Here's another approach to the voltage divider problem. Step 1. Find the current. I = V/Rseries = 6/2+6 = 6/8 = 0.75 Amps Step 2. Use Ohm's Law(V = IR) to find the voltage drop or voltage difference for each resistor. V1 is the voltage drop for R1 and V2 is the voltage drop for R2. V1 = IR1 (2a) = (0.75A)(2ohms) V1 = 1.5V V2 = IR2 (2b) = (0.75Amps)(6ohms) V2 = 4.5V Step 3. Kirchhoffs Voltage Law states that the sum of the voltage differences in a given circuit must be equal to zero. Use Kirchhoffs Law determine if the circuit is valid. We have a 6V battery connected to two resistors in parallel. The first resistor has a resistance of 2 ohms and the second has a resistance of 6 ohms. When the current from the battery travels across R1, the voltage drop is 1.5V (6V - 1.5V = 4.5V). Reference 2a in step 2 for math. When the current runs across R2, the voltage drop is 4.5V(4.5 - 4.5 = 0). Reference 2b in step 2 for math. Since we started with 6V and the circuit "used" all of it, we have determined that this circuit does follow Kirchhoff's Law.
@@Newbport849 Eh whatever, you already have names like "AC/DC", which is actually an abbreviation for "alternating current/direct current". Also, with "division" I don't just mean "inverse multiplication", I also mean "team" or "group".
I don't think this one was done by khan, the voice is different. but yeah khan can teach a lot of subjects very well. physics, chemistry and especially math. I don't know about other areas like computer science but if I'm not mistaken he read computer science at Stanford so im sure he makes great videos of cs too. However most of the videos are high school or junior college level only I suppose, I don't think you'll find anything higher than that. They aren't experts in everything :)
When V out is used to power the gate for example on a small transistor that barely sinks 10 milliamps its not very relative to speak of. But when V Out is sinking current into a substantial load of for example Amps. Now you have the current passing thru R2 and whatever is in parallel to R2 (load on V out) which is completely relevant to the current passing thru R1. So if my memory serves me right you Thevenize R2 and the Load attached to V Out and use the parallel value of the two as the actual number in the current equation. Usually when I want to calculate how many volts are going to drop across a voltage divider I just use the percentage ratio and multiply the voltage by that number. But I remember using this method and its accepted and easy to understand. 8)
Thank you. I need videos on Engineering Electronics circuits too. Stuffs including frequency response. Transistor biasing. Small and large signals. IC 555 timer. Oscillators etc.
you sometimes referred to v_in and v_out as "v one" and "v zero" and you wrote 6 6/8 as in it's a mixed fraction instead of 6 * 6/8, made me less confident that you're focused on what you're saying but overall good video and I understood the concept, thank you ^^
If a resistor is connected across the battery for a long time, then won't it drain off the battery/cell. Can you please let me know ? (3.7V cell 4.7K X 2 connected across as shown in this video)
Is this same for parallel circuits? i saw my teacher do this a different way wtf his way was inverted the other way, both add up on top and at the bottom the divider is the voltage in R that youre trying to find
I know it's been 2 years, but I'm answering just in case someone will also need an explanation regarding what you asked, so here it is : 1)Why is V(out) only affected by one resistor ? Because it's only applied across one resistor. If it was affected by both resistors, then you'd get the exact same voltage as you applied V(in). 2)V(out) in this case is a predictable fraction of V(in) as the output V(out). Roughly speaking, you take an applied voltage V(in) and make it smaller as an output V(out) using a divider.
6:12 if i wanna know the voltage for R2 i would just go 6V/(R1+R2) X R2 Its easier in my head that way. First i divide the voltage over the total resistance so now its spread out evenly, and i have like a base value i guess and then i multiply it by the resistor value i want t find the voltage for and i get the right answer. Can someone tell me if there is anything wrong with doing it this way?
Okey we know that in order to get the Vout we need to pick R2 as our reference and it would be i = Vo / R2. Because R2 at the below. But what about the sensor that can change the value of resistance and its installed at the R1.. how do we know the value of Vo for that? Can u please help me out?
These lessons have been very beneficial for me but this one needs redoing please. There's little to no explanation of anything. It's all a bit random. The math at 4:00 made no sense to me.
