Main content

## Physics library

### Course: Physics library > Unit 12

Lesson 1: Ohm's law and circuits with resistors- Introduction to circuits and Ohm's law
- Basic electrical quantities: current, voltage, power
- Resistors in series
- Resistors in parallel
- Example: Analyzing a more complex resistor circuit
- Analyzing a resistor circuit with two batteries
- Resistivity and conductivity
- Electric power
- Kirchhoff's current law
- Kirchhoff's voltage law
- Kirchhoff's laws
- Voltmeters and Ammeters
- Electrolytic conductivity

© 2023 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# Analyzing a resistor circuit with two batteries

An example of simplifying a seemingly complicated resistor circuit. Created by Willy McAllister.

## Want to join the conversation?

- What would happen if I had two batteries in parallel, and they had different voltages, say, 6V and 12V, what would be the voltage from the point they combine into one wire and another point on that wire ?(30 votes)
- Hello Feraru,

If you did this with real batteries I recommend you run away! There is a real risk of explosion. The 6 VDC battery will be charging at an excessive rate. The 12 VDC battery will be discharging at a rapid rate. Both will get hot. Both will liberate hydrogen (assuming lead acid chemistry). Both could explode!

I made this video that address this very topic:

https://www.youtube.com/watch?v=rWYgoL-wzNE

Regards,

APD(45 votes)

- At6:30, don't you need to take the inverse of that expression to get the equivalent resistance of the 2 resistors in parallel?(12 votes)
- R1R2/(R1+R2) is actually already the inverse of the expression 1/R total.(20 votes)

- Sorry, I'm not clear about the section at 3.06-3.26 when he talks about nodes (in orange). I understand why all three have the same voltage but he then says the ones on the right are 11 volts lower. Why 'lower'?(10 votes)
- They are lower because they are on the negative side of the battery. The bigger side of the battery ie. the node drawn longer is 11V but the negative side of the battery is 0V and the current flows from an area with high voltage to low voltage.(13 votes)

- Why is it that you can draw the imaginary wire between the two batteries to make them one big battery? Why doesn’t current flow through it?(6 votes)
- These batteries are treated as ideal voltage sources. This means that they will source whatever current is pulled by a load and sink any current flows in to keep their output voltage at 11V.(4 votes)

- At8:15, how did he know that the total voltage, or the two batteries together, would still equate to 11V? I'm confused. Won't they add up or something? I don't have the intuition to understand this right now. Can someone please explain?(2 votes)
- Two batteries connected in parallel have the same voltage as one battery, but twice the capacity to deliver current.

Two batteries connected in series (like in a flashlight) have sum of the voltage of the two batteries. Two AA batteries (1.5v) in series produces 3v from end to end.

Batteries in parallel are kind of like two big bricks standing up side-by-side. The top of one brick is about one foot off the floor. That corresponds to a voltage of "1 foot". If you put an identical brick next to it, the top is still one foot off the floor (so it is the same voltage). But now that there's two bricks you can put twice as much weight on top of them before they crumble.(12 votes)

- What happens if two batteries of different voltage are connected in parallel? In series they would add up right?(6 votes)
- Hello Ayushi,

This video may help: https://www.youtube.com/watch?v=cYGHibYmzb0

Keep in mind that batteries are NOT ideal voltage sources. But the results are similar.

Please forgive the crude nature of the video. I'm still trying to figure out how to use the tools...

Regards,

APD(2 votes)

- are the batteries connected in parallel always at the same voltage?if they are then why cant the batteries in parallel be at diff. voltages? if these batteries were connected in series they would have been additive right?(2 votes)
- Hello Sidra,

If you accidentally connect batteries of different voltages together your should run away especially if the batteries are large there is a real risk of explosion!

In this case there will be an uncontrolled current flow as the high voltage battery discharges and the low voltage battery charges.

I made this video to help explain the situation:

https://www.youtube.com/watch?v=rWYgoL-wzNE

Please leave a comment below if you would like to continue the conversation.

Regards,

APD(6 votes)

- So 2 parallel batteries with 11 Volts produce the same amount of current as a single 11V batterie? I would get same results with just one batterie...(2 votes)
- The 2 parallel batteries of 11 Volts produce the same amount of
**voltage**as a single 11V battery. The two batteries have twice the amount of stored energy as one battery, so as a pair they will last longer. The amount of current provided by both the 1-battery or 2-battery setup is not determined by the batteries, but rather by the components the batteries are connected to (using Ohm's Law on the resistors in the circuit). The batteries maintain 11V on their terminals, by providing whatever current is "demanded" by the connected components. If the circuit demands a lot of current, the single battery will run out of energy and die sooner than the double battery setup.(4 votes)

- What would happen if the voltage of the batteries is different?(3 votes)
- Hello Ananya,

Congratulations on posting your first question to KA!

If the voltages are different then a current will flow between the batteries. Depending on the batteries this could lead to a spectacular explosion!

I made this video to help demonstrate the situation: https://www.youtube.com/watch?v=rWYgoL-wzNE

Observe that many mathematical tools are applied in this situation including Kirchhoff, Thevenin, and Ohm. Know that It take time to learn the tools and how to apply them. Have fun experimenting.

