Dude this guy teaches everything I am so astounded. I remember trying to get help for chemistry, then some for calculus, and he is even here in an electrical system and circuit lesson. What can't this guy do?
Clear, detailed and overall. Easy to understand the gist of each topic! I found this channel when I was a freshman. It helped me to understand chemistry. Now I'm a senior. I can't believe I'm here again! He is exactly Mr. Know Everything!
Wow! Before watching this video I watched and read many other tutorials on voltage dividers, but still was confused and had many question. Now I finally understand. Thanks.
At 17:53, does he know that the voltage around R2 is 3V due to the fact that the device needs a voltage of 3V and voltage is the same across parallel branches?
The bigger the internal resistance, the closer it is to the original output voltage. The internal resistance, as he stated, is recommended to be at least a hundred times more than R2.
@@Imkicelee it wouldn't be as close to 3V. If R2 was 5000ohms the total R would be: (1/5000+1/5000)^-1 + 15000 = 17500ohms, wich means the voltage out is: V in * (R2/R1+R2) = 12 * (2500/2500+15000) = 1.71 V and not even close to the 3V we wanted.
For the question at 4:23, it should be written as. A voltage divider circuit, where R2 has an output of 12V, and the circuit has a current of 50mA, using a 20V battery. Find the resistance of R1, and R2.
Just a Word of thank you for your efforts to make this simple, I came across a voltage divider problem in lab and saw the efficiency was terrible with parallel device to R2 , this video made my day thank you from all my heart.
When I have 20vt, and know vo is 12v that tells you to get r1 is 8v (20v-12v) x .05a is 160 ohms. r2 = 12v /.05a =240. You seemed to do more work to get the same answer.
7:40 for this question i solved it slightly different but got the same answer. i did 12/20 times 100 which equals 3/5. i know that R2 must be 3/5 of 400 and that leaves R1 to be 2/5.
Another easier way is that the voltage in series adds up so you can subract the given voltage against the total voltage to find the missing voltage across 1 resistor
@10:51 Why is the numerator Req? If we're finding the voltage across the device, shouldn't the numerator be just 1000 Ohms (resistance of the device)? Similarly, if we were to find the voltage across R2, shouldn't the equation be vout = vin ( R2/R1+R2+R3 ) and not vout = vin (Req/Req + R1)?
at 11:43 , the voltage calculated to be 9.52, that's across the two parallel resistors right? So the voltage drop across R1 is 11.48 since 20-9.52=11.48 right?
At 06:00 Why u considered only 2 resistors to design vdr circuit..why not 3or4? and at 09:04 current must enter in to 100ohm coz..current always chooses low resistance path? If reply..very thankful...
1) VDR circuit only has 2 resistors. It is used to divide the output voltage in any desired ratio from the input voltage . 2) Higher resistance means lower current flow. Lower resistance means higher current flow. Current would still flow through both resistors. Just that there would be a high current flow through the 100ohm resistor.
0. Construct a dual potential divider circuit with a common voltage source as supply. The potential drop should produce an output which is 5 times greater than the cut in voltage of a PN junction diode. Can u please tell us the solution
I have a transformerless LED driver from a China LED Bulb but it's voltage output is more than 300V DC. Can I reduce the V-out to 45V DC using this Voltage Divider Method? Will it be safe to used to turn on several SMD LED in series??
found kind of a hacky way to find the resistances. If you divide 12v/20v you get 0.6 and 0.6x400Ω is 240Ω, so you get r2 from that equation. no clue how that worked out but somehow it did.
I was following with everyone until we go the internal resistor. Which he made it make sense. But my test coming up we can’t use calculators and then I gave up on using my paper of my scratch work😊
For #5, I think the current that being delivered from the battery is not 60.2mA because in his calculation, he did not allow for the the current that being consumed by R1 = 150 ohm.
I know it’s been two years, but for anyone else wondering the same thing. No, he is correct. Current does not drop in a series circuit. So the current before resistor 1 is the same as current after resistor 1
3 beers and a shot of whiskey, watching tube. You make more sense. Definitely better understanding for me , rather than by my highly qualified engineering mates in the pub. Ta.
