Lessons should just be like this where there is 1 great teacher per lesson who does a video like this for everyone to do its a lot cheaper than supporting schools
It doesn't matter at all. As long as you write the equations of the circuits consistently - applying the same rule for every voltage drop or rise - you will get exactly the same answer. It is common for different books and sources to use different convention for this. I use the one the makes the most sense for me and the one that is the most common for ENGINEERING. Physics sometimes likes to use the opposite convention as you pointed out.
I'm a Electronic Engineering Technologist and I Like Yo Come Up With EASY Solutions To Problems Without unnecessary Loop Equations!! As Stated Before More Than 1 source Requires us to Pay Mr. Kirchoff a Visit!!! This is still A Great Video To Prepare You For More Complex Circuits
Find the equivalent resistance of the two in parallel, then find the total current flow of 25 A. Boom! Your almost done! 25A through a resistor that has a drop of 50 V across it gives you the answer. Done. As for Kirchoff's voltage law, the voltage drops around the loop will add up to the source voltage. That's all you need to know. It really doesn't matter which direction you go. From the positive, you have 50 V + 100 V = 150V (arriving at the negative terminal). From the negative terminal you have 100 V + 50 V = 150V (arriving at the positive terminal).
I love how this guy explains everything, I learn a lot from him I'm only in eighth grade and I even know about Quantum physics from him thanks for your help.
You're lucky to have RUclips and videos like these while still in school. All I had was a heavy textbook and a professor who just confused us more. (If I had access to these videos back in my time, maybe I wouldn't get scared away from engineering lol.) Definitely take advantage of these videos!
You are right! 2 voltages add up to be 150. Once you solve for the parallel resistor,system will become a series circuit. The ratio equation is same as formula for i=v/r.In the series circuit, currents are always same. i=i is same as v1/r1=v2/r2
It does not matter which convention you use for the sign of the current as long as you are consistent, you will get the right answer either way. It does not matter that there is another way to solve this, even if you think it is an easier method, this lesson is about using KVL and KCL and imparting that skill ready for use in other cases where it is much more complex. Neither is the time he spends on the algebra manipulation wasted. Who knows just what the viewer is capable of, or not. He does not have a live audience to signal to him that they do or do not understand. A skilled teacher can tell by the reactions of the class.
Both the 'university physics' book and wikipedia says that the signs are opposite yours. According to those, the voltage law says: When you travel from - to +, then the emf is positive. When you travel from + to -, then the emf should be wrote negative. The same with resistors. If you travel from + to - with the current, then its negative, and the opposite. But it will give same answer, i think. Anyway, thanks for the video and the problem solving guide. Got exams next week. Great video again:)
I saw other teachers saying that from + to - there is a voltage drop so it must be - , and from - to + there is a voltage gain so it must be + . He is saying the opposite.
Im in my second half of my junior year of Electrical Engineering. I just now realized on this day that KVL is essentially ohms laws over and over again !
You are changing me current law since l know that when a current flow out of a junction is negative and the one that flows in is positive, but it is a new chalange now.
Thank you for this video. You are an excellent, relatable instructor and your teaching style is just what I think a lot of us need. This is an excellent refresher of training I had 30+ years ago in tech school in the USAF.
It's good intro to Kirchoff which is essential when the circuits get more complex, but for the example it's simpler to find the reciprocate of the two parallel conductivities (1/(0,05+0,2))=4 Ohms and divide that into the given 100V which gives you current through the unknown resistor of 25A, and since the voltage drop is given as 50, the resistor is 2 Ohms.
Good job, though it would be easier to make an equivalent series circuit in my opinion. I teach the opposite convention of what you teach in the video, but since you used the same convention throughout it works just fine. Good presentation
Thanks Joe, yes this circuit is simple and can be solved much faster by using equivalent resistance, as you indicated. But, the point here is to understand Kirchhoff's laws. Its easier to understand those if you start with simpler circuits which is what I have done here. Also, the sign convention that I used is the one used in most engineering textbooks. You can of course use the opposite convention as it doesn't really matter - you will get the same answer. Thanks! Jason
What you did here confused me. What I did was say we know the voltage across r2 and we know it is 20 ohms, so the current through that resistor is V/R, 100/20 = 5 A. The voltages across resistors in a parallel section are the same so the voltage across r3 is 100V. We know r3 is 5 ohms so, V/R, 100/5 = 20A. Now, the sum of currents through a parallel section must equal the total current in the circuit so 5A+20A = 25A. From the start we are given that the voltage across r2 is 100V and we know that the voltage across r3 is also 100V. Vs is 150V so the voltage across r1 must be 50V. R= V/I and we know the current through r1 must be the total current, 25A and the Voltage across r1 is 50V... 50/25 = 2 Ohms.
