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# Parallel resistors (part 3)

Learn to calculate parallel resistors using the product-over-sum formula for efficient circuit analysis. Delve into the special case where R1 equals R2, resulting in Rp being half of the resistor value. Grasp the concept that parallel resistance is always smaller than the smallest resistor, as it provides multiple current paths, effectively reducing the overall resistance in the circuit. Master these techniques to enhance your understanding of electrical engineering concepts. Created by Willy McAllister.

## Want to join the conversation?

• So, we can see from this video (starting at around ) that two resistors in parallel with the same resistance will produce an equivalent resistance that is half of the original resistance. What happens, though, if we keep increasing the number of resistors? Let's say, for example, we have three resistors in parallel with the same resistance. Will the equivalent resistance be one-third of the original resistance?

So what I'm wondering can be summarized like this:

R = resistance of all resistors in circuit
n = number of resistors
Rp = equivalent resistance
So, will Rp = R^n/nR?
• With Req = R/N, this means that one could find an integer fraction of a specific resistance given said integer amount of resistors having said specific resistance.

This means if you want a 40 Ohm resistor, you can wire three 120 Ohm resistors in parallel.

40 Ohm = 120 Ohm/3 Resistors
(1 vote)
• If you use the Rp=R1*R2/R1+R2 equation presented here how do you solve it if there are more than 2 parallel resistors? I tried Rp=R1*R2*R3/R1+R2+R3 and I didn't come up with the right answer.
• Hello Phelps,

Sorry, the Rp=R1*R2/R1+R2 is only applicable when you have two resistors. When you have more than two you should use 1/Rp = 1/R1 + 1/R2 + .... + 1/Rn

Regards,

APD
• Hello, at you say that (1) the equivalent resistance of a bunch of parallel resistors will always be smaller than the smallest resistance in the bunch. You argument this (2) "because you have two current paths that allow current to go two different ways". I can't really get my head around that, why does (2) imply (1)? How does the current having two different paths (or more) determine a smaller equivalent resistance than the smallest resistance of one of the paths?
• Can I use the part3 equation when I have more than two resistors
• What happened to Parallel resistors (part 2)?
• i do not understand what did you meant by "Because you have two current paths that allow current to go two different ways, the effective resistance is always smaller than the smallest original path, because there's a way for current to go around another way" could you explain more the reason ?
• When you have two resistors in parallel they share the same voltage, and have their own individual currents. You find the current using Ohm's Law, I1 = V/R1, and I2 = V/R2.

We can ask the analytic question, "What is the 'equivalent' resistance?" or asking another way, "If you replace the pair of resistors with a single resistor, what resistance would you chose for Rp such that the total current would be the same? That is, I(Rp) = I1 + I2.

You were asked to derive or memorize an equation to find the precise value of Rp...
Rp = (R1*R2)/(R1+R2)

It is also illuminating also ask a qualitative question, "Is Rp bigger, smaller, or in between R1 and R2?, and why is that?"

If you look at the math it tells you Rp is always smaller than the two original resistors. Can you see why? The current in Rp is Ip = the sum of the currents in R1 and R2. That means Ip is always bigger than both I1 and I2. When you put Ip into Ohm's Law along with the original V, you see that Rp has to be smaller than R1 and R2 to get Ohm's Law to come out true.

Rp = V/Ip = V/(I1 + I2)

That's what the math says. What is the physical reason?

I think of electric current like water in a creek. The tilt of the creek provides the pressure to make water flow, corresponding to the voltage value V. Suppose you have a narrow creek with three stones blocking the water, named Stone1, MiddleStone, and Stone2. If you lift up the stone on the left you will get a current flow proportional to the width of the opening. That's R1 and the flow is I1. If you replace Stone1 and lift up Stone2 on the far right you will see I2 flowing proportional to the width of the opening left by Stone2.

If you lift up both Stone1 and Stone2 there are two openings and total flow is greater than both I1 and I2. That's because the two gaps are open and there is more space for water to flow. There's no arrangement of stones that will make less total water flow.