Good video. I feel that the wording in the outputs voltage and amperage is deceiving. You can get a fixed voltage using this method but the amperage through the output of our new source will depend on the resistance of the load. I think the wording would be better if you stated that the circuit without the load connected in the voltage divider would be (blank)mA
Why didn't you use the 100 ohm from R1 when calculating Req, so that R1 @ 100 ohm will be in series with R2 + R Device to give 190.9 ohms? Was it Req of the entire circuit or just the part around R2 + R Device?
Thanks a lot, this video It has been very useful. On the min: 9:01 Would it be possible to find the value of R2, in case Vin, Vout is unknown? Here is an example of the same circuit, only adding an additional parallel to the resistor r1: Vin = ? Vout = ? R1=100 || R2=2k R3= ? || R4=1k
You need to know more. For example: I build a circuit to turn on an appliance using an LDR (light dependent resistor). The specs said that it has a resistance between 5k and 500K (depending on amount of light). So I choose the minimum to work which is 5k. Next I looked at my relay. It said that it needs minimum 2V to turn on. So no I know that my Vout is 2V. I also know that my R1 the (LDR) is 5k. I also know my Vin because I used a 5V supply. With all this info I was able to build the circuit using this voltage divider here and I turned on the appliance with a flashlight over my LDR resistor. All I had to do was to find R2 to give me the 2V at Vout (which was the relay).
Is Vin referring to the total Voltage? I'm not entirely sure about this but otherwise it wouldn't make sense. Assuming that Vin doesn't refer to Vtotal, then I assume its the voltage "coming in" from the first point we measure it, right before Req. But after assuming this, Vin shouldnt be equal to 20V which is Vtotal since R1 should cause a voltage drop before the starting point that we're measuring. Can someone explain? I feel like I understood part of it but didn't get the main point. **If** my assumption is right, then shouldnt we say Vtotal instead of Vin?
i think the voltage drop accross r1 would be taken across it and not at the point before it. if you measure with a multimeter you should get 12V before the resistor but if measured across r1 you would have 4V if I am not mistaken
@@a_potat Coming back to this, and looking at the exercise in the middle of the video, it's just the way to get the V drop across Req. Same equation is used for V drop across R2. A setup with 3 resistances would be different, unless you can simplify them to 2 as shown in the vid. And of course, when I said voltage at a point, I meant it relative to the ground, or the point with lowest voltage, in this case the negative terminal of the source. When I left that comment I assumed that "Vtotal" and "Vin" are two separate concepts. Doesn't matter what it's called, Vin, or Vtotal, it's the voltage of the source. Why? Cuz the math works..
Normally the equation for resistance of a parallel circuit is 1/(1/R1+1/R2). So as you add more resistors in parallel the current is shared among the resisitors in parralel decreasing the resistance. So this guys just replaced the top level 1/ with ^-1 as they are equivilant.
Hello, Great videos. But I have a question about your calculations for problem 2. If you want the output voltage to be 12V, wouldn't you want r1 to be 240ohms and r2 be 160 ohms?
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Dude this guy teaches everything I am so astounded. I remember trying to get help for chemistry, then some for calculus, and he is even here in an electrical system and circuit lesson. What can't this guy do?
Ya this guy is a complete beast
And he does it brilliantly
He can't excrete stuff and take it back
Guarantee 💩💩🤯🥶
@@youtubeshorts3711 lmao
Haha same I’ve used him for all of these. Great dude
Clear, detailed and overall. Easy to understand the gist of each topic! I found this channel when I was a freshman. It helped me to understand chemistry. Now I'm a senior. I can't believe I'm here again! He is exactly Mr. Know Everything!
Wow! Before watching this video I watched and read many other tutorials on voltage dividers, but still was confused and had many question. Now I finally understand. Thanks.
