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### Course: Class 10 Physics (India) > Unit 3

Lesson 5: Series and parallel resistors# Parallel resistors (derivation)

Discover the fascinating world of parallel resistors in electrical circuits! Learn how to identify parallel resistors, understand the concept of shared nodes and voltage, and apply Ohm's law to calculate equivalent resistance. Master the formula for parallel resistors and enhance your circuit analysis skills with real-life examples. Created by Willy McAllister.

## Want to join the conversation?

- May be asking this question prematurely, but the obvious questions is this: why does the net resistance actually DECREASE when using 2 resistors in parallel?(11 votes)
- Think of it this way, imagine you are in a supermarket and a bunch of people want to check stuff out. If you only have one cashier, the lines going to go pretty slowly. But if you add another cashier, more people can get their stuff checked out and the line would go faster. The more cashiers you add, the faster the line will go. Parallel circuits work similarly. The more resistors you add, the more paths through which current can flow. Hope this helps!(37 votes)

- Hi, I know how to calculate resistors in parallel but I've never fully understood the math behind it. To be more specific, I don't understand why we take 1/R1 + 1/R2. Can anyone please explain this to me or link to a useful video? Thanks.(11 votes)
- Hello Ben,

Here you go - hot off the YouTube presses. In this video I show how the equation may be derived by analyzing the current in the circuit.

https://www.youtube.com/watch?v=GEVjKzmr39E

Regards,

APD(7 votes)

- I might just be misunderstanding something, but I always thought that the current took the path of least resistance, so if you have parallel resistors with different amounts of resistance such as at7:43, wouldn't the current just choose the route with the 20 ohm resister?(5 votes)
- Imagine a narrow stream with water flowing. Drop a rock in the stream about 1/3 of the way from one edge. The water will split and flow around the rock on either side. The wide side corresponds to the lower resistor. Most, but not all, of the water goes on the wide side. There is always some that flows on the narrow (high resistance) side. Electric current does exactly the same thing.

If the rock is right in the center of the stream you would expect a 50-50 split of water on either side. If you move the rock slightly to one side you wouldn't expect all the water to suddenly 100% snap over to the slightly wider side.(10 votes)

- at1:23, Willy said that current divides itself. since current don't have any brain it would divide itself in a random way. then how can we find the relation of division of current?(3 votes)
- A mental image that might help is to think of water flowing in a stream. If there is a rock in the stream, the water flow splits in two, with some flowing to one side of the rock and some flowing around the other side. The water doesn't have a brain, but the process is not random. The flow of electrical current splits in the same way. It is not random, because it is based on known principles. For resistors in parallel, that principle is Ohm's Law. If you know Ohm's Law, and you realize the voltage on the two resistors is the same, you can solve for both currents.(12 votes)

- How does sharing the same node mean sharing the same voltage?(5 votes)
- Check out the definition of "node" and "distributed node",

https://www.khanacademy.org/science/electrical-engineering/ee-circuit-analysis-topic/modal/a/ee-circuit-terminology

or

https://www.khanacademy.org/science/electrical-engineering/ee-circuit-analysis-topic/modal/v/circuit-terminology(4 votes)

- hi .

why did we take the voltage as V itself in finding parallel resistance whereas for finding series resistance , we took the voltage for each resistor as V1, V2, V3 etc. ?(4 votes)- We should always take Voltage as V in parallel resistors and V1, V2 and V3 in Series.

We take it so because, voltage splits in series while the voltage is constant in parallel.(2 votes)

- At8:46, what is the significance of using the reciprocal of 1/Rp when we established that i = V * (1/Rp)?(3 votes)
- Your point is valid; there is no mathematical need to work out R since we use the reciprocal directly in the next step to calculate i, the current. Taking the reciprocal is just a habit. I feel more comfortable knowing the resistance value as well as the 1/R value (the conductance). It is easier for me to notice an error if the resistance value is way off, compared to noticing an incorrect conductance (often a number way less than 1). Lots of engineers work with parallel circuits using conductance (1/R) values, it is a matter of personal choice.(3 votes)

- Why is voltage in a parallel circuit the same as the voltage of the source? I thought voltage was added up across components which would be equal to the voltage of the source.(3 votes)
- If you have two resistors in parallel that means their two terminals are tied together (perhaps think of the resistors as holding hands). There are only two wires. There can only be one voltage between two wires.

You are perhaps describing how the voltages add up for resistors in*series*.(1 vote)

- 6:52

shouldn't V * (1/R1 + 1/R2)

equal

V * 2/Rp?

because 1/n + 1/n = 2/n generally, or is it this way because you reduced it?(0 votes)- Check your denominators. They are not all the same.(7 votes)

- I've been taught this equation for Rp=(R1×R2×...×Rn)/(R1+R2+...+Rn). For some reason i find it easier instead of taking 1 over the Rp and the rest of the Rn s' ) . But will it affect in some other way the solution of any problem ?(1 vote)
- That formula does not work for 3 or more resistors. Use the version with all the reciprocals.

