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- [Voiceover] Hi, my name's Willy.

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I'm the Khan Academy
Electrical Engineering

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Content Fellow,

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and we got a Tweet in from Jane,

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and Jane asks, "Could you take a video,

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"could you make a video
on solving complex circuit

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"problems in physics?

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"For example, something
similar to this question."

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So, Jane sent us a circuit here,

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here it is over here,

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and it's an interesting looking circuit.

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We're going to take a
shot at solving it here,

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and I'll tell you how
I would approach this.

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Let's take a quick look at it.

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We have two batteries in
it, which is a bit unusual.

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There's two resistors here.

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They happen to have the same value,

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1.4 ohms.

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There's a third resistor connected up,

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and the question we're
being asked is to find,

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what's the current in
this resistor right here?

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So that's what we're looking for.

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So the way I approach
these kind of questions,

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first, I sort of just live with
this schematic a little bit.

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I look at it and see if I
can imagine how the currents

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are flowing and what's going on,

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and then a really good
way to do that is to,

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I'm going to draw this schematic myself,

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I'm just going to draw it
over and let me own hand

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kind of learn this circuit as we go,

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so I'll draw the battery.

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Here's the two batteries.

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And I see that they're connected together.

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Okay, that's interesting.

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Each battery then goes to

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a resistor,

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this goes to a resistor,

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and those are connected, let me check,

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yeah, those are connected together,

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and then they go to a
third resistor like this.

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And now if you notice what I'm
doing is I'm taking all the

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funny corners and dots and
curves and things like that

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and I'm just going to
draw a square schematic

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that I can understand.

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Now, I'm going to label
it just to make sure

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I got everything right,

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so, this is 11 volts.

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And I know from the battery symbol,

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that the short line is the minus side,

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and the long line is the plus side,

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and same here.

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So, these two batteries
are hooked up exactly

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the same way in the same direction.

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This one's also 11 volts.

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And that's going to prove
out to be interesting.

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We'll call this one R1

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because the problem had i1 here.

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We'll call this one R2.

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And we'll call this one R3, of course.

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And the current that we're
looking for is this current

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right here, what's that?

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Okay.

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So, before I start doing
mathematics on this,

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what I'm going to do is
see if I can simplify

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the diagram, and what
I'm going to notice here,

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this is kind of a trick
that was done here.

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Let's look at these batteries.

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These two ends of the
batteries are connected

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together, so they are at the same voltage.

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We don't know what that
voltage is, but we know

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it's the same, and we know
that they're both 11 volts

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and that means that this node here,

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this node here, and this node here

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are at exactly the same voltage

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with respect to this node here.

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This one is 11 volts lower in voltage

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and this node is 11
volts lower in voltage,

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so what I'm going to imagine in my head,

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if I connected a wire
between these two nodes,

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this end would be at 11 volts,

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this end would be at 11 volts.

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So no current would
flow through that wire.

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It would have no effect on the circuit.

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It wouldn't,

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there would be no current
like diverted from here

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over to here.

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I'm actually going to draw that wire in.

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I'm going to draw that wire in,

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and I didn't change the
circuit when I did that.

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The same current goes
through this wire as before

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because zero goes this way,

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the same current goes through this wire

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because zero current goes this way

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because I know they're both at 11,

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at the same potential above here.

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Right, so this is a good
time to redraw this again.

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If you look here, I have two batteries

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that are hooked up,

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their inputs,

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their positive side is hooked up together

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and their negative side
is hooked up together,

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so they're actually just
acting like one, big battery,

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so let me draw that.

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I'm going to draw the circuit
again so it looks like this.

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Here's my combined big battery.

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And it goes to...

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Relabel these again so
we don't get mixed up.

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Okay.

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This is R1.

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This is R2.

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And this is R3.

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So this circuit looks a little simpler,

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and I'm gonna look at it again,

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see if I can do any more simplification,

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so what I recognize right
here, right in this area right

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here, R1 and R2 are in parallel.

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They have the same voltage
on their terminals.

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That means they're in parallel.

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I know how to simplify parallel resistors.

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We'll just use the
parallel resistor equation

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that I have in my head,

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and that looks like this,

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let's go to this color here.

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Okay, so parallel resistors, R1

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in parallel with R2.

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I made up this symbol, two vertical lines,

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that means they're in parallel,

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and the formula for two
parallel resistors is R1

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times R2,

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over R1 plus R2.

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Now I'll plug in the values.

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We know that R1 and R2
are actually, if we look

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over here at our schematic,

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they're the same value,

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and that has a special thing
when in parallel resistors

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so it's actually R R

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over 2R.

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Because those resistors are the same.

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And you can see I can cancel
that and I can cancel that

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and two parallel resistors,

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if the resistors are equal,

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is equal to half the resistance.

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And let's plug in the real values,

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1.4 ohms over 2

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equals

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0.7 ohms.

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That's the equivalent resistance
of these two resistors

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in parallel.

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So this is a good time to
redraw this circuit again.

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Let's do it again.

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Here's our battery.

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This time I'm going to draw
the equivalent resistance.

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Then we have R3,

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we'll put in R3 down here.

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Nothing's changed there,

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and that goes back to our voltage source.

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Again, do some labels,

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so this is R1 in parallel with R2.

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This is R3.

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This is the current we
want to know, right there,

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and this is 11 volts,

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that's the plus side and
that's the minus side.

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Okay, we got a pretty
simple circuit here now.

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Let's fill, well,

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let me fill in the values

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for

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our resistor here.

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R1 in parallel with R2 is 0.7 ohms.

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R3, let's go look that
up, remind ourselves,

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2.7 ohms.

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Now I can almost do this in my head.

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These are both now just
two series resistors.

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So, Ohm's law equals

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V equals iR.

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This is the form of it I always memorized.

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If I want to know i, I'm
solving for a current,

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so i equals V over R.

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We can start filling in values,

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i equals 11 volts over,

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what's the resistance?

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It's the series resistance here,

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so it's the sum of these two resistors,

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which is 0.7

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plus 2.7.

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If I work that out,

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it equals 11 over 3.4

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equals,

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okay let me actually work that out.

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It's 11 volts

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divided by 3.4 ohms

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and that equals

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3.23 blah-blah-blah

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something something something.

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And that will be in Amps.

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So, this right here is our answer.

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Right there.

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So find the current i,

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i equals 3.2 Amperes.

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And now I'm going to check,

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actually can check to
see if that's the answer.

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Let's see what Jane sent.

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She actually sent in the answer.

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So thank you, Jane, for doing that,

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and, there it is.

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So that's it.

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Thank you very much for listening.