Though you didn't mention practical examples of voltage divider usage
Orgo tutor teaching me physics---- absolutely brilliant!!!!
I've been watching these videos for years I really appreciate the great work you do!
Am I glad I found this video. I find it refreshing for it to bring me back to the basics. Thank you JG!
from confused to confident, I am grateful for your help!
please also add something about Thevenin theorem
I love this guy :) can y'all watch the ads completely the only way to support him.thanks
Am getting it together here,,God bless u sir
Best tutor ever. Deeply explained for better understanding. Thumbs up👍👍
Bro u are a lifesaver thanks a lot man. Have a nice day
YOU MADE THIS LOOK SO EASY TYSMM!
At 17:53, does he know that the voltage around R2 is 3V due to the fact that the device needs a voltage of 3V and voltage is the same across parallel branches?
This video series is awesome! Very specific and helpful problems for real world examples. Thank you for sharing.
Really is it awesome? What resistors did he use in his first example? He just puts 10 and 20 are that doesn't tell anybody anything
Lagi akong nanunuod ng mga posted video mo, salamat ( I've always watch your posted videos, thankyou) ☺
Cant explain how helpful this channel is
God of mathematics and science
aw jeez you explained it far better than my physics teacher
I so happy this video is so much helpful
Thank you so much for this class i got it now keep the good work teacher
For #5 how does he know that R2 should be approx 100times less to get 3V?
The bigger the internal resistance, the closer it is to the original output voltage. The internal resistance, as he stated, is recommended to be at least a hundred times more than R2.
Universal Narcissist what about if the internal resistance wasn’t as big?
@@Imkicelee it wouldn't be as close to 3V. If R2 was 5000ohms the total R would be: (1/5000+1/5000)^-1 + 15000 = 17500ohms, wich means the voltage out is: V in * (R2/R1+R2) = 12 * (2500/2500+15000) = 1.71 V and not even close to the 3V we wanted.
For the question at 4:23, it should be written as.
A voltage divider circuit, where R2 has an output of 12V, and the circuit has a current of 50mA, using a 20V battery. Find the resistance of R1, and R2.
You are required to design the circuit..
We really love you sir
Your a real one for this
You all the tutorials so easy....even with the math in between..
Thank you
Amazing!! Thank you!
Just a Word of thank you for your efforts to make this simple, I came across a voltage divider problem in lab and saw the efficiency was terrible with parallel device to R2 , this video made my day thank you from all my heart.
When I have 20vt, and know vo is 12v that tells you to get r1 is 8v (20v-12v) x .05a is 160 ohms. r2 = 12v /.05a =240.
You seemed to do more work to get the same answer.
Thanks for this
Thanks tomorrow is my exam😎
2:11 BREAKING NEWS: THE LEGENDARY ORGANIC CHEMISTRY TUTOR HAS USED A CALCULATOR
Thank you so much! Sir your lectures help me alot
In question 5....how did you get R2 as 50 ohms
7:40 for this question i solved it slightly different but got the same answer. i did 12/20 times 100 which equals 3/5. i know that R2 must be 3/5 of 400 and that leaves R1 to be 2/5.
You are an excellent teacher. Thanks.
4:10 if u want to decrease the voltage why do the resistors have to be the same? thank you so much xx
thats to divide in half, they have to be the same to divide in half
Another easier way is that the voltage in series adds up so you can subract the given voltage against the total voltage to find the missing voltage across 1 resistor
why did you place the internal resistor in parallel and not series @8:39? I mean what's the thought process behind it?
i think bcz it said across it
thanks for including CURRENT!!! i blew my fuse on my multimetre so i couldnt find out what it was...
@10:51 Why is the numerator Req? If we're finding the voltage across the device, shouldn't the numerator be just 1000 Ohms (resistance of the device)? Similarly, if we were to find the voltage across R2, shouldn't the equation be vout = vin ( R2/R1+R2+R3 ) and not vout = vin (Req/Req + R1)?
