explanation won't fit here and needs some maths.
An exam question: '' For a single turn rotational pot. What is the resolution of this angular displacement measurement method if the outputs range is 0.5 to 2.5V?''
+Andreas G. That's a poor exam question as the 'resolution' of a potentiometer depends on its design and construction (grain-size of the conductive layer, wiper size).I think the examiners may be asking about precision, not resolution. But I’m guessing. The range (0.5V to 2.5V) has end-points specified to a precision of 0.1V, so the precision of a reading should be taken as 0.1V.In angular terms (assuming 1 full rotation of 360º is possible), 360º is equivalent to (2.5V - 0.5V=) 2.0V. So by proportion 0.1V is equivalent to (0.1/2.0) x 360 = 18º. So possible answers could be 0.1V or 18º. Apart from that, I can’t guess what the question is asking for.
At 03:05, you said that when the Slide is positioned at the top, middle, and bottom, the Voltmeter reads 2.0, 1.0, and 0.0, respectively. So, at 05:20, shouldn’t the answer be 0.6 V?
Can you please explain to me how the number of output voltage increases, when i increase the number of input voltage in a potential divider circuit (i need scientific explanation ).........
The voltage across the 70cm section is 2 x 70/100 volts. It’s just like having a 2V supply connected in series with 30Ω and 70Ω resistors. The voltage across the 30Ω resistor is 2 x 30/100 volts. The voltage across the 70Ω resistor is 2 x 70/100 volts.
@@Steve4Physics Hello Sir, Thank you for your reply i am close to understanding what you are saying ,my last question to you was that ,the electron are not supposed to travel throw that 70 cm resistive write that why are we comparing it with a 70 Ohm resistor ?
You said “the electrons are not supposed to travel through that 70 cm resistive wire” Oh yes they are! The electrons travel through the entire 100cm wire - not through the voltmeter because the (ideal) voltmeter has infinite resistance. To help explain, we can - for convenience - *pretend* that each 1cm of wire has a resistance of 1Ω. The 100cm wire’s resistance would then be 100Ω. The wire can then be thought of as 30Ω and 70Ω resistors in series. First make sure you undestand the potential divider.
@@Steve4Physics Thank you so much ,Sir .You have been a Life Saver to me and Always will be .I've searching for the answer everywhere ,My exam is in a week and i started pretty late in October when i decided to give the exam ,i had an year study gap in between due to some financial reason so my concept got a bit blurry ,THANK YOU SO MUCH ,SIR.
Can't believe a 13 year old video is helping me this much. Thanks mate 👌
Thank you!! There aren’t that many videos on potentiometer for a levels on RUclips so this was very useful!!
Great video. I've spent a while looking around the internet to help with an assignment I have and after watching yours I was finally able to figure it out. You have a good way of explaining things.
ok.
what are you doing as a job now
hello bro pls tell me if physics a level make u rich??????????????????????????????????????????????????????????????????????????????????????????????????????????????????????
Hi. Pot’s are usually used in a device to vary the voltage to different parts of the device. See example near the end of the video (volume control with a preamplifier/power amplifier). In audio-visual systems, rotary or linear potentiometers are often used to control volume, tone, balance, brightness, etc. (though this is less true in digital circuits). In servomechanisms, potentiometers can be used to monitor the position of an object (by attaching the object to the sliding contact).
Absolutely amazing video, have been looking for the 2 resistor analogy everywhere (3:44). That made everything click together. I appreciate it.
same :-) 3 days till my paper 1 !!
The output voltage (across R1 say) is given by:
Output voltage = (R1/(R1+R2) x Input voltage.
(this is the potential divider formula as explained in the video).
Providing R1 and R2 are fixed, then R1/(R1+R2) is constant. This means the output voltage is proportional to the input voltage. E.g. if you double the input voltage, you will double the output voltage.
Hi. The video is correct.