He confused me too! He really made it complicated! Why? It's so easy to understand without all that formula writing and re-writing. After I got out of the Navy, i took some additional electronics. I spent many years after as a bench tech, then a field service tech, finally ending up in marketing and sales for an oceanographic instruments company.
Can you comment on what the choice of loops should be in a given circuit generally. Should no loop be the combination of any of the other loops chosen. For example, would choosing the outer loop also here just yield the same information as the two inner loops. Also, can you give any intuition whether Kirchoff rules always lead to a solution: ie that we always get as many independent equations as variables. Thanks.
The majority of viewers pretty much agree to use an equivalent resistor value to find R. Although useful for this simple circuit, you'll find knowing KCL/KVL comes in handy when dealing with far more complex circuits especially those with multiple voltage sources.
Here is something simpler with few equations and no new variables: 20 ohm & 5 ohm in parallel are equivalent to (20x5)/(20+5) = 4 ohm. Total current = 150/(R+4) = 100/20 + 100/5 => R+4 = 150/25 = 6, So, R = 2 ohms.
Hi yes of course you could solve it that way, but that’s not the point. The point is to learn kirchhoff’s laws so you can use this method to solve more difficult circuits. Jason
Current flowing into a node has a positive value not negative while current leaving a node is written as negative .So KCL @ top node should it not be i1 - i2 - i3 =0???
You get the same equation either way. Typically, the convention is how I have it in the video for engineering students. You can use the opposite convention and you will get the exact same node equations.
mitnavnvirkerikke It doesn't really matter because we're talking about electrical potentials which can be defined either way just like you choose yourself when solving other physics problems with coordinate systems e.g. what direction should be positive or negative.
Hello i don't understand the polarity as expressed. As a beginner it seems like the voltage should be +150 and then -100. The arrows make the current appear to flow clockwise
Hello.i think voltage of 20 ohm and 5 ohm is the same and there is no need to have a loop in the second side.just you need to calculate the equal of two parallel resistances.it would be: (20*5)/(20+5)=4 ohm Then: IR=100/4=25 A VR=50 V So=50/25=2 ohms I think its the simple solution.
another way to do it is combine 20||5 and you get 4 ohm. You know its 100v on V4 because it was given. That means R has to be 50 V (rest of the power supply) Do a KVl and you will get -150+I1R1+4I1=0 . What do you know? I1R1=50V. 4I1=100V and I1=25 A. R=V/I => 50/25 = 2 ohm
Absolutely. I chose to illustrate these concepts using a simpler circuit. But later you will have to apply these concepts to more complex circuits that you can't do by inspection. Thanks again. Jason
lol i dont understand a lick of this and I solved for i1 just by using common sense lol... im so lost :) thanks for the soft serve on KVL and KCL though, really helps
We had an assignment to be solved using KCL and KVL but our instructor failed us coz she said that we used mesh analysis. Now, I don't see the difference between the two. Would you please explain? thanks a lot
is it not easier to divide to get the total current going through 20 and 5 with same voltage of 100 then divide 150 over that number and you will get total resistance and just get the difference between that and the parralle resistance (20 and 5)??? final answer is 2
Simplify this Circuit down to a equivalent Series Circuit!!! The equiv 20 ohm in parallel with 5 ohms equals 4 ohm ( Rt=R1xR2/R1+R2)=100/5=4 ohms We know R unknown drops 50 volts(150-100=50volts) Set up a RATIO EQUATION R/Vr =R/Vr,,,,,,,R/50 = 4/100.....100R=200 therefore R=2 ohms.....No Need To write all these loop Equations Unless You have more than 2 or more Current or Voltage sources That's what My Engineering Technology Teacher told ME!!(Who Happens to be a ELECTRICAL ENGINEER!!!)
why are you making it hard? Just parallel the 5-ohm resistor with a 20-ohm resistor and that is=4. then you can find the current there which is 100/4=25 and we know the current will be same at the R since they connected in series, therefore, -150+25R+100=0 so R=50/25=2 ohm
You’re missing the point. The point here is to solve this simple circuit using Kirchhoff laws, so that you can use this more powerful method for more complex circuits.
