How To Solve Any Resistors In Series and Parallel Combination Circuit Problems in Physics

This physics video tutorial explains how to solve any resistors in series and parallel combination circuit problems. The first thing you need to do is calculate the equivalent resistance of the circuit which you can use to calculate the current flowing from the battery based on ohm's law. Using the current and the voltage drops across each resistor, you can calculate the electric potential at every point in the circuit which will allow you to calculate the current flowing through any resistor. This technique is useful if there's not multiple batteries in different loops of the equation. Keep in mind, the current flowing in two resistors that are in series with each other is the same. The voltage is the same across resistors connected in parallel. This video contains plenty of examples and practice problems including a current division formula for resistors connected in parallel.
Schematic Diagrams & Symbols:
• Schematic Diagrams & S...
Resistors In Series:
• Resistors In Series - ...
Resistors In Parallel:
• Resistors In Parallel ...
Series and Parallel Circuits - Light Bulb Brightness:
• Series and Parallel Ci...
Equivalent Resistance of Complex Circuits:
• Equivalent Resistance ...
_________________________
Voltage Divider Circuit:
• Voltage Divider Circui...
Current Dividers:
• Current Dividers Expla...
Parallel Circuit Challenge Problem:
• Finding The Current In...
Kirchhoff's Current Law:
• Kirchhoff's Current La...
Kirchhoff's Voltage Law:
• Kirchhoff's Voltage La...
DC Circuits Review:
• Series and Parallel Ci...
____________________________
KCL and KVL Circuit Analysis:
• Kirchhoff's Law, Junct...
Thevenin's Theorem - Circuit Analysis:
• Thevenin's Theorem - C...
Norton's Theorem - Circuit Analysis:
• Norton's Theorem and T...
Superposition Theorem:
• Superposition Theorem
Maximum Power Transfer:
• Maximum Power Transfer...
Physics PDF Worksheets:
www.video-tuto...

[@TheOrganicChemistryTutor]

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update us

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for some reason i was never taught to redraw the circuit at all. I just watched this video and another one where you both redraw the circuit and my mind is blown omfg. it makes everything so much easier to understand

[@gravityhealer]

Exactly, I'm like wtf. My prof just told us to simplify the circuit in our head instead of redraw it. I'm like wut?

[@MrDoneboy]

Absolutely NO redraws are necessary!. Find total resistance from right to left, back to the source voltage. Then figure the total current by Ohms law. Now, going from left to right...figure your voltage drops across R1 and R6. Then subtract those voltages from the source voltage. Divide the 15 volts by R2 resistance to get 1.5 amps, which is half of the total current. The other 1.5 amps goes thru the rest of the circuit, or just divide the 15 volts by the total resistance of the three resistors in the far right part of the circuit. Bingo, problem solved people!

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How did you solve without redrawing??😱😱

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@@adabujiki it's taught

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For some reason this has conceptually been one of the hardest topics for me to grasp at the beginning.. thanks for helping clear it up!

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man you made my basics on this topic sky-rocketing. I couldn't even solve a simple circuit now I'm able to do some circuit problems. thanks, isn't enough for you man

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[EDIT 2 YEARS LATER PASSED WITH A B!] You’re going to be the reason why i’ll pass my final tomorrow, thank you so much!

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I got finals too. Trying to figure this out cuz my instructor full of shit .

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did u pass

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ravi kiran rly 1 year later lol.

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@@theuntormented6560 no

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So how was the final?

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Since we take this in high school. And assuming you were 16 then. You are 68 now. Of course I believe 50 years is rounded.

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@@damnit8904 uhhh. okay? lol

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Wow the second example was almost exactly like one of my tricky homework problems and it helps a lot! Thanks!

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Dan Probably someone who failed their exam

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[@didiroxforever]

In physics class, to combine parallel repressors, we have been using the formula 1/Rt =1/R1 +1/R2, but that doesn't seem to be in your video. I'm still kind of struggling to understand circuits as a whole, and it would be very helpful to me if you could explain that rule/ its relevancy to circuits. The right hand rule also has been confusing me as well. I love your videos, they've helped me loads in AP Chem and in physics. I'm really glad that there are people out there who make videos to help others, rather than click-baited videos.

