Parall resistors - Page 1
Forum section: electronics / beginners

author: Conny73
author_rank: Newbie
timestamp: September 05, 2017, 07:34:20 pm
content: Hi.Can anyone explaine why there is a drop in Ohm when you put tow resistors in parall Connection?Should it not be the same as the lowest resistor value?Cant relly find out why you can reduce te Ohm value in this type of Connection.If you are connecting speakers this is important but in an normal Circuit can it be useful or like Dave shown in the Zener video a Zener has a resistor in it and if the load is constant should one take this in consideration when calculation the right value of the resistor?//Conny

author: Vtile
author_rank: Super Contributor
timestamp: September 05, 2017, 07:45:10 pm
content: It is because there is "twice as much" room for the current to flow through that causes the power to double (in case of regid power supply / source and equal resistances). The explanation of reduced combined resistance comes from the idea of the allowance of current flow, usually coined as conductance (swe, konduktans).

author: suicidaleggroll
author_rank: Super Contributor
timestamp: September 05, 2017, 07:46:36 pm
content: Current (I) through a circuit is equal to the voltage (V) divided by the resistance (R). So take one voltage source, put a resistor on it, and you can calculate the current through the resistor as I=V/R. Now take a second, identical circuit, both of them will act the same way, both will have a current of I through their resistor.Now merge these two circuits so they share a voltage source. Nothing changes, you still get the same amount of current through each resistor, right? So what's the total amount of current leaving the voltage source? 2*I, since you have two resistors, each one with a current of I. Ok, so now if you have a voltage of V, and a current of 2*I, what is the effective resistance? It's R/2. If you get double the current for the same voltage, it means you have half the resistance.There's a popular phrase, "current takes the path of least resistance". It's 100% wrong. Current does not take the path of least resistance, current takes ALL AVAILABLE paths, all the time. The current will be distributed according to the relative resistance (actually impedance) for each path, but they all work together. This is why adding a second resistor lowers the effective resistance, the current now has two paths it can take instead of just one.

author: Ice-Tea
author_rank: Super Contributor
timestamp: September 05, 2017, 07:49:33 pm
content: Imagine a basin of water with a single tube dumping water out of it. If you add another, identical, tube the second tuve will dump exactly as much water. In other words, the resistance the entire 'solution' presents to the basin of water is half of the original.Same thing.

author: Brumby
author_rank: Unknown
timestamp: September 06, 2017, 02:10:48 am
content: This is, so far, the best (simplest) answer to the question that was asked (Not that the others had anything wrong with them)Quote from: Ice-Tea on September 05, 2017, 07:49:33 pmHowever, the tubes don'thaveto be identical. You can have a large diameter tube and a small diameter tube andbothwill allow water to flow - just at different rates.... and you can have more than 2, if you want.

author: rstofer
author_rank: Super Contributor
timestamp: September 06, 2017, 04:33:18 am
content: Quote from: Conny73 on September 05, 2017, 07:34:20 pmThe internal resistance of a zener diode is quite low, a few Ohms, perhaps. It is much to low to use as the only resistance in the circuit as it is unlikely the diode can handle the possible current.So, we add another resistance of some type in series with the reverse biased zener diode.Here's a discussion of the zener diode (after a long discussion of the ordinary diode):https://www.allaboutcircuits.com/textbook/semiconductors/chpt-3/zener-diodes/

author: Beamin
author_rank: Super Contributor
timestamp: September 06, 2017, 04:39:24 am
content: Quote from: rstofer on September 06, 2017, 04:33:18 amI thin he getting confused with resistance of the device and a resistor. A cap has ESR but it doesn't have resistor or inductor inside of it. A zener has resistance but no resistor inside it.