Parallel Circuits merge with whatever Series Circuits they're apart of, so since the 1,000 Ohms Resistor is Parallel to R2, it's considered part of R2
at 11:43 , the voltage calculated to be 9.52, that's across the two parallel resistors right? So the voltage drop across R1 is 11.48 since 20-9.52=11.48 right?
The channnel name is misleading. This teaches everything not justOrganic Chemistry
you lost me @ 17:40 ... Does anyone know the logic behind this other than looking at other examples?
At 06:00 Why u considered only 2 resistors to design vdr circuit..why not 3or4? and at 09:04 current must enter in to 100ohm coz..current always chooses low resistance path? If reply..very thankful...
1) VDR circuit only has 2 resistors. It is used to divide the output voltage in any desired ratio from the input voltage .
2) Higher resistance means lower current flow. Lower resistance means higher current flow. Current would still flow through both resistors. Just that there would be a high current flow through the 100ohm resistor.
@@samsungnote2683 thankyou
Why is his voice so *SOOTHING* ?! 😍
😂this guy is literally saving my electrical engineering degree
Also mine
according to kvl, the voltage exiting the circuit should be 0 but at 4:08 you said that the voltage output is 6v. Should not 6v be between R1 and R2?
Nice😎 helpfull for me
Thank you sir. You are just so good.
The largest resistor in the voltage divider will drop More Voltage. That resistor can get very very hot! So watch your power rating of that resistor.
You are awesome. Finally I understand. Please make a video of problems of strain gauge.
I'm here again. Thanks for sharing it with us daddy
Thank you for all your content!
Thank you so much.
On the 5th problem how did you know that R2 should be 100x less than the device resistor?
did you ever get an answer?
Why did you divide r2 by a hundred in 18:02 ?
wow sir you are everywhere..
If a ckt contains r1,r2,r3,r4,and ,r5 resistors then what resistor value we used as numanetor
0. Construct a dual potential divider circuit with a common voltage source as supply. The
potential drop should produce an output which is 5 times greater than the cut in voltage of a
PN junction diode.
Can u please tell us the solution
good video
I have a transformerless LED driver from a China LED Bulb but it's voltage output is more than 300V DC. Can I reduce the V-out to 45V DC using this Voltage Divider Method? Will it be safe to used to turn on several SMD LED in series??
found kind of a hacky way to find the resistances.
If you divide 12v/20v you get 0.6
and 0.6x400Ω is 240Ω, so you get r2 from that equation.
no clue how that worked out but somehow it did.
I was following with everyone until we go the internal resistor. Which he made it make sense. But my test coming up we can’t use calculators and then I gave up on using my paper of my scratch work😊
Hello sir . Sir i cant beleive this because you can solve or know about every topic .from every subjects how you can do this
bhai he is an all rounder
For #5, I think the current that being delivered from the battery is not 60.2mA because in his calculation, he did not allow for the the current that being consumed by R1 = 150 ohm.
I know it’s been two years, but for anyone else wondering the same thing. No, he is correct. Current does not drop in a series circuit. So the current before resistor 1 is the same as current after resistor 1
hi thank you for the information you din't show us how you calculated Resistor equivalent on qsn 4
Really nice lessons.
4:33 but the output to the load on the divider is 50mA and not from the battery. I think that is the real practical problem
3 beers and a shot of whiskey, watching tube. You make more sense. Definitely better understanding for me , rather than by my highly qualified engineering mates in the pub. Ta.
Good video. I feel that the wording in the outputs voltage and amperage is deceiving. You can get a fixed voltage using this method but the amperage through the output of our new source will depend on the resistance of the load. I think the wording would be better if you stated that the circuit without the load connected in the voltage divider would be (blank)mA
Are voltage divisor and current divisor concepts used in capacitors, batteries etc ? Just like in resistors
Please what app did u use in writing and drawing the circuits......it made explanations easier....let me know the name of the app please
shouldnt number 4 be R2/90.9+R2? if we want to solve the voltage across R2 specifically?