In a timed exam environment, I agree that your method is more efficient. He purposely made this example "hard" so we can understand the concepts of Kirchhoff's Laws (KCL and KVL).
I don't get it. it took me 5 seconds watching only the circuit to see it would be 2 ohm. because? 100 Volts over a 20 Ohm gives 5 Amp, 100 Volt (as they are parallel) over a 5 ohm gives 20 Amp. Both currents run through R, which has the remaining 50 Volts. R = U / I = 50 / 25 = 2 Amp. A (!) simple example I guess.
you think you are smart or something? this is a tutorial on Kirchhoff’s Rules. i think anyone who has basic knowledge will know the answer in 5 seconds using simple ohms law
And that is the idea of my comment. There is no challenge to use a calculation on a simple circuit. A difficult circuit would get our minds working, just wanting to solve it. It is not about being smart, but about teaching our brains to the most possible extend. Your view on the matter is -as good- as mine. No offense meant.
Exactly. I don't know what he's trying to prove. Guess he likes to confuse people. I learned electronics in the Navy electronics school at Treasure Island in the 60's. We learned the easy way. We never would have gone through all that. But we did learn Kirchoff's laws as well.
I was following along fairly well until, when analyzing the right part, you converted a passive resistor into an active source - this completely threw me. In the actual circuit the voltage at the + of the 5 ohm resistor must be > 100 volts else the value of the other resistor can be calculated. If the voltage at 5 ohm resistor is > 100 then the current flow is not correct when calculated using 100V???
Why move the 50 over at the end?? Even if you move it over why is it not still -50 You just stated the final answer is 2 ohems why is it?? What equation did you do to get 2ohmes ?
You made us understand what our teachers failed to do in 6 months
Lessons should just be like this where there is 1 great teacher per lesson who does a video like this for everyone to do its a lot cheaper than supporting schools
OMG THANKS TO YOU I ACTUALLY LEARNED SOMETHING IN THIS SUBJECT!!!! My lecturer doesn't know how to teach at all. U r a life saver 😍
Have you graduated yet?
@@ofimportance5458have YOU graduated yet?
It doesn't matter at all. As long as you write the equations of the circuits consistently - applying the same rule for every voltage drop or rise - you will get exactly the same answer. It is common for different books and sources to use different convention for this. I use the one the makes the most sense for me and the one that is the most common for ENGINEERING. Physics sometimes likes to use the opposite convention as you pointed out.
This video series is quite good. I'm taking circuit analysis classes and these videos are better than the lectures from my professor!
same as my condition now
I agree 💯
The world needs for teachers/professors like you. Great job
This guy's videos are excellent. He's a great teacher.
He's a great confuser. Better to learn the easy way. Why confuse people?
@@wno1043 you have to know what your doing just go from basics then head on here it will make so much more sense
Who
My teacher needs to take lessons from you.
I'm a Electronic Engineering Technologist and I Like Yo Come Up With EASY Solutions To Problems Without unnecessary Loop Equations!! As Stated Before More Than 1 source Requires us to Pay Mr. Kirchoff a Visit!!! This is still A Great Video To Prepare You For More Complex Circuits
You are a very good teacher. The approach to the problem is very good.
Thanks for that it helps a lot
This teacher makes the problem become something easy since the difficulty becomes something very simple.
Thank you. Im getting there slowly but surely
Find the equivalent resistance of the two in parallel, then find the total current flow of 25 A. Boom! Your almost done! 25A through a resistor that has a drop of 50 V across it gives you the answer. Done. As for Kirchoff's voltage law, the voltage drops around the loop will add up to the source voltage. That's all you need to know. It really doesn't matter which direction you go. From the positive, you have 50 V + 100 V = 150V (arriving at the negative terminal). From the negative terminal you have 100 V + 50 V = 150V (arriving at the positive terminal).
I love how this guy explains everything, I learn a lot from him I'm only in eighth grade and I even know about Quantum physics from him thanks for your help.
You're lucky to have RUclips and videos like these while still in school. All I had was a heavy textbook and a professor who just confused us more. (If I had access to these videos back in my time, maybe I wouldn't get scared away from engineering lol.)