If the voltmeter were connected between the TOP of the red wire and the slider, the voltmeter would give the voltage across the upper (30cm) section of wire. This would be 2.0V x (30cm/100cm) = 0.6V.
But in the video, he voltmeter is connected between the slider and the BOTTOM of the red wire, i.e. across the lower (70cm) section of wire. The reading is therefore 2.0V x (70cm/100cm) = 1.4V
(Note that the 2 voltages add up to the supply voltage: 0.6V + 1.4V = 2.0V.)
mashaallah thank you very much for your contribution to the society with your useful lessons
Thanks for making this video :) I had missed the lesson on this in college, but I feel like I understand the topic now
I'm doing this in 8th grade
THIS VIDEO IS REALLY NICE LIKE YOUR ANSWER ON YAHOO.BOTH HELPS A LOT.THANKS.
U are the best Teacher ever
Hi. Good question. A 'good' voltmeter has a very high impedance (resistance), so only a negligible curent flows through it. The same applies to a 'good' amplifier.
If the input imepedance (of voltmeter or amplfiier) are small, this is no longer true but this doesn't prevent the operation of the potentiometer, but makes it non-linear in that the voltage from it is not proportional to the slider's distance from the end. This is not usually important.
My video on potential dividers might help.
Thanks very much I understand this topic fully now I thought it was difficult I appreciate you thank you🙌
Hi cpdp999. Unfortunately time is limited. I will eventually do some waves videos but not for a while I'm afraid.
The battery's voltage is exactly divided in 2 because the resistors (each half of the wire) are equal.
The resistors could be 100Ω or 999Ω each for example - it makes no difference.
This is because the resistors have the same current (as they are in series). I and R are the same for both halves of the of the wire. V=IR, so V is the same for both halves. Since the 2 voltages must add up to 2volts, each part has 1volt.
Increasing R would reduce I, so IxR remains 1volt.
Hope that makes sense.
A low resistance potentiometer wire (PW) would produce a large current, which causes overeating of the power supply and the PW. It is basically a short-circuit, so it is avoided.
There are also technical reasons for avoiding a low resistance PW (to minimise the effect of slider contact resistance and resistance-change due to heating).
You CAN have a low resistance, but usually a small current is adequate. Potentiometers vary from a few ohms to millions of ohms depending on the application.
Q1 (Volume control)
.If:
- the volume control is a simple pot’ between a mic’ and preamp, and;
- the input impedance of the preamp is much larger than the mic’s, and;
- by ‘ground’ you mean the common connection (usually grounded for safety and noise reduction);
then, in answer to your question, yes.
But the current is tiny. It is generally better to think in terms of voltages at the input stage.
very educative. Thank you Regards
The syllabus is easily found by doing a search on “OCR Physics A Specification”.
RUclips blocks links in comments, so I can’t send you it.
In section 2.3.2, potentiometers are mentioned in passing. But potential dividers ARE expected. Since a potentiometer is simply a type of potential divider, I would advise being familiar with it, if only as a way of helping your understanding of potential dividers.
It is well worth looking through your syllabus by the way!
appreciate the work you put into this video
Good explanation
To TheMichaeljohnny.
The cell (or battery) can be either way around. It would only matter if the polarity (which way around + and -are) was important, e.g. if the circuit included a diode.
I finally get it. Thank you for this video!
To ZerkosXD. Yes - GCE physics isn't as hard as people make out. But students with weak maths, tend to find it hard going. When choosing GCE subjects, you need to think ahead to what you want to do after the GCEs are over. But physics is a great choice if you’re planning to do anything technical/scientific.
Q4 (Impedance matching for guitar.)
Maybe this will help.. Referring to high and low impedances is not clear - actual figures are needed,
A guitar pickup’s impedance is quite high - typically around 50kΩ. It should really be connected to a much higher input impedance amplifier (say 500kΩ) to minimise voltage-loss (see answer to Q3).