@@Technovore88 yeah perhaps i have a gap in my math skills somewhere but how vout/r2 = vin/(r1+r2) then goes to vout = vin * r2/(r1+r2) makes no sense to me I get hes multiplying both sides by r2 but his answer is confusing and he didn't explain it at all
@@HamedAdefuwa Gotcha, I do agree it would have helped if he had stated the given current in the circuit so it could be proof checked with ohm's law. Just be clear you have grasp of the mathematics up to his actual answer of 4.5 volts for the most part?
umm so i see that this video is for college students pursuing electrical engineering... but but but i am a student of grade 8 nd studying it..so unfair!!!!
Here's another approach to the voltage divider problem.
Step 1.
Find the current.
I = V/Rseries
= 6/2+6
= 6/8 = 0.75 Amps
Step 2.
Use Ohm's Law(V = IR) to find the voltage drop or voltage difference for each resistor.
V1 is the voltage drop for R1 and V2 is the voltage drop for R2.
V1 = IR1
(2a) = (0.75A)(2ohms)
V1 = 1.5V
V2 = IR2
(2b) = (0.75Amps)(6ohms)
V2 = 4.5V
Step 3.
Kirchhoffs Voltage Law states that the sum of the voltage differences in a given circuit must be equal to zero.
Use Kirchhoffs Law determine if the circuit is valid.
We have a 6V battery connected to two resistors in parallel.
The first resistor has a resistance of 2 ohms and the second has a resistance of 6 ohms.
When the current from the battery travels across R1, the voltage drop is 1.5V (6V - 1.5V = 4.5V). Reference 2a in step 2 for math.
When the current runs across R2, the voltage drop is 4.5V(4.5 - 4.5 = 0). Reference 2b in step 2 for math.
Since we started with 6V and the circuit "used" all of it, we have determined that this circuit does follow Kirchhoff's Law.
"Voltage division" would be a pretty badass name for a heavy metal group, to be honest.
No, No it wouldn't.
@@Newbport849
Eh whatever, you already have names like "AC/DC", which is actually an abbreviation for "alternating current/direct current".
Also, with "division" I don't just mean "inverse multiplication", I also mean "team" or "group".
@@Peter_1986 Anything engineering has no place in heavy metal.
Newbport you must be fun at parties.
@@JCarrollTV bold of you to assume he gets invited to any
I wish I had this when I was studying in uni. Very useful and informative. Keep up the good work.
I am grateful to you that you are teaching the students with great passion and making the courses easy and interesting.
This is perfect! now I finally now how to calculate and get voltage divider correctly! thanks forever.
Hmmm. This would have been handy last semester... Oh well.
Un-Professional Duck lol same thoughts here...ducky
that what she said
i really have to say that I'm grateful for this video i have been trying to figure it out for the past six hours
Got ya degree yet? Asking for a friend
@@h.k3260 lol i cant believe i clicked on this and reliving the moment. But yes, yes I did. and def not in physics.
Dats doooooooopee
sooooo.....did the video help? :D
does Khan do all these lectures? does he literally know everything?
I don't think this one was done by khan, the voice is different. but yeah khan can teach a lot of subjects very well. physics, chemistry and especially math. I don't know about other areas like computer science but if I'm not mistaken he read computer science at Stanford so im sure he makes great videos of cs too. However most of the videos are high school or junior college level only I suppose, I don't think you'll find anything higher than that. They aren't experts in everything :)
Video is done by Willy MacAllister,
Written in description.
Great! I like when you show the derivations! 🎉
Thanks for the explanation !Trying to understand electronics *-*
When V out is used to power the gate for example on a small transistor that barely sinks 10 milliamps its not very relative to speak of. But when V Out is sinking current into a substantial load of for example Amps. Now you have the current passing thru R2 and whatever is in parallel to R2 (load on V out) which is completely relevant to the current passing thru R1. So if my memory serves me right you Thevenize R2 and the Load attached to V Out and use the parallel value of the two as the actual number in the current equation. Usually when I want to calculate how many volts are going to drop across a voltage divider I just use the percentage ratio and multiply the voltage by that number. But I remember using this method and its accepted and easy to understand. 8)
Couldn't understand the formula that I got, thank you whoever made this.
I really do appreciate this video .
And thanks for telling the case where our assumed current wasn't negligible.
Thankyou khan
Thanks for the video and explanation
can you also make a video on current divider
i0=it(r1/r0+r1)
Thanks for such an easy explanation of this topic
Thank you so much!
It would've been helpful to see an explanation of the math
The video is really useful and understanding
Thank you for sharing. Helps immensely.
if you understand this you also know what a pontentiometer is
I wrote it.
thank you
best and most simple explanations always ◉‿◉
THANKS A LOT
Hey ,shouldnt the arrow direction be up to down for the 6kOhm resistor? because the current is flowing up to down through the resistor right?
excellent
Cool story, bro !
