IB Physics: Circuit Analysis

Explains the basic principals of circuit analysis including Kirchoff's Laws, and then illustrates the principals with several examples.
IB Physics Topic 5.2
Click JOIN to become a member of C. Doner’s IB Physics RUclips Channel, or become a patron by visiting / cdoner .
Level/Tier 1: $1.99 per month gives you access to a google sheet with a full updated, well organized, linked listing of all the videos and other resources such as notes, worksheets and solution sets.
Level/Tier 2: $4.99 per month gives you access to new videos. The new videos include IB style multiple choice and problem solving questions with full explanations, previews/reviews of the course content, and updates to the video lessons.
After becoming a member, go to bit.ly/donervidlist and request access to the video listing.
Many students have written to thank me for the success they had with their IB Physics exams and their IA or Extended Essay. Though the style is that of Khan Academy, the videos are more specific to the IB content, contain more simulations and interactive examples, and are performed by someone who has 25 years of experience teaching IB Physics, an MSc in Physics from a top University, and who was an NSERC scholar.
"I would just like to say a massive thank you to you for uploading your videos. I started the IB [and]decided to try HL Physics as it seemed interesting. It turned out to be very far from my forté and I was consistently getting 3s and 4s in tests and mock exams. So a few months before the exam period of this year, I began to panic...Then I discovered your channel. For four weeks prior to paper one, I virtually worshipped your videos, and worked along to them in my 'Big Book of Revision.' Last week, I received my results. Incredibly and almost unbelievably, I obtained a 6 in HL Physics. Not only that, but I was 3 marks off a 7."

[@Emzo99]

I actually love you

[@unesco2433]

the last two question showed up consecutively on my test last week. I love how the ib questions in ur vids come up on my exam

[@snjn5914]

THE MOST HELPFUL PERSON ON THE INTERNETTTT

[@sogandgolshahian8662]

Hello Mr. Doner. Firstly, I wanted to thank you for your incredible videos on the IB Physics syllabus. And secondly, I just had a quick question about the last IB question in your video at time 17:25. Since the resistances of all the resistors are the same, and Z and Y are part of one loop, wouldn't their potential differences be equal to each other? And the same question goes for X and Z, as they are also in one continuous loop? I am having a bit of trouble knowing when to use the loop rule, while the junction rule makes a little more sense to me. Answer B is talking about Y only, but you are talking about the equivalent resistance of X and Y in your video, which is why I am confused as to why the answer would be B and not C. Thank you for your time and for your help.

[@donerphysics]

Z and Y are part of a loop so their voltages would add to that of the battery, not be equal. X and Y are in parallel so they must have equal voltages. We could go around the loop with X and Y and then claim that X and Y are equal in voltage since there is no battery in that loop.

[@sogandgolshahian8662]

@@donerphysics Thank you so much for your reply, Mr. Doner. That makes a lot more sense. Thank you.

[@hamza9198]

These videos are extremely helpful, thank you!

[@guanhantan2136]

Hello sir, for the question at 17:28, what if i use kirchoff law instead? So i=i1+i2. But since resistance of X and Y are the same so current i1=i2. So we can say that i1=i2=0.5i. When we add up the voltage of x and y, which is (0.5i×R)+(0.5i×R), we get i×R which is the same voltage as Z(i×R). So the answer should be C right?

[@donerphysics]

You can't add up the voltages in parallel, only in series. (Kirchoff's loop rule). The two resistors in parallel have less equivalent resistance than the first resistor. Answer is B.

[@guanhantan2136]

@@donerphysics thank you sir

[@mohammedshukri2648]

Hello Mr. Doner. What would we do if two batteries were involved?

[@donerphysics]

+Mohammed Shukri
Involved in what?
In series, the battery voltages add. In parallel the battery voltage is the same but the internal resistance decreases.

[@mohammedshukri2648]

+Chris Doner (C. Doner's IB Physics) If by using Kirchoff's law, two batteries (with different voltages) are involved and we have three resistors in parallel, what we do in this case?

[@donerphysics]

Beyond this course, but you would write equations for each loop and current junction, and solve using algebra.

[@arhankamath871]

Are you sure its beyond the course, sir?, since the K.A. Tsokos textbook has several questions using such a set up, but hasn't explained it very well

[@shashankgarikipati6272]

Hello Mr. Doner,
why isn't the answer to the last question at 17:27 option C.

[@donerphysics]

Beccause the equivalent resistance of the two in parallel is less than a single resistor.

[@shashankgarikipati6272]

Okay, thanks

[@julienh3965]

Hello Mr Doner,
At 10:50 you say that there is 6 ohms over 2 paths. Would there not be 12 ohms from the three resistors? Why is there only 6 when there are definitely 3 resistors that add up to a resistance of 12 ohms? I'm just a little confused about the math.

[@donerphysics]

If the resistors were one after the other, in series, we would simply add them up to get 12 Ohms. But we have two equal paths (4+2 ohms and 6 ohms) for the current to flow. These paths are in parallel. We have two equally good roads to get to the same destination, and that means that twice the traffic can flow, or we could say the resistance is half as much.

[@spifuntastic621]

13:46 for this question can't we also say that the voltage is 6V across the 2 ohm resistors cuz it's a continuous piece of wire?

[@donerphysics]

We could say it is 6V across the entire configuration because we have continuous wire, but once we hit the resistors, we can't say the wire is continuous.

[@spifuntastic621]

@@donerphysics Alright makes sense... Thank you

[@denzeltracy]

this was amazing thank you

[@artemkrazhan6778]

Thank you

[@donerphysics]

You're welcome

[@leenkhasawneh9189]

Thank you for all your videos on here. I have one question regarding the calculations on 14:45: Wouldn't 1/25 +1/35 = 1/ 60 ~ 0.01667? And so R ~ 60 Ω?

[@donerphysics]

You need to have a common denominator to add fractions.

[@johnjones8454]

Leen Khasawneh denominators aren’t equal

[@georger.4427]

What does equivalent resistance mean?

[@donerphysics]

See my video on equivalent resistance.

[@MK-wk4vm]

your canadian accent :)))))))))))))))