author: Ice-Tea
author_rank: Super Contributor
timestamp: September 06, 2017, 08:41:01 am
content: Quote from: Brumby on September 06, 2017, 02:10:48 amOff course. But simplifying the example often helps to understand

author: Brumby
author_rank: Unknown
timestamp: September 06, 2017, 09:10:08 am
content: I agree about simplification - but a constraint was added that was not necessary.Well ... that's how I felt.Anyway, we can just hope the OP has gleaned something useful from everyone's efforts.

author: John B
author_rank: Frequent Contributor
timestamp: September 06, 2017, 09:26:34 am
content: As vtile touched on, the key concept is conductance which is the reciprocal of resistance. If you have multiple conductive channels for current to flow, the total current is increased, therefore the resistance is less.The key mathematical point is that the total resistance in a parallel bank of resistors will be less than any single resistance.

author: retrolefty
author_rank: Super Contributor
timestamp: September 06, 2017, 11:31:42 am
content: Quote from: Ice-Tea on September 06, 2017, 08:41:01 amThere are also simplifying that can be confusing for beginners. How does the 'rule' current takes the path of least resistance' help, probably doesn't?

author: Conny73
author_rank: Newbie
timestamp: September 06, 2017, 03:37:57 pm
content: Hi GutsThanks for all of yous replays.If i understan all off yours replaysCorrect me if im wrong but if i do a Circuit White only two resistors of 10 ohm`s each connected in parallel i only get 5 ohm on my meter due too the meter is sending out the same test voltage and current and my resistors shere the load and my meter doesen know there are too or more resistors i reads the value wrong.. So the math do add upp but it is not the right way too messure it..Intresting too learn...BRConny

author: Beamin
author_rank: Super Contributor
timestamp: September 06, 2017, 05:23:33 pm
content: Quote from: Conny73 on September 06, 2017, 03:37:57 pmThe value is right. Two resistors in parallel half the measured resistance. For capacitors the opposite is true two in parallel double the capacitance but half the rated voltage they can handle. Two resistors in series add the resistance of each one. Two caps in series keep capacitance the same but double the voltage they can handle.

author: Brumby
author_rank: Unknown
timestamp: September 06, 2017, 05:26:54 pm
content: Yes, the meter is providing a voltage and measuring the current that the resistor(s) allow through.When there is one resistor, the current going through it is a certain value. When you add another resistor (which is the same) then it, too, allows the same amount of currentas well, so the meter only sees that there is double the current.The only deduction the meter can make is that if there is double the current, then there must be half the resistance. It doesn't know if you have added two 10 ohm resistors in parallel - or a single 5 ohm resistor, since both of these will allow the same amount of current to pass through, for a given voltage.

author: Brumby
author_rank: Unknown
timestamp: September 06, 2017, 05:29:26 pm
content: Quote from: Beamin on September 06, 2017, 05:23:33 pmNot too much too quickly...!That's thenextlesson.

author: suicidaleggroll
author_rank: Super Contributor
timestamp: September 06, 2017, 07:28:18 pm
content: Quote from: Beamin on September 06, 2017, 05:23:33 pmI'm not sure why you're bringing up caps and voltage ratings, but what you've said here is very wrong. Two caps in parallel does double the capacitance, but it doesn't change the rated voltage at all. Two caps in series drop the capacitance in half and double the rated voltage.

author: suicidaleggroll
author_rank: Super Contributor
timestamp: September 06, 2017, 07:30:49 pm
content: Quote from: Conny73 on September 06, 2017, 03:37:57 pmNothing is being read wrong, it's perfectly correct. You have to remember that the meter is measuring the resistance of the entire circuit. If that circuit consists of two resistors in parallel, then it will measure the resistance of the parallel combination. If you wanted to measure the resistance of a single resistor, then the only thing "wrong" is the test setup, you shouldn't be connecting the meter to two resistors in parallel, since that's not the circuit you're interested in.If you have two milk jugs and you want to know how much one of them weighs, you only put one of them on the scale. You don't put both of them on the scale and then claim that the scale is wrong.

