Current due to closing a switch: worked example | DC Circuits | AP Physics 1 | Khan Academy

- [Instructor] We are
asked, "How does the current

"going through R1," so, this resistor,

"when the switch is open,"

this switch, "compare to
the current through R1

"when the switch is closed?"

Pause this video and see
if you can figure that out.

Alright, so let's just think
about the two scenarios.

So, we can view the current as this,

right over here, this
current that we care about.

We could either measure it there

or you could measure it right over there,

and let's first think about the scenario

where the switch is open.

So, our current when our switch is open

is going to be equal to the voltage

across the resistors, and
that's going to be our 12 volts,

12 volts,

divided by the equivalent
resistance of these resistors.

When the switch is open, essentially,

we just have R1 and R2 in series,

and so this is just gonna be R1+R2.

If you have two resistors in a series,

their equivalent resistance is just

the sum of the resistances.

Fair enough.

Now, let's think about the situation

where the switch is closed.

Closed.

So here, our current at
this point of our circuit,

or the current going through R1,

so I sub closed,

is, once again, it's going
to be equal to 12 volts,

the voltage across the resistors,

but what are we gonna divide by now?

When we close the switch, what happens?

Well, these lines where
we see no resistors

in circuit diagrams, that's
assumed to be resistance-less,

so all of the current will
actually flow that way.

So, by closing this switch,

you're essentially removing
R2 from the circuit.

The current will just go through R1,

and then follow the path of
least resistance, literally.

And so, in this situation,
our current is going

to be 12 volt divided
by, you essentially just

have one resistance, divided by R1.

So, when you closed the circuit,

you've essentially taken a resistor out,

and so if you took a resistor out,

you're going to increase the current,

so you could just write it as,

the current when the switch is open

is going to be less than,

is going to be less than the current

when the switch is closed.

Once again, why is that?

Well, just look at the denominators here.

When the switch is open,

you're dividing by a larger number

than when the switch is closed.

Or, another way of thinking about it,

when the switch is open, the
R2 resistance is factored in.

When the switch is closed,
the R2 resistance essentially

becomes a non-factor and
you have less resistance,

which would mean you
would have higher current.