Best wishes,

APD(1 vote)

- How did you get the equation (R1*R2)/(R1+R2)at5:23.(2 votes)
- It's just algebraic manipulation of 1/R1 + 1/R2. Get a common denominator to add the fractions, then take the reciprocal of both sides.(3 votes)

## Video transcript

- [Voiceover] Hi, my name's Willy. I'm the Khan Academy
Electrical Engineering Content Fellow, and we got a Tweet in from Jane, and Jane asks, "Could you take a video, "could you make a video
on solving complex circuit "problems in physics? "For example, something
similar to this question." So, Jane sent us a circuit here, here it is over here, and it's an interesting looking circuit. We're going to take a
shot at solving it here, and I'll tell you how
I would approach this. Let's take a quick look at it. We have two batteries in
it, which is a bit unusual. There's two resistors here. They happen to have the same value, 1.4 ohms. There's a third resistor connected up, and the question we're
being asked is to find, what's the current in
this resistor right here? So that's what we're looking for. So the way I approach
these kind of questions, first, I sort of just live with
this schematic a little bit. I look at it and see if I
can imagine how the currents are flowing and what's going on, and then a really good
way to do that is to, I'm going to draw this schematic myself, I'm just going to draw it
over and let me own hand kind of learn this circuit as we go, so I'll draw the battery. Here's the two batteries. And I see that they're connected together. Okay, that's interesting. Each battery then goes to a resistor, this goes to a resistor, and those are connected, let me check, yeah, those are connected together, and then they go to a
third resistor like this. And now if you notice what I'm
doing is I'm taking all the funny corners and dots and
curves and things like that and I'm just going to
draw a square schematic that I can understand. Now, I'm going to label
it just to make sure I got everything right, so, this is 11 volts. And I know from the battery symbol, that the short line is the minus side, and the long line is the plus side, and same here. So, these two batteries
are hooked up exactly the same way in the same direction. This one's also 11 volts. And that's going to prove
out to be interesting. We'll call this one R1 because the problem had i1 here. We'll call this one R2. And we'll call this one R3, of course. And the current that we're
looking for is this current right here, what's that? Okay. So, before I start doing
mathematics on this, what I'm going to do is
see if I can simplify the diagram, and what
I'm going to notice here, this is kind of a trick
that was done here. Let's look at these batteries. These two ends of the
batteries are connected together, so they are at the same voltage. We don't know what that
voltage is, but we know it's the same, and we know
that they're both 11 volts and that means that this node here, this node here, and this node here are at exactly the same voltage with respect to this node here. This one is 11 volts lower in voltage and this node is 11
volts lower in voltage, so what I'm going to imagine in my head, if I connected a wire
between these two nodes, this end would be at 11 volts, this end would be at 11 volts. So no current would
flow through that wire. It would have no effect on the circuit. It wouldn't, there would be no current
like diverted from here over to here. I'm actually going to draw that wire in. I'm going to draw that wire in, and I didn't change the
circuit when I did that. The same current goes
through this wire as before because zero goes this way, the same current goes through this wire because zero current goes this way because I know they're both at 11, at the same potential above here. Right, so this is a good
time to redraw this again. If you look here, I have two batteries that are hooked up, their inputs, their positive side is hooked up together and their negative side
is hooked up together, so they're actually just
acting like one, big battery, so let me draw that. I'm going to draw the circuit
again so it looks like this. Here's my combined big battery. And it goes to... Relabel these again so
we don't get mixed up. Okay. This is R1. This is R2. And this is R3. So this circuit looks a little simpler, and I'm gonna look at it again, see if I can do any more simplification, so what I recognize right
here, right in this area right here, R1 and R2 are in parallel. They have the same voltage
on their terminals. That means they're in parallel. I know how to simplify parallel resistors. We'll just use the
parallel resistor equation that I have in my head, and that looks like this, let's go to this color here. Okay, so parallel resistors, R1 in parallel with R2. I made up this symbol, two vertical lines, that means they're in parallel, and the formula for two
parallel resistors is R1 times R2, over R1 plus R2. Now I'll plug in the values. We know that R1 and R2
are actually, if we look over here at our schematic, they're the same value, and that has a special thing
when in parallel resistors so it's actually R R over 2R. Because those resistors are the same. And you can see I can cancel
that and I can cancel that and two parallel resistors, if the resistors are equal, is equal to half the resistance. And let's plug in the real values, 1.4 ohms over 2 equals 0.7 ohms. That's the equivalent resistance
of these two resistors in parallel. So this is a good time to
redraw this circuit again. Let's do it again. Here's our battery. This time I'm going to draw
the equivalent resistance. Then we have R3, we'll put in R3 down here. Nothing's changed there, and that goes back to our voltage source. Again, do some labels, so this is R1 in parallel with R2. This is R3. This is the current we
want to know, right there, and this is 11 volts, that's the plus side and
that's the minus side. Okay, we got a pretty
simple circuit here now. Let's fill, well, let me fill in the values for our resistor here. R1 in parallel with R2 is 0.7 ohms. R3, let's go look that
up, remind ourselves, 2.7 ohms. Now I can almost do this in my head. These are both now just
two series resistors. So, Ohm's law equals V equals iR. This is the form of it I always memorized. If I want to know i, I'm
solving for a current, so i equals V over R. We can start filling in values, i equals 11 volts over, what's the resistance? It's the series resistance here, so it's the sum of these two resistors, which is 0.7 plus 2.7. If I work that out, it equals 11 over 3.4 equals, okay let me actually work that out. It's 11 volts divided by 3.4 ohms and that equals 3.23 blah-blah-blah something something something. And that will be in Amps. So, this right here is our answer. Right there. So find the current i, i equals 3.2 Amperes. And now I'm going to check, actually can check to
see if that's the answer. Let's see what Jane sent. She actually sent in the answer. So thank you, Jane, for doing that, and, there it is. So that's it. Thank you very much for listening.