Definitely take advantage of these videos!
@@epochseven4197 same here
@@epochseven4197 o
Thank you sir, I really appreciate this teaching
You are right! 2 voltages add up to be 150. Once you solve for the parallel resistor,system will become a series circuit. The ratio equation is same as formula for i=v/r.In the series circuit, currents are always same. i=i is same as v1/r1=v2/r2
yes that is different method of finding the answer. good job
It does not matter which convention you use for the sign of the current as long as you are consistent, you will get the right answer either way. It does not matter that there is another way to solve this, even if you think it is an easier method, this lesson is about using KVL and KCL and imparting that skill ready for use in other cases where it is much more complex. Neither is the time he spends on the algebra manipulation wasted. Who knows just what the viewer is capable of, or not. He does not have a live audience to signal to him that they do or do not understand. A skilled teacher can tell by the reactions of the class.
One in every crowd
I encourage you guys to buy his DVD series he makes things easier to grasp :)
You are the best teacher thank you so much
Was struggling to understand the Loop Law for a Lab assignment, Thank you!
Excellent! simple and easy to understand. Thank you!
Thank you! Sincerely!
from negetive to positive there is a potential gain so it must be positive not negetive as you said
Both the 'university physics' book and wikipedia says that the signs are opposite yours. According to those, the voltage law says: When you travel from - to +, then the emf is positive. When you travel from + to -, then the emf should be wrote negative. The same with resistors. If you travel from + to - with the current, then its negative, and the opposite.
But it will give same answer, i think. Anyway, thanks for the video and the problem solving guide. Got exams next week. Great video again:)
I saw other teachers saying that from + to - there is a voltage drop so it must be - , and from - to + there is a voltage gain so it must be + . He is saying the opposite.
It doesn’t matter as long as you’re consistent
Im in my second half of my junior year of Electrical Engineering. I just now realized on this day that KVL is essentially ohms laws over and over again !
Thanks so much for hosting these. It's such a valuable resource, given that online courses in 2020 are kind of crap.
You are so much better than my EE98 Professor at SJSU... I wish you were the prof...
THANKS . YOU HIT THE GOALS EASILY...
when im looking for a tutorial then i see this guys automatic this is very niceeee!
I understood the way you explained.
Thank you!
You are changing me current law since l know that when a current flow out of a junction is negative and the one that flows in is positive, but it is a new chalange now.
Thank you sir I appreciate your time ❤
Excellent Job. Very thorough. Enjoy all the videos!
Thank you for this lesson you are awesome
Thank you for this video. You are an excellent, relatable instructor and your teaching style is just what I think a lot of us need. This is an excellent refresher of training I had 30+ years ago in tech school in the USAF.
I really appreciate the kind words!
learning this over quarantine
It's good intro to Kirchoff which is essential when the circuits get more complex, but for the example it's simpler to find the reciprocate of the two parallel conductivities (1/(0,05+0,2))=4 Ohms and divide that into the given 100V which gives you current through the unknown resistor of 25A, and since the voltage drop is given as 50, the resistor is 2 Ohms.
Thanks for your service......dan
Thank you, you have explained everything clear and concise.
Thank You Sir, Great Help!
Clear explaination😄helps me a lot
Nicely and methodically explained.
With RUclips and all the other great social media platforms around, schools will soon become artifacts in the museums. 😮
If I would have come across this Tutor when I was In school. I would have been an Electronics and Communication engineerc instead of IT professional
A question: At 10:43, shouldn't I1 be positive because it is going inside the node and I2 be negative because it is exiting the node?
This guy is amazing!! Does he still upload ??
THANK YOU... SIR...!!!
Good job, though it would be easier to make an equivalent series circuit in my opinion. I teach the opposite convention of what you teach in the video, but since you used the same convention throughout it works just fine. Good presentation
Thanks Joe, yes this circuit is simple and can be solved much faster by using equivalent resistance, as you indicated. But, the point here is to understand Kirchhoff's laws. Its easier to understand those if you start with simpler circuits which is what I have done here. Also, the sign convention that I used is the one used in most engineering textbooks. You can of course use the opposite convention as it doesn't really matter - you will get the same answer. Thanks!