[@doctorluzbubs258]

1/Rt = [1/R1 + 1/R2] and don't forget it's to the power of -1

[@doctorluzbubs258]

You only combine them when u are trying to figure out the current encompassing the whole device (circuit). Once you do that you can troubleshoot your device with a multimeter and see where its failing you.

[@sea5205]

didiroxforever that method still applies, in the first question he did he said the total resistance for the the two ten ohm resistors in parallel was 5. He got that by adding 1/10 + 1/10 then finding the inverse of that answer

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Nice job. Have not done this since the 1990s, but you brought it back to life with clear and concise explanations. Good job. Thank you.

[@successthruknowledge]

You do such a great job explaining multiple ways to solve the same problem! In earlier lectures you showed how to do it by summing the resistors in series on right side and then solving the parallel resistors and now you explained the alternate way to solve to find the current through the middle resistor by using the potential difference between two voltage points on either side of the middle resistor and solving for the current through the middle resistor and then subtracting that current to find the remaining current flowing through the resistors in series on the right side of the circuit! Just what I needed explained clearly!

[@johnnyfoute3517]

I always come back to this video. As I progress through class more and more things in this video make sense. It’s all here. Excellent material here. Thank you.

[@user10247]

Got confused about the current you wrote down towards the end. At 28:09 Should the current going from C to E be 1.7295 and not 0.5765 which is the current going from C to D to E?

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[@ibrahimengineeringtutor9369]

I believe you had it correct you are just slightly confused. The current from C to E from the top going right and down is 0.5765A (this is the current going through the 12ohm series combination which is the 3 and 9 ohm resistor combination). But the current going the 4 ohm resistor (diagonal section) is 1.7295. I hope this helps!

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[@grimjogaming878]

@ 28:13 , it should be 1.7295 A that runs through 4-ohm resistor right? Why is it the same as current is 0.5765A??

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[@armandecastro]

To compute 2 resistor in parallel with different value just multiply the Numerator then add the Denominator and Devide the value that you get in 15:31 minutes 12 * 4 = 48 (Numerator) then 12 + 4 = 16 therefore 48/16 = 3ohms

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Sir to find Potential at F we are moving from A to F by this way it will be Va - Vf and Vf will be -2

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[@Harminder_Saini]

There is also another way you could solve the current going through the 3 branches @ 28:00 , you could find the equivalent resistance of r1 and r2, then use that equivalent resistance and the current to find the voltage once you have the voltage you could then find the individual current of r1 and r2.

[@Randomperson11121]

How do you find the voltage of one with the equivalence of all resistors

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Great work! when calculating the currents at each resistor, i get lost due to the resistors not being labelled.

[@magnumopus1415]

Hi sir, i am sure the calculation is wrong at 28:14, the current passing the 4 ohm resistor must be 1.76. It cannot be 0.576 because then the total current won't be 4.615, and so you violate KCL, and we calculated the current across the resistor is 1.76A

[@andrewynter5828]

Yeah he mixed them up. But switch the current for i1& i2 respectively and you'll get them right

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[@edymarkonthego4096]

The current flow across 4ohms is not 0.5765amp. It is 1.729Amp. Finally that you’ve corrected it.

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The method is pretty simple tq

[@false_is_real]

28:11 I think you made a mistake assigning the current as 0.5765amps when it should be 1.7295mps.... Correct me if im wrong however:)

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Nice thankz for your effort....its really help me lot...thank you so much....

[@WaleedTariqSethi]

One question?? How to know which branch (series or parallel) appears next when simplifying the resistors.?? E.g. when solving 3 ohm with 9 ohm, why we got a series branch and not a parallel one??

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[@nephartuzim8879]

28:05 Shouldn't the current flowing through the 4 Ohm resistor be 1.7295 A?

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He fixes it later, in case you never watched all of the video.

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