Many amplifiers do not have a high enough input impedance to suit a guitar pickup, so you then use a DI unit.
Ohh, I see. We gotta note at which point in the circuit the encircled V is connected. Got it, thanks.
thanks a lot. u are one good tutor keep it up.
Thank you! Very helpful.
Very useful... Thanks from Palestine
Thank you for this video!!
Hi. I think you have asked 4 different questions. I’m no expert on audio systems but I’ve tried to answer. I’ve spread the answers over several messages due to message-size limits.
Would it be possible for you to do any vides on waves?
Hi. Unfortunately I have a large backlog of videos I need to make, but lack of time is preventing me from working on them. However, if you search RUclips for ‘emf potentiometer’ you will find a couple of videos that explain it.
Very informative, thanks!
Great vid. Thank you sir
For the OCR A physics exam do you know wheather I will need to know about this?
Great, this really helped - one thing, are we changing the resistance in order to vary the voltage being input to a device?
if you are still there What did you become now ? Haha
@@gardenmenuuu a jazz musician LOL
@@annachaimusic really ?what does that mean
@@gardenmenuuu Jazz is a genre of music and this person makes music (as a profession) in this genre
@@annachaimusic musician here too😂Taking physics... Studying psychology in a year or two
Idk what I'm doing in life
How are you?
THANK YOU MR STEVE
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Thanks for this! Very useful
Pretty good! Very good explanation!
hello bro pls tell me if physics a level make u rich??????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????
thank you!
Hi, do I have your permission to use some of the concept described in this video for a video on controller aiming? I will put citations in the description and a link to this video if you would like!
Yes. Please feel free to use any of the material. A brief citation/link would be appreciated. Steve.
Since we assume the current flows form positive to negative, the in you example circuit, the current will only flow through the 30cm part and will not flow through the 70cm part. So shouldn't the voltage be: 2*0.3 instead of 2*0.7?
+Mohammed Ali
You asked about the question at about 5 minutes in.
A voltmeter has a very high (ideally infinite) resistance. That means negligible current (basically zero current) flows through the voltmeter.
Therefore all the current flowing through the 30cm section of wire also flows through the 70cm section.
The current flows from the cell's positive terminal, through the 30cm length, then through the 70cm length, then back to the cell's negative terminal. No current flows through the voltmeter.
The 30cm and 70cm wire sections are simply 2 resistors in series
carrying the same current.
Pretend the 2 resistors are 3Ω and 7Ω for example. Total resistance =
3+7 = 10Ω.
Current I = emf/R = 2/10 = 0.2A
Voltage across 3Ω = IR = 0.2x3 = 0.6V
Voltage across 7Ω = IR = 0.2x7 = 1.4V
It gets more complicated if the voltmeter (or anything else in the same
place) doesn't have a very large resistance. In this case the current
splits. Some goes through the voltmeter and the rest goes through the
70cm wire section. But these two currents then combine and flow back
to the cell's negative terminal.
Hope that helps.
@@Steve4Physics thanks so much that makes sense
excellent teaching
hello bro pls tell me if physics a level make u rich??????????????????????????????????????????????????????????????????????????????????????????
@@alexandrucoca9770 Well I'm doing OK for myself I guess haha. I studied Electrical engineering at uni. I'm now an engineer now for the railways.
If you want to get rich do computer science
thanks bro :) great video
Q2 (Poweramp)
Yes, But it’s more usual to think about poweramps in terms of power-gain (output power /input power) than in terms of current-gain.
Q3 (Voltage transfers from lo-Z to hi-Z)
Yes. Low-Z to high Z minimises voltage loss, because the current is kept small by the high total impedance. By keeping the current small, voltage losses are minimised (V=IR or V=IZ; small I means small V).
Hi.
How do we measure the emf of a battery by using a potentiometer?
Would be kind if you illustrate it with a video.
hello bro pls tell me if physics a level make u rich??????????????????????????????????????????????????????????????????????????????????????????