Jason
Yep very true, if my students have trouble ill send them to this channel
Thank you That made so much sense
What you did here confused me. What I did was say we know the voltage across r2 and we know it is 20 ohms, so the current through that resistor is V/R, 100/20 = 5 A. The voltages across resistors in a parallel section are the same so the voltage across r3 is 100V. We know r3 is 5 ohms so, V/R, 100/5 = 20A. Now, the sum of currents through a parallel section must equal the total current in the circuit so 5A+20A = 25A. From the start we are given that the voltage across r2 is 100V and we know that the voltage across r3 is also 100V. Vs is 150V so the voltage across r1 must be 50V. R= V/I and we know the current through r1 must be the total current, 25A and the Voltage across r1 is 50V... 50/25 = 2 Ohms.
He confused me too! He really made it complicated! Why? It's so easy to understand without all that formula writing and re-writing.
After I got out of the Navy, i took some additional electronics. I spent many years after as a bench tech, then a field service tech, finally ending up in marketing and sales for an oceanographic instruments company.
thank you ,you just saved me
Welcome!
Can you comment on what the choice of loops should be in a given circuit generally. Should no loop be the combination of any of the other loops chosen. For example, would choosing the outer loop also here just yield the same information as the two inner loops.
Also, can you give any intuition whether Kirchoff rules always lead to a solution: ie that we always get as many independent equations as variables. Thanks.
Perfect, thank you!
Thanks ,i learned alot:)
One of the best !
very well explained, great job!
excellent explanation
thank you very much it was really helpful.
The majority of viewers pretty much agree to use an equivalent resistor value to find R. Although useful for this simple circuit, you'll find knowing KCL/KVL comes in handy when dealing with far more complex circuits especially those with multiple voltage sources.
Here is something simpler with few equations and no new variables: 20 ohm & 5 ohm in parallel are equivalent to (20x5)/(20+5) = 4 ohm. Total current = 150/(R+4) = 100/20 + 100/5 => R+4 = 150/25 = 6, So, R = 2 ohms.
Hi yes of course you could solve it that way, but that’s not the point. The point is to learn kirchhoff’s laws so you can use this method to solve more difficult circuits. Jason
Thanks sir
Thanks you teacher
Current flowing into a node has a positive value not negative while current leaving a node is written as negative .So KCL @ top node should it not be i1 - i2 - i3 =0???
You get the same equation either way. Typically, the convention is how I have it in the video for engineering students. You can use the opposite convention and you will get the exact same node equations.
mathtutordvd Oh I see ,I got a little bit confused ,many thanks for the explaining.:)
Thanks so much
mitnavnvirkerikke It doesn't really matter because we're talking about electrical potentials which can be defined either way just like you choose yourself when solving other physics problems with coordinate systems e.g. what direction should be positive or negative.
Amazing job
thank you
sir is there any priority between kvl and kcl ? i mean when you are writting the equation kvl should be first or kcl .thank you
Hello i don't understand the polarity as expressed. As a beginner it seems like the voltage should be +150 and then -100. The arrows make the current appear to flow clockwise
Hello.i think voltage of 20 ohm and 5 ohm is the same and there is no need to have a loop in the second side.just you need to calculate the equal of two parallel resistances.it would be:
(20*5)/(20+5)=4 ohm
Then:
IR=100/4=25 A
VR=50 V
So=50/25=2 ohms
I think its the simple solution.
Thank you sir
Ur Amazing..... thank you
So happy you liked it!
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Thank you sir..!
I really appreciate it!
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another way to do it is combine 20||5 and you get 4 ohm. You know its 100v on V4 because it was given. That means R has to be 50 V (rest of the power supply) Do a KVl and you will get -150+I1R1+4I1=0 . What do you know? I1R1=50V. 4I1=100V and I1=25 A. R=V/I => 50/25 = 2 ohm
really nice topic
..............Iunderstand well
why negative for i1, when it’s gaining? is it not a positive?
emil lagman That's not the point. It's demonstrating a concept that can be applied to more complicated circuits.