You are perhaps thinking of the special case for 2 resistors in parallel, Rp = (R1 R2)/(R1 + R2). This only works for 2 resistors.(3 votes)

## Video transcript

- [Voiceover] In this
video, we're gonna look at another familiar pattern of resistors called parallel resistors. And I've shown here two
resistors that are in parallel. This resistor is in
parallel with this resistor. And the reason is it shares nodes. These two resistors share the same nodes. And that means they have the same voltage. And they are called parallel resistors. So if you share a node. Share the same node. Then you share the same voltage. And you are in parallel. That's what that word means. Now, if we go look closer here, we'll see some interesting things. It's hooked up, we have a battery here, some voltage v. And because there's a path,
a complete path around here, we're gonna have a current. We're gonna have a current i flowing in this circuit. Let's label these resistors. Let's call this one R
one and this one R two. Those are our parallel resistors. When the current reaches this point here, when a current reaches this node, it's gonna split. It's gonna split into
two different currents. That current and that current. We'll call that one i one,
because it goes through R one. And we'll call this one i two. And that goes through R two. Now, we know any current
that goes into a resistor comes out the other side. Otherwise it would collect
inside the resistor, and we know that doesn't happen. This one comes here. And they rejoin when
they get to this node, and flow back to the battery. So the current down here is again i, the same one as up here. Now, what I want to do is I want to replace these two resistors with an equivalent resistor,
one that does the same thing. And by "the same thing," we mean causes the same current to
flow in the main branch. And so that's what's drawn over here. Here's a resistor here. We'll call this v again. And we'll call this R parallel. R P. And this resistor causes the
same current i to flow here. And now we're gonna work
out an expression for that. We want to figure out
how do we calculate R P in terms of the two
parallel resistors here. Okay, so let's go at it. What we know, let's see what we know
about this over here. What we know is the voltages on the two resistors are the same. We know there's two different currents, assuming that these are two
different-valued resistors. And now, with just that information, we can apply Ohm's law. And we use our favorite thing, Ohm's law. Which says that voltage
on a resistor equals the current in the resistor
times its resistance. So let's write down Ohm's law for R one and R two. Okay, we know the voltage,
we'll just call the voltage v. This is for R one. v equals i one times R one. And for R two, we can
write a similar equation, which is v, same v, equals i two R two. Now there's one more fact that we know, and that is that i one
and i two add up to i. And these are the three facts that we know about this circuit. What I'm gonna do now is come up with an expression
for i one and i two based on these expressions, plug them into this equation. Okay, I can rewrite this equation as i one equals v over R one. I can write this one as i two equals v over R two. And now I'm gonna plug
these two guys into here. Let's do that. i equals i one, which is v over R one plus v over R two. Let me move the screen up a little bit. Okay, now we're gonna continue here. I just want to rewrite this a little bit. i equals v times one over R one plus one over R two. Okay, so here we have an expression. It actually sort of looks like Ohm's law. It has an i term, a v term, and this R term here. Let me go back up here. Here's our original Ohm's law. I'm gonna write this, I'm gonna solve for i in terms of v, just to make it look
a little more obvious. I can say i equals v over R. And what I hope you see here is the similarity between this equation and this one down here. So I have this R here. And what's happening is this term is playing the role of that, resistance. So I'm gonna bring this
equation down here, and write it right down here. Times one over R. I'm gonna call this R P. Because what I want is for this expression and this expression, I'm gonna set those equal. Same i, same v. These guys are equal. I can write it all, I'll
just write it over here. One over R P equals one over R one plus one over R two. This says we have a resistor, we're gonna call it R P or R parallel, that acts like the parallel
combination of R one and R two. So this is the expression for a parallel resistor. If you want to calculate a replacement for R one and R two in parallel, you do this computation and you get R P. So let's do one of these for real. Here's an example. Here's an example where I've actually filled in some numbers for us. So I have a 20-ohm
resistor in parallel with a 60-ohm resistor, driven
by a three-volt battery. And what I want to do is I want to combine these two parallel resistors and find out what is the current right here. Find out what is the current,
that's my unknown thing here. I know everything else about this. So let's use our equation. We said that one over R P was equal to one over R one plus one over R two. And let's just fill in the numbers. One over R P equals one over 20 plus one over 60. That equals, let's just make
60 the common denominator. So I have to multiply this one by three. Three over 60 plus one over 60. And that equals four over 60. And so now I'm gonna
take the reciprocal here. R P equals 60 over four or R P equals 15 ohms. So what this is telling us is if we have two resistors in parallel, 20 ohms and 60 ohms, that is, for the purposes of
calculating the current here, that's the same as 15 ohms. It took the three volts. Just like that. Let's check what the current is. The current is i equals v over R equals three volts over 15 ohms. That's equal to 0.2 amps. Or you can say it's the
same as 200 milliamps. So we actually have now
simplified our circuit from two resistors to one resistor. And we were able to
compute the current here, which is 0.2 amps. And I would invite you to check this by going back and computing
this current up here to make sure it's the same. And the way you would do that is you would calculate the voltage. The voltage here is three volts. Three volts across 20,
three volts across 60. You'll get i one and i two. And if you add those together, you'll get the total i, and it should come out the same as this. And I think that's a good
exercise for you to do, to prove that the expression
for a parallel resistor, one over R parallel can be computed from one over R one plus one over R two.