So where is the required voltage taken from? Across r2? Would you connect a component across it?
yes + across and - to negative
I hope I understand you correctly the voltage is what you select you choose a voltage and you choose the current then you figure out the resistor size
why is the current in milliamps instead of just amps at 21:30
im on reading week and totally forgot i had an assignment and had a massive brain fart, i could kiss you! thanks bro!
Why didn't you use the 100 ohm from R1 when calculating Req, so that R1 @ 100 ohm will be in series with R2 + R Device to give 190.9 ohms? Was it Req of the entire circuit or just the part around R2 + R Device?
Thanks a lot, this video It has been very useful.
On the min: 9:01 Would it be possible to find the value of R2, in case Vin, Vout is unknown?
Here is an example of the same circuit, only adding an additional parallel to the resistor r1:
Vin = ?
Vout = ?
R1=100 || R2=2k
R3= ? || R4=1k
Usefull tips
Does that mean that if I want to limit the current I should add R2 again?
thanks you legend
What if we had more resistances could we still do (r1/(r1+r2+r3)) or we’d need to do it two at a time
can just add more
Thank u for the vedio, but i can't understand how we decide resistors for circuit with knowing Am .
You need to know more. For example: I build a circuit to turn on an appliance using an LDR (light dependent resistor). The specs said that it has a resistance between 5k and 500K (depending on amount of light). So I choose the minimum to work which is 5k. Next I looked at my relay. It said that it needs minimum 2V to turn on. So no I know that my Vout is 2V. I also know that my R1 the (LDR) is 5k. I also know my Vin because I used a 5V supply. With all this info I was able to build the circuit using this voltage divider here and I turned on the appliance with a flashlight over my LDR resistor. All I had to do was to find R2 to give me the 2V at Vout (which was the relay).
Is Vin referring to the total Voltage? I'm not entirely sure about this but otherwise it wouldn't make sense. Assuming that Vin doesn't refer to Vtotal, then I assume its the voltage "coming in" from the first point we measure it, right before Req. But after assuming this, Vin shouldnt be equal to 20V which is Vtotal since R1 should cause a voltage drop before the starting point that we're measuring. Can someone explain? I feel like I understood part of it but didn't get the main point. **If** my assumption is right, then shouldnt we say Vtotal instead of Vin?
i think the voltage drop accross r1 would be taken across it and not at the point before it. if you measure with a multimeter you should get 12V before the resistor but if measured across r1 you would have 4V
if I am not mistaken
@@a_potat Coming back to this, and looking at the exercise in the middle of the video, it's just the way to get the V drop across Req. Same equation is used for V drop across R2. A setup with 3 resistances would be different, unless you can simplify them to 2 as shown in the vid. And of course, when I said voltage at a point, I meant it relative to the ground, or the point with lowest voltage, in this case the negative terminal of the source.
When I left that comment I assumed that "Vtotal" and "Vin" are two separate concepts. Doesn't matter what it's called, Vin, or Vtotal, it's the voltage of the source. Why? Cuz the math works..
Thank u
I can u dis in my modification circuit
So nice
why did we raise it to power minus 1 at 10:27
Normally the equation for resistance of a parallel circuit is 1/(1/R1+1/R2). So as you add more resistors in parallel the current is shared among the resisitors in parralel decreasing the resistance. So this guys just replaced the top level 1/ with ^-1 as they are equivilant.
I saw this after my exam 😄
how should calculate input voltage
Thankyou bro
If multimeter shows Vin with respect to Vout to be a lesser value than expected what could be the reason?
Nice one
Please how do you calculate for the input and outputt
Thanks
love you sir...
I don't know if I can get clarification on how the external resistance across R2 is 50 ohms while the internal resistance across it is 5000 ohms.
bro reply i will explain
Why was question four raised to the -1 power?
Legend!
Is this For IGCSE or A-levels?
Hello, Great videos. But I have a question about your calculations for problem 2. If you want the output voltage to be 12V, wouldn't you want r1 to be 240ohms and r2 be 160 ohms?
I thought so too.