Great! Thanks!
There are different types of DI unit but the simplest is a transformer. If you want to understand how it works, you will have to search on ‘impedance matching transformer’ as the explanation won't fit here and needs some maths.
An exam question: '' For a single turn rotational pot. What is the resolution of this angular displacement measurement method if the outputs range is 0.5 to 2.5V?''
+Andreas G.
That's a poor exam question as the 'resolution' of a potentiometer depends on its design and construction (grain-size of the conductive layer, wiper size).I think the examiners may be asking about precision, not resolution. But I’m guessing. The range (0.5V to 2.5V) has end-points specified to a precision of 0.1V, so the precision of a reading should be taken as 0.1V.In angular terms (assuming 1 full rotation of 360º is possible), 360º is equivalent to (2.5V - 0.5V=) 2.0V. So by proportion 0.1V is equivalent to (0.1/2.0) x 360 = 18º.
So possible answers could be 0.1V or 18º.
Apart from that, I can’t guess what the question is asking for.
Amazing
Hi steve, i have a question: why do you want to use a high resistance wire if you want current to flow through the wire?
hello bro pls tell me if physics a level make u rich??????????????????????????????????????????????????????????????????????????????????????????
At 03:05, you said that when the Slide is positioned at the top, middle, and bottom, the Voltmeter reads 2.0, 1.0, and 0.0, respectively.
So, at 05:20, shouldn’t the answer be 0.6 V?
No, because in the one it was a potential difference across the wire and the other it was another "cell"
Can you please explain to me how the number of output voltage increases, when i increase the number of input voltage in a potential divider circuit (i need scientific explanation ).........
aren't the formulae supposed to be 2×(30/100) selections through the wire travelling less distance?
The voltage across the 70cm section is 2 x 70/100 volts. It’s just like having a 2V supply connected in series with 30Ω and 70Ω resistors. The voltage across the 30Ω resistor is 2 x 30/100 volts. The voltage across the 70Ω resistor is 2 x 70/100 volts.
@@Steve4Physics Hello Sir, Thank you for your reply i am close to understanding what you are saying ,my last question to you was that ,the electron are not supposed to travel throw that 70 cm resistive write that why are we comparing it with a 70 Ohm resistor ?
You said “the electrons are not supposed to travel through that 70 cm resistive wire”
Oh yes they are!
The electrons travel through the entire 100cm wire - not through the voltmeter because the (ideal) voltmeter has infinite resistance.
To help explain, we can - for convenience - *pretend* that each 1cm of wire has a resistance of 1Ω. The 100cm wire’s resistance would then be 100Ω. The wire can then be thought of as 30Ω and 70Ω resistors in series.
First make sure you undestand the potential divider.
@@Steve4Physics Thank you so much ,Sir .You have been a Life Saver to me and Always will be .I've searching for the answer everywhere ,My exam is in a week and i started pretty late in October when i decided to give the exam ,i had an year study gap in between due to some financial reason so my concept got a bit blurry ,THANK YOU SO MUCH ,SIR.
@@boboom4657 You are welcome. I wish you good luck in your exam!
i think u should place the battery of known Emf E in opposite direction..isn't it????
Great!!!!!
In practice if you use a potentiometer to divide voltage you'll burn it
Super !
hello bro pls tell me if physics a level make u rich??????????????????????????????????????????????????????????????????????????????????????????
A volt meter has infinite resistance hence no current flow through it
plz recomend easy book for these electricity topics
I know it's a long time ago, but the new syllabus, Cambridge AS and A level book explains it well
@@emilyesnyman Ye think he graduated by now
Gosh, that is really obvious. Sorry, I must have been sleepy or something when I watched this last time.
nice
AS exam
I'm writing this 8 years later and this video is still relevant😂
hello bro pls tell me if physics a level make u rich??????????????????????????????????????????????????????????????????????????????????????????
2018?
2021😂
I like pot