Resistors in Series and Parallel

[@rakanmohamed8733]

This video just made me believe I will pass my physics test tomorrow. Thanks! It was very easy to understand and it was a fun video to watch

[@daz41262010]

this series is so cool to keep on book mark to view again and again :)

[@Good-Enuff-Garage]

I've been looking for a proper resistor explanation for some time, and I am so glad I found yours, extremely well done, thank you

[@BenFinio]

Thanks! Glad you found it helpful.

[@cpace123]

Excellent Video, for a less then great mathematician, and hobby electronics person

[@InverCollegeTech]

I find the best way to explain resistors in parallel vs series is to use the anology of turnstiles at a football stadium. 1 turnstile = 1 resistor e.g. 10,000Ω. So if you had two turnstiles, one after another (in series) it would be 20,000Ω which slows the crowd entering the stadium even more. However, if you've two turnstiles in parallel, it allows the crowd into the stadium quicker, so half the resistance. Then I explain resistors of different values in parralel, which is when you've got two turnstiles, one that can quickly scan tickets from a QR code on your phone (low resistance), while the other is old man with bad eyesight that has to read the ticket and mark it off a list of the tickets sold (high resistance) i.e. resistors of different values.

[@thanktheknife]

very well explained and easy to follow. Gratitude to you my dude!

[@BurgerBeatz]

Great explanation, to the point and easy to follow.

[@Sarge7711]

You are the freaking best, Straight to the point I love it THANK YOU!!!!!

[@Papa-ro7ch]

Awesome tutorial thanks. I will try and keep up.

[@neharawat08]

Thank you for this video! Simple and easy to learn! Thanks a ton!

[@RC-Heli835]

Thanks for sharing Ben! This helped me to solve the answer to this question.......... If Joe used 40 2k ohm 5 watt resistors to build a 50 ohm 3000 watt RF dummy load submersed in oil in what configuration did he install the resistors? You have to wind up with 50 ohms, that's the key.

[@markholden3216]

Thank you so much! This was clear and made sense of a topic I am only beginning to study. Thank you too for the fritzing link.

[@zacharyriggs8681]

Awesome. Anyone know of any videos that explains these equations better?

[@AMPROEngineering]

Thank you for your videos. They are very helpful but I was curious if wattage says the same. If you have 2 1/4 watt resistors in parallel, are the now a single 1/2 watt resistor or does that have no bearing on it?

[@BenFinio]

You need to calculate the power for each resistor individually to make sure none of them exceed their individual power rating.

[@08abreur]

The exact answers I was looking for

[@sampam263]

very easy rules to use, I may have an even more simple formula for resistors: (R1 x R2) / (R1 +R2)

[@dharmakissoon]

Hi how do I put resistors in aluminum case

[@pufango4059]

Easy when you know ! Excellent video A+++++

[@reedy8585]

how is the power handling distributed across resistors in series , natturally you would expect 2 2w resistors in parallel to be capable of handling 4w however does this also apply in series or is the first resistor in line exposed to the full load and still only capable of 2w ?

[@BenFinio]

Hi - this is kind of a lot to answer in a single comment, you might want to check out my intro to circuits playlist: kzbin.info/aero/PLKL6KBeCnI3U6KNZEiitdtqvrxkBhpuOp.

[@eaoke3383]

What a lad- thank you for your help

[@sanjogkumarbarik7460]

sir,what about watt...?? watts increase or decrease in series and parallel ckt ?

[@jimcatanzaro7808]

Combination of resistors in a audio circuit could be bad a resistors value drift especially in tube amplifier designs only in a power supply would work

[@pauldavis7703]

Perfect. Thank you!

[@edwardhugus2772]

Thanks! All thumbs up!

[@enriqueemilio8990]

What about the watt resistance?

[@seeharvester]

Thanks Doc!

[@shsftech3792]

Thank vety helpful video and easy to understand!

[@clairejones314]

SUPER HELPFUL!!

[@JM.MEL_]

Thank you!

[@0GHatMak3r]

Example ps: there is no subscript q so I'll be using "ₚ" R2 = R1 Series Rₑₚ ≍ r1 x 2 Parallel Rₑₚ ≍ r1 ÷ 2

[@torukojinnn6331]

Great video dude

[@BenFinio]

Thanks!