author: Brumby
author_rank: Unknown
timestamp: September 07, 2017, 03:19:38 am
content: Quote from: suicidaleggroll on September 06, 2017, 07:28:18 pmI didn't even look that closely. I just saw "caps" and wanted to put the brakes on.Mea culpa.This is correct:Quote

author: Ice-Tea
author_rank: Super Contributor
timestamp: September 07, 2017, 06:33:56 am
content: Quote from: retrolefty on September 06, 2017, 11:31:42 amIt helps for understanding short circuits. For other cases, it's too ambiguous.

author: amyk
author_rank: Super Contributor
timestamp: September 07, 2017, 08:56:22 am
content: I agree the water analogy is the best way to think of this.Quote from: Beamin on September 06, 2017, 04:39:24 am...and a zener also has capacitance and inductance!

author: kalel
author_rank: Frequent Contributor
timestamp: September 07, 2017, 09:06:46 am
content: Here are two different example circuits. Just a 5v power supply and:- 2 resistors in parallel on the top circuit- 1 resistor (the 2nd is disconnected) on the bottom circuitYou can see how much current goes through each wire.We can see on the top circuit, 500mA goes through both of the resistors.So, on the bottom wire of the top circuit, we get 1A total current (500mA through each resistor, so 500mA*2 resistors=1000mA).On the bottom circuit, one resistor is disconnected, so we have 500mA traveling only through one resistor.That also makes the total that returns on the bottom wire 500mA (the other "path", the second resistor is disconnected, that's why we lost the extra 500mA).I hope I didn't make some mistakes in my description. You can experiment with the example here if you wish:http://tinyurl.com/yajka884This also makes the total resistance less. A single resistor still has 10ohm of resistance. But because you have 2 resistors connected to the same power source in parallel, the total resistance is 5ohms (because more current can travel through the circuit, having 2 paths), while each individual resistor still has 10ohms. If you disconnect the second resistor, you will get 10ohms reading. If you reconnect the second transistor, again 5ohms.The same way, if you connect two 10ohm resistors in series, you will get 20ohms, because the current will have two "barriers" to go through, or two "hills" to climb on, which is more difficult than going through just a single barrier.

author: Beamin
author_rank: Super Contributor
timestamp: September 19, 2017, 12:34:58 am
content: Quote from: Brumby on September 07, 2017, 03:19:38 am

author: Brumby
author_rank: Unknown
timestamp: September 19, 2017, 04:25:44 am
content: Quote from: Beamin on September 19, 2017, 12:34:58 amYou need to do some reading on capacitors. The first thing you need to put to one side is the voltage rating of a capacitor. This is a value where the mechanical design of the capacitor is intended to be capable of functioning. Intheory, there is no essential link between the voltage rating and capacitance, but in practice there are, because of the finite limitations of the materials used.The next thing is that the capacitance isinverselyproportional to the distance separating the plates of capacitor. That is, capacitanceincreasesas the distance between them gets smaller.This is where we find voltage ratings come into play - not because of capacitance, but because of the distance.The closer the plates, the easier it is for charge to jump across the gap - just like a spark plug. So we have to address this somehow if we want big capacitances in a small space. We do this by putting a dielectric material between the plates. This enables the charges between the plates to have influence on each other, but prevents electrons from jumping across - to the limit of their ability.Better quality dielectrics can withstand higher voltage differences across a given distance, but there isn't a perfect dielectric (that I know of) that will withstand huge voltages over very short distances. This is the reason why a 10uF 16V capacitor can be a lot smaller than a 10uF 450V one. It is also why a 1F super capacitors the size of a couple of CR2032s will only run to 5V.When you put capacitors in series, it is like having the plates separated in two stages. Using two identical capacitors, the mid point can be effectively ignored and the top and bottom plates have a total gap between them that is twice as large as a single capacitor. This halves the capacitance.The voltage capability of each capacitor is not changed and with two of them, each can withstand the same voltage as before, which means the series pair can withstand double.