Main content

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

Lesson 6: Solving a circuit with series and parallel resistors# Example: Analyzing a more complex resistor circuit

Master the art of solving complex circuit problems with this guide. Learn how to simplify circuits by finding equivalent resistances, understand the concept of resistors in series and parallel, and apply Ohm's law to find the current. Created by Sal Khan.

## Want to join the conversation?

- Can we calculate the current in each of the resistors ?(54 votes)
- its quite easy infact if you just remember these two basic concepts....

1). current is same across all resistors in series..

2). to find current in a particular resistance (r1) connected in parallel to others (equivalent resistance = R.Eq )...

current through r1 = * [ eq. resistance of all (connected in parallel )except r1 ] X {total current flowing through " R.Eq " ] / [ R.Eq ] *

P.S. - derivation through V = IR or krihoff voltage law..

`code`

`if (x < 0) {`

return;

}(1 vote)

- Why in a real life application would we use multiple resistors in both series and parallel configurations together?(20 votes)
- How can we calculate the internal resistance inside a battery?(6 votes)
- You can't calculate it, but you can measure it by measuring how the terminal voltage changes as the current changes, and then applying ohm's law.(3 votes)

- what i think here is that when the circuit is in parallel the equivalent resistance is less than the least resistance in the parallel circuit .. and when in series... the equi. resistance is greater than the largest resistance in the series circuit.. am i right??(4 votes)
- You are absolutely correct. The series circuit, I need not answer. The parallel circuit, I will prove it intuitively. A wire is also a resistence, and a good example. When we see the electrons bumping into each other and into the atomic nuclei inside the wire, we call that property resistence. Kind of like viscosity in fluids. So if a wire gets longer, it has more chances to bump into something. Also, if a wire gets thicker, an electron will have more chance to AVOID collisions. So a thicker wire means less resistence. In analogy, if one has few wires in parallel, it will have same effects as making one wire thicker. So multiple wires would mean less resistence than the least of one(5 votes)

- If you call these complex, Idk what the hell I'm solving(4 votes)
- can u please make it clear how to determine the direction of the current flow then it would be helpful(3 votes)
- in a simple DC circuit, the direction of current is from positive terminal to negative(2 votes)

- what is e.m.f and terminal voltage of a cell?how are they related to internal resistance?(1 vote)
- EMF is electromotive force, and it's another way to refer to voltage.

Terminal voltage is the voltage that the battery can deliver into a working circuit. If you have a battery, you might think of it as say a 9 volt battery, but that 9v is what it can deliver when it is not having to put out much current - in other words, when the resistance in the rest of the circuit is very high.

When you connect a batter to a circuit where the resistance is low, then you have a situation where the battery is trying to put out a lot of current. If you want to calculate how much current there will actually be, now you can't ignore the resistance that is internal to the battery.

To make the math easy, let's say we have a 10v battery connected to a 10k ohm resistor. The current is going to be very small, only 1 mA, but technically that is only an approximation, because the battery has resistance, too. So, say the battery has internal resistance of 1 ohm. Now the real resistance of the circuit is 10.001k ohm, and you can see, that's not going to make any meaningful difference when you calculate the current. It's still really, really close to 1 mA. The voltage drop throughout the circuit is going to occur almost entirely across the 10k resistor. The battery terminal voltage therefore will be really close to 10 V.

Now, connect the battery instead to a resistor that is only 10 ohm. If you had a batter with no internal resistance, you have 10 V/ 10 ohm = 1 amp. But now you can see that the internal resistance matters. The real resistance around the circuit is not 10, but 11. So the current is closer to 0.9A (I am rounding 10/11) And now you can see that the voltage drop that occurs across the resistor will be 0.9A x 10 ohm = 9V, and that's the terminal voltage of the battery. Where is the other 1 V we expected to see, since the voltage all the way around the loop is supposed to sum to 0 (+10 from battery, -9 lost across resistor, 1 more volt needs to be accounted for). That voltage drop occurs across the internal resistance of the battery itself.(5 votes)

- The circuit in the video was 2D and therefore could be analyzed step by step. However what would happen if resistors were connected in a 3D circuit. For example, a circuit is made in the shape of a cube where there is a resistor of 1 ohm on each edge of the cube. In this case, what would be the total resistance.(2 votes)
- how do you solve for the current through each individual resistor in the circuit?(1 vote)
- If it is a series circuit, the current is the same through all the resistors. Find the total resistance, divide the voltage by that, and you will have the current.

If it is a parallel circuit, then just do V/R for each of the resistors, and you will have the currents.(4 votes)

- Is the conventional symbol for the resistor a zigzagged line or a kinda like rectangle cause I`m confused.