Resistors in parallel are exactly the same. If you provide more pathways for electric current to flow, the equivalent resistance is always lower than either original resistor.
• why is that when the current was split, it became doubled, with a much higher resistor? how does that happened?
• Imagine a four-lane highway splitting into an eight-lane highway. No matter how much resistance there is on the new path, adding another path will always allow more electrons to pass
(1 vote)
• Is their a part 2 or is it mislabeled part 3?
• would there be a difference in the answer if u use the 1/RP FORMULA instead of the RP= R1R2/R1+R2 formula
(1 vote)
• What about if you add in an extra resistor, so 1 resistor before the 2 in parallel. Do you add all 3 fractions or do you just stick to the 2 resistors in parallel ?
• If you have 3 resistors in parallel you can simplify them in one step by using the general form of the parallel resistor formula,

1/Rp = 1/R1 + 1/R2 + 1/R3

As an exercise, draw three resistors in parallel. Combine two of them, then combine that result with the third resistor. You get to use the 2-resistor special form of the parallel resistor formula,

Rp = (R1 * R2)/(R1 + R2)

Compare this 2-step method with combining all three in one calculation. Should come out the same, right? Try combining different pairs of resistors, like start with R1 and R3.

The math should work great any way you do it!
(1 vote)

## Video transcript

- [Voiceover] In this video we're gonna talk even some more about parallel resistors. Parallel resistors are resistors that are connected end to end, and share the same nodes. Here's R one and R two, they share the same nodes. That one and that one. And that means they share the same voltage. And we worked out an expression for how to replace that with a single resistor. R parallel, and we found that one over R parallel is one over R one plus one over R two. So in this video I'm gonna actually start working with this expression a little bit more. And we'll just change it around a little bit to an easier, an easy way to remember it, and then we're gonna do a special case, where R one and R two are the same value and we're gonna see what happens. So right now I wanna do just a little bit of algebra on that expression, one over R p is the same as one over R one plus one over R two. What I wanna end up with here is an expression that on this side says R p equals something, and on this side I just want one expression, not two fractions. So we're gonna go about that by combining these two fractions first. So the least common denominator here, LCD, equals R one times R two, and I'm gonna convert both these to convert this to that proper denominator I have to multiply it by R two over R two, so we'll do it. We'll do all the steps. We'll multiply it by R two over R two. This expression we have to multiply it by R one over R one. And this equals, of course, one over R p. Continuing on, one over R p equals R two over R one R two plus R one over R one R two, and now I can combine them together. Let's move up here. One over R p equals, we'll keep everything in numerical order, R one plus R two over R one R two. And now I'm gonna take the reciprocal of both sides of the expression so I get an expression in R p, R p equals, and just flip over this expression, R one R two over R one plus R two. That is a way you can remember how to combine parallel resistors. The parallel equivalent resistor, R p, is the product of the two resistors over the sum. So it's the product over the sum. That's how I remember it. Now this arithmetic, this expression, is exactly the same as the original one that we had, these are the same, and it's just a question of which one do you want to remember, which one's easier to remember and which one's easiest to calculate. I like to remember this one here. Let's do a quick example using it. I'll move the screen up a little bit. Let's leave that there so we can see it. Let's say we have two parallel resistors. And we'll say the first one is 1,000 ohms, and the second one is 4,000 ohms. And the question is, what is the parallel combination of those things. How can I replace these with one resistor so the same current flows. And we'll use this expression here. So R p equals a product which is 1,000 times 4,000 divided by the sum 1,000 plus 4,000 and that equals, oh my goodness there's a lot of zeroes here. Four and six zeroes, one, two, three, four, five, six over, that's easy, 5,000. All right? And let's say, let's take out three zeroes out of this one. Let's knock off three zeroes here, and three zeroes here. And I have 4,000 divided by five and that equals 800 ohms. So that is this and that's how we use this expression. Now just something to notice here, notice that R parallel, the equivalent parallel resistor, is smaller than both of these. And that happens every time, that happens every time. The parallel resistance is smaller than the smallest resistor. So here it was 1,000, it's gonna be smaller than that. And that's a property of parallel resistors. Because you have two current paths that allow current to go two different ways, the effective resistance is always smaller than the smallest original path, because there's a way for current to go around another way. So this is now the expression for two parallel resistors, and that's a good one to remember. Now I'm gonna show you one special case, and we'll do this in this color. What if R one equals R two? What is R p? And for this special case, we use the same expression, we say R p, we say R p is the product, we'll just use R because R is, it's the same value. R times R over R plus R, and that's multiplied. And so that is R squared over two R, and one of these Rs cancels, so let's cancel that R and that squared, and we end up with R over two. So for the special case, if R one equals R two, then R p equals R over two. It's just half, and that should make sense. If we do something like this, if we draw two resistors in parallel, like this. and we say this is 300 ohms and this is 300 ohms, that means the effective, the effective parallel resistor, 150 ohms. And that's got a very pleasing symmetry to it because these resistors are the same, they have the same voltage, they're going to have the same current, basically twice as much current is gonna flow in this circuit as would flow if there was just one of these guys, so that's where the divide by two comes from. So for two resistors in parallel, if the resistors are the same value, the effective parallel resistance is just half.