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Class 10 Physics (India)

Unit 3: Lesson 5

Series and parallel resistors

Parallel resistors (part 3)

When two resistors are in parallel, the equivalent resistance is the product of the two resistors divided by their sum. When both resistors are the same value, the equivalent parallel resistance is exactly half of the original resistance.  Created by Willy McAllister.

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• 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
• Can I use the part3 equation when I have more than two resistors
• why is that when the current was split, it became doubled, with a much higher resistor? how does that happened?
• What happened to Parallel resistors (part 2)?
• 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)
• 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.
(1 vote)
• 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?