Absolutely. I chose to illustrate these concepts using a simpler circuit. But later you will have to apply these concepts to more complex circuits that you can't do by inspection. Thanks again. Jason
lol i dont understand a lick of this and I solved for i1 just by using common sense lol... im so lost :)
thanks for the soft serve on KVL and KCL though, really helps
Sir could you please help me with all the equations of Kirchhoff’s law in a closed circuit
We had an assignment to be solved using KCL and KVL but our instructor failed us coz she said that we used mesh analysis. Now, I don't see the difference between the two. Would you please explain? thanks a lot
is it not easier to divide to get the total current going through 20 and 5 with same voltage of 100 then divide 150 over that number and you will get total resistance and just get the difference between that and the parralle resistance (20 and 5)??? final answer is 2
an equation with a high degree of difficulty is converted into an easy solution
awesome but how the negative affects the values I don't get that
it just take 30 sec :) Req= 20//5 = 4 ohm V=150V Vr=100V Vr=(Req/R+4)*V => 100=(4/R+4)*150 => R=2 Ohm
So this is kind of like mesh analysis?
Simplify this Circuit down to a equivalent Series Circuit!!! The equiv 20 ohm in parallel with 5 ohms equals 4 ohm ( Rt=R1xR2/R1+R2)=100/5=4 ohms We know R unknown drops 50 volts(150-100=50volts) Set up a RATIO EQUATION R/Vr =R/Vr,,,,,,,R/50 = 4/100.....100R=200 therefore R=2 ohms.....No Need To write all these loop Equations Unless You have more than 2 or more Current or Voltage sources That's what My Engineering Technology Teacher told ME!!(Who Happens to be a ELECTRICAL ENGINEER!!!)
why are you making it hard? Just parallel the 5-ohm resistor with a 20-ohm resistor and that is=4. then you can find the current there which is 100/4=25 and we know the current will be same at the R since they connected in series, therefore, -150+25R+100=0 so R=50/25=2 ohm
You’re missing the point. The point here is to solve this simple circuit using Kirchhoff laws, so that you can use this more powerful method for more complex circuits.
In a timed exam environment, I agree that your method is more efficient. He purposely made this example "hard" so we can understand the concepts of Kirchhoff's Laws (KCL and KVL).
I don't get it. it took me 5 seconds watching only the circuit to see it would be 2 ohm. because? 100 Volts over a 20 Ohm gives 5 Amp, 100 Volt (as they are parallel) over a 5 ohm gives 20 Amp. Both currents run through R, which has the remaining 50 Volts. R = U / I = 50 / 25 = 2 Amp. A (!) simple example I guess.
Sure, that is one way to solve it. But the prof solves it using KCL and KVL, which is a different approach entirely.
you think you are smart or something? this is a tutorial on Kirchhoff’s Rules. i think anyone who has basic knowledge will know the answer in 5 seconds using simple ohms law
And that is the idea of my comment. There is no challenge to use a calculation on a simple circuit. A difficult circuit would get our minds working, just wanting to solve it. It is not about being smart, but about teaching our brains to the most possible extend. Your view on the matter is -as good- as mine. No offense meant.
Whoa we got a badass over here.
Exactly. I don't know what he's trying to prove. Guess he likes to confuse people. I learned electronics in the Navy electronics school at Treasure Island in the 60's. We learned the easy way. We never would have gone through all that. But we did learn Kirchoff's laws as well.
Thank u
why does my schaums basic electricity book state that if voltage flows negative to positive its a positive voltage, and u say its a negative voltage
how we takeposive negative from our circuit?
I was following along fairly well until, when analyzing the right part, you converted a passive resistor into an active source - this completely threw me. In the actual circuit the voltage at the + of the 5 ohm resistor must be > 100 volts else the value of the other resistor can be calculated. If the voltage at 5 ohm resistor is > 100 then the current flow is not correct when calculated using 100V???
A perfect example of the danger of a handwaving argument.
Is this part 4 to your presentation?
Honestly I knew how to go about solving this, but if I were to solve it on my own I would have definitely screwed up those plus/minus and directions.
Why kcl is not applied on series circuit and kcl on parallel circuit
Sorry, I meant 2 ohms at the end!
googd job
Why move the 50 over at the end??
Even if you move it over why is it not still -50
You just stated the final answer is 2 ohems why is it??
What equation did you do to get 2ohmes ?
More problems sir...upon sies pallel
Hey sir,i have a problem with my payments on your site and blocked my ip on the specific site,can you help me please,i really want to study them :)
Hi, please email jason.mathtutor@gmail.com and we will get everything going for you.
@@MathAndScience already did,gracias :)
Ok great I see it. I sent that to my office manager who will reach out to you later today. Thanks! Jason