[@xavierlozano9459]

I have a 100 wattsiren (8 ohms) with amp (11 ohms)... Im trying to lower the volume of the siren by adding resistors. I tried 4 900 ohm resistors in series parallel but the volume dropped to low. I'm afraid to damage the amp, any suggestions? Will it be better to use a 100 watt attenuator?

[@BenFinio]

Hi - it's common to use current-limiting resistors with LEDs (kzbin.info/www/bejne/e5ammWlulpZ6e8k) but they are not typically used in this manner with other types of loads. For example with a motor you would either decrease the voltage or use pulse-width modulation (PWM) to decrease the speed, not put it in series with a resistor. I'm not that familiar with sirens so I'm not sure of the best way to control the volume.

[@CerebralDad]

Hey Ben! Do you know how to mathematically predict how a pull down resistor will affect a 5v pwm signal. I have a controller board that has a floating pwm pin where I needed to add a pull down. I chose 1k ohm- because “the internet”, however I don’t think I’m getting 100% signal any longer. How do I find the minimal resistor value to pull down the output without affecting the value. It’s a poor Chinese board design where having a floating pwm pin is a safety issue and when powered on defaults to 100%. This is for a laser.

[@BenFinio]

Hi Keith! You would have to try and look up the specs of the board and see what the "output impedance" or "internal resistance" of the pin is. A mathematically ideal pin has zero impedance when used as an output (i.e. voltage does not drop at all when it's loaded) and infinite input impedance (does not draw any current from whatever it's measuring). In reality, the pin has some finite impedance in both cases. So when used as an output, and you add a pull-down resistor, you actually have two resistors in series forming a voltage divider, which will drop your output voltage. If you can find the internal resistance of your pin, then in theory you can calculate that voltage from this equation (where R1 is the internal resistance, R2 is your pull-down resistor, Vin is the ideal voltage from the board, e.g. 5V, and Vout is the actual output voltage): en.wikipedia.org/wiki/Voltage_divider#Resistive_divider (note that technically, this is assuming that whatever else you have your PWM pin connected to is not drawing any current, and all the current flows through R2). So, depending on what your minimum acceptable voltage is, you could calculate the minimum resistor value you need. However, that's all a very long-winded way of saying "just try a bigger pull-down resistor." I think 10k is a pretty common value. I don't think it *hurts* to oversize the resistor.

[@CerebralDad]

This is Keith Brown

[@panzznine713]

i've been wondering, is there a significant effect on using a multiple resistor than using 1 resistor, for example i have 1 10k ohms resistor and also have 2 5k ohms resistor, but i only need 10k ohms for my simple circuitry. so does using more resistor than using 1 resistor have an effect on the circuit?

[@BenFinio]

It depends on the circuit, but in general for most beginner/hobby circuits it probably won't matter. Two resistors take up more physical space than one resistor, so it could matter if you were designing a printed circuit board (PCB). However, assuming the resistors all have the same power rating (1/4 watt is the most common), then the two resistors would be able to dissipate 1/2 watt total. So if for some reason you need to pass a lot of current through the resistor but don't have a 10K resistor rated at 1/2W, then you could use the two 5K resistors.

[@panzznine713]

@@BenFinio i see, ty ty that was bugging me from the start lol

[@Delta_li_gifs]

Thanks

[@kenya_3364]

I don't get the parallel equation :( Can someone tell me how to do the equation

[@Efrancoskun]

thanks mate

[@soupinnile4268]

Very helpfull.

[@dogbarbill]

3:55 I hate to be a dummy but, explain the math for the 100 and the 47 in parallel to get 31.97

[@BenFinio]

1/(1/100+1/47)=31.97

[@steve-paul]

1÷(1÷100+1÷47)=31.97

[@feteheademe8746]

where have you been, now I know my electronics instructor suck's

[@jhoanstickmosqueravargas9998]

Entonces quien gano?

[@Unplugged_Reality]

Keep it up :D

[@jimcatanzaro7808]

Funny I just got a bunch of 100ohms R today

[@ebrahimkarimi1504]

Easy

[@bobbyd1961]

unfortunately Fritzing is NOT free they are asking for $10 to download it but thanks anyway

[@BenFinio]

Hm - is the "donation" mandatory, or can you just close that and download anyway? It's been a while since I downloaded it but it was free at the time. Tinkercad is a good alternative that runs in a browser and is free (for now at least).