Thank you. I need videos on Engineering Electronics circuits too. Stuffs including frequency response. Transistor biasing. Small and large signals. IC 555 timer. Oscillators etc.
What about the accuracy? stability as long as R is function of T, temperature?
you sometimes referred to v_in and v_out as "v one" and "v zero" and you wrote 6 6/8 as in it's a mixed fraction instead of 6 * 6/8, made me less confident that you're focused on what you're saying but overall good video and I understood the concept, thank you ^^
Thanks sir
Thanks a lot sir
I think you keep using "v1" when you mean to be saying "vi" it is confusing
i think hesays v in
what is the meaning of v_out if the output voltage changes significantly when a load is connected to it??
awesome man
that Tom Hanks voice XD
thank you for the video
So Vout will always have little to no current? What's the use of it?
Fairy Tail nope it is because the vi in the video is also at little to no current
HEYYYYYY FAIIIRYYYYYYY TAILLLLLLLL!!!!
There is no current passing through horizon 0 but voltage is Vo, how?
If a resistor is connected across the battery for a long time, then won't it drain off the battery/cell. Can you please let me know ? (3.7V cell 4.7K X 2 connected across as shown in this video)
Isn’t current in the wrong direction?
Why is gonna v0 be affected only by R2? it ran through both R1 and R2
Not sure of the math @3:58 isolating Vo, etc
Do you have any examples for bridge circuits?
Is this same for parallel circuits? i saw my teacher do this a different way wtf his way was inverted the other way, both add up on top and at the bottom the divider is the voltage in R that youre trying to find
Why's V out only affected by one resistor? What is v out in the first place, and what's the point? idgi
I know it's been 2 years, but I'm answering just in case someone will also need an explanation regarding what you asked, so here it is :
1)Why is V(out) only affected by one resistor ? Because it's only applied across one resistor. If it was affected by both resistors, then you'd get the exact same voltage as you applied V(in).
2)V(out) in this case is a predictable fraction of V(in) as the output V(out). Roughly speaking, you take an applied voltage V(in) and make it smaller as an output V(out) using a divider.
Raidom im about to have my final exam in like 1 hour and all i dont get is voltage divider like what the hell is vout i still dont understand
what software you use to teach??
6:12 if i wanna know the voltage for R2 i would just go 6V/(R1+R2) X R2
Its easier in my head that way. First i divide the voltage over the total resistance so now its spread out evenly, and i have like a base value i guess and then i multiply it by the resistor value i want t find the voltage for and i get the right answer. Can someone tell me if there is anything wrong with doing it this way?
You wrong 6v and didn’t use that to find the current to use in the equation. You just plugged it in as if it was a current value.
Hi ? Please if I use variable résistance 8k can I find 4,5 volt and thank you v much
Okey we know that in order to get the Vout we need to pick R2 as our reference and it would be i = Vo / R2. Because R2 at the below.
But what about the sensor that can change the value of resistance and its installed at the R1.. how do we know the value of Vo for that?
Can u please help me out?
so what if there was a current and it wasn't just 0,how do we solve it ??
Then the resistors would be parallel and have the same voltage
who else is here cuz of his igcse exams
Why does Vo depend upon R2 and not R1
Because Vo, as you can see at around 6:00 , is measured across R2, therefore you need to take that resistance in account.
Explain how you got 4.5 please.
6*6/8 = 36/8 = 4.5. Simple elementary math
These lessons have been very beneficial for me but this one needs redoing please. There's little to no explanation of anything. It's all a bit random. The math at 4:00 made no sense to me.
Ok, so do you understand the 4 prior equations and what they solve for before that ?
@@Technovore88 yeah perhaps i have a gap in my math skills somewhere but how vout/r2 = vin/(r1+r2) then goes to vout = vin * r2/(r1+r2) makes no sense to me
I get hes multiplying both sides by r2 but his answer is confusing and he didn't explain it at all
@@HamedAdefuwa Gotcha, I do agree it would have helped if he had stated the given current in the circuit so it could be proof checked with ohm's law. Just be clear you have grasp of the mathematics up to his actual answer of 4.5 volts for the most part?
something more advanced?
Current through R1 and R2 = I, not i !!!!!!!
doesnt' matter, theyre used interchangebly
This is wrong... Current flows from negative to positive
Electrons flows from -ve to +ve...
But conventional current flows from positive to negative
umm so i see that this video is for college students pursuing electrical engineering... but but but i am a student of grade 8 nd studying it..so unfair!!!!
Thank You
thank you so much