I drew that on a test and failed(HORRIBLY)

ANY HELP WOULD BE APPRECIATED,THANKS!(1 vote)- As much as I have come across, the "zig-zagged line" is the conventional symbol for a resistor. However, on some informal or school level practical physics experiments, and related observational notes, a
*Resistance box*is often represented with a rectangle with an 'R' written inside it.

Cheers!(3 votes)

## Video transcript

Let's see if we can apply
what we've learned to a particularly hairy problem
that I have constructed. So let me see how I can
construct this. So let's say in parallel, I have
this resistor up here. And I try to make it
so the numbers work out reasonably neat. That is 4 ohms. Then I have
another resistor right here. That is 8 ohms. Then I have
another resistor right here. That is 16 ohms. And then, I
have another resistor here, that's ohms. Actually, I'm now
making it up on the fly. I think the numbers
might work out OK. 16 ohms. And let's say that now here in
series, I have a resistor that is 1 ohm, and then in parallel
to this whole thing-- now you can see how hairy it's getting--
I have a resistor that is 3 ohms. And let's say
I have a resistor here. Let's just make it
simple: 1 ohm. And just to make the numbers
reasonably easy-- I am doing this on the fly now-- that's
the positive terminal, negative terminal. Let's say that the voltage
difference is 20 volts. So what I want us to do is,
figure out what is the current flowing through the wire
at that point? Obviously, that's going to be
different than the current at that point, that point, that
point, that point, all of these different points, but it's
going to be the same as the current flowing
at this point. So what is I? So the easiest way to do this
is try to figure out the equivalent resistance. Because once we know the
equivalent resistance of this big hairball, then we can just
use Ohm's law and be done. So first of all, let's just
start at, I could argue, the simplest part. Let's see if we could figure out
the equivalent resistance of these four resistors
in parallel. Well, we know that that
resistance is going to be equal to 1/4 plus 1/8
plus 1/16 plus 1/16. So that resistance-- and
now it's just adding fractions-- over 16. 1/4 is 4/16 plus 2/16 plus 1
plus 1, so 1/R is equal to 4 plus 2 is equal to 8/16-- the
numbers are working out-- is equal to 1/2, so that equivalent
resistance is 2. So that, quickly, we just
said, well, all of these resistors combined is equal to
2 ohms. So let me erase that and simplify our drawing. Simplify it. So that whole thing could now
be simplified as 2 ohms. I lost some wire here. I want to make sure that
circuit can still flow. So that easily, I turned that
big, hairy mess into something that is a lot less hairy. Well, what is the equivalent
resistance of this resistor and this resistor? Well, they're in series, and
series resistors, they just add up together, right? So the combined resistance of
this 2-ohm resistor and this 1-ohm resistor is just
a 3-ohm resistor. So let's erase and simplify. So then we get that combined
resistor, right? We had the 2-ohm that we
had simplified and then we had a 1-ohm. So we had a 2-ohm and a 1-ohm
in series, so those simplify to 3 ohms. Well, now this is getting
really simple. So what do these two resistors
simplify to? Well, 1 over their combined
resistance is equal to 1/3 plus 1/3. It equals what? 2/3. 1/R is equal to 2/3, so R is
equal to 3/2, or we could say 1.5, right? So let's erase that and
simplify our drawing. So this whole mess, the 3-ohm
resistor in parallel with the other 3-ohm resistor is equal
to one resistor with a 1.5 resistance. And actually, this is actually
a good point to give you a little intuition, right? Because even though these are
3-ohm resistors, we have two of them, so you're kind of
increasing the pipe that the electrons can go in by a
factor of two, right? So it's actually decreasing
the resistance. It's giving more avenues for the
electrons to go through. Actually, they're going to be
going in that direction. And that's why the combined
resistance of both of these in parallel is actually half
of either one of these resistances. I encourage you to think about
that some more to give you some intuition of what's
actually going on with the electrons, although I'll do a
whole video on resistivity. OK so we said those two
resistors combined-- I want to delete all of that. Those two resistors combined
equal to a 1.5-ohm resistor. That's 1.5 ohms. And now all
we're left with is two resistors in parallel, so the
whole circuit becomes this, which is the very basic one. This is a resistor: 1.5
ohms, 1 ohm in series. Did I say parallel just now? No, they're in series. 1.5 plus 1, that's 2.5 ohms.
The voltage is 20 volts across them. So what is the current? Ohm's law. V is equal to IR. Voltage is 20 is equal to
current times our equivalent resistance times 2.5 ohms. Or
another way to write 2.5 five is 5/2, right? So 20 is equal to I times 5/2. Or I is equal to 2/5 times
20, and what is that? 2/5 is equal to I
is equal to 8. 8 amperes. That was not so bad,
I don't think. Although when you saw it
initially, it probably looked extremely intimidating. Anyway, if you understood that,
you can actually solve fairly complicated circuit
problems. I will see you in future videos.