### Video Transcript

Consider the circuit shown
below. What is the reading on the
ammeter? (A) Two ๐ผ over three, (B) ๐ผ over
two, (C) ๐ผ, (D) ๐ผ over three.

To find the reading on the ammeter,
we will need to calculate the current in that branch of the circuit. To do this, we will need to use
Kirchhoffโs laws. We will begin by labeling all the
components on the circuit as follows. Weโve labeled the resistors as ๐
one, ๐
two, ๐
three, and ๐
four, and weโve also labeled currents ๐ผ one and ๐ผ
two in addition to the current ๐ผ.

Now, recall that Kirchhoffโs first
law states that the sum of the currents into a junction or node in a circuit must be
the same as the sum of the currents out of the junction. We can see that the current ๐ผ
comes into the junction highlighted in blue and the currents ๐ผ one and ๐ผ two go
out of this same junction.

So, by using Kirchhoffโs first law,
we find that ๐ผ is equal to ๐ผ one plus ๐ผ two. The reading on the ammeter will
equal the current ๐ผ one. We can make ๐ผ one the subject of
the equation by subtracting ๐ผ two from both sides. This gives us that ๐ผ one is equal
to ๐ผ minus ๐ผ two. Weโll label this as equation
one. To find the values for these
currents, we will also need Kirchhoffโs second law. Recall that Kirchhoffโs second law
states that the sum of the potential differences across each of the components in a
loop is equal to zero.

To begin with, weโll consider loop
one in this circuit. Weโll label the potential
difference across resistor ๐
one as ๐ sub ๐
one. Similarly, weโll label the
potential difference across resistor ๐
two as ๐ sub ๐
two. Then, by applying Kirchhoffโs
second law to loop one, we find that ๐ minus ๐ sub ๐
one minus ๐ sub ๐
two is
equal to zero. Notice that the potential
difference provided by the cell is positive, while the potential difference across
each of the resistors is negative.

We can get expressions for ๐ sub
๐
one and ๐ sub ๐
two by using Ohmโs law. Recall that Ohmโs law can be
written as ๐ is equal to ๐ผ times ๐
, where ๐ is the potential difference, ๐ผ is
the current, and ๐
is the resistance. The current in resistor ๐
one is
the current ๐ผ, so the potential difference across resistor ๐
one is ๐ sub ๐
one
is equal to ๐ผ times ๐
one. The current in resistor ๐
two is
the current ๐ผ one, so the potential difference across resistor ๐
two will be ๐
sub ๐
two is equal to ๐ผ one times ๐
two. Then, by substituting these
equations into the equation we obtained from Kirchhoffโs second law for loop one, we
have the equation ๐ minus ๐ผ times ๐
one minus ๐ผ one times ๐
two is equal to
zero.

We are given the resistances of
both resistors, with ๐
one is equal to 12 ohms and ๐
two is equal to nine
ohms. Substituting these values into the
equation, we get that ๐ minus 12๐ผ minus nine ๐ผ one is equal to zero. Weโll label this as equation
two.

Now we can consider loop two in
this circuit. We can label the potential
difference across resistor ๐
three as ๐ sub ๐
three and the potential difference
across resistor ๐
four as ๐ sub ๐
four. Applying Kirchhoffโs second law to
loop two, we find that ๐ minus ๐ sub ๐
one minus ๐ sub ๐
three minus ๐ sub ๐
four is equal to zero. The current in resistor one is
still the current ๐ผ.

So Ohmโs law tells us that the
potential difference across resistor ๐
one is ๐ sub ๐
one is equal to ๐ผ times ๐
one. Then, the current in resistor ๐
three is the current ๐ผ two. Ohmโs law tells us that the
potential difference across resistor ๐
three will be ๐ sub ๐
three is equal to ๐ผ
two times ๐
three. Similarly, the current in resistor
four is also equal to ๐ผ two. So, Ohmโs law tells us that the
potential difference across resistor ๐
four will be ๐ sub ๐
four is equal to ๐ผ
two times ๐
four.

Substituting these expressions into
the equation we obtained from Kirchhoffโs second law for loop two, we have the
equation ๐ minus ๐ผ times ๐
one minus ๐ผ two times ๐
three minus ๐ผ two times ๐
four is equal to zero. We are given the values of these
resistances. ๐
one is equal to 12 ohms, ๐
three is equal to seven ohms, and ๐
four is equal to 11 ohms. Substituting those values into this
equation, we get that ๐ minus 12๐ผ minus seven ๐ผ two minus 11๐ผ two is equal to
zero. We can then simplify this to get ๐
minus 12๐ผ minus 18๐ผ two is equal to zero. Weโll label this as equation
three.

We can now compare equations two
and three. We have two simultaneous
equations. So by subtracting equation two from
equation three, we get this equation here. We can simplify this by expanding
out this bracket to get ๐ minus 12๐ผ minus 18๐ผ two minus ๐ plus 12๐ผ plus nine ๐ผ
one is equal to zero. This then simplifies to negative
18๐ผ two plus nine ๐ผ one equals zero, or nine ๐ผ one is equal to 18๐ผ two. Dividing both sides by 18, we find
that ๐ผ two is equal to ๐ผ one divided by two.

We can now substitute ๐ผ one over
two in place of ๐ผ two in equation one, which we found using Kirchhoffโs first law,
in order to find the value of ๐ผ one. This gives us that ๐ผ one is equal
to ๐ผ minus ๐ผ one over two. Adding ๐ผ one over two to both
sides of the equation, we get that three ๐ผ one over two is equal to ๐ผ. Dividing both sides by three over
two, we get ๐ผ one is equal to two ๐ผ over three.

Weโve already seen that ๐ผ one is
the current that passes through the ammeter, and so the reading on the ammeter will
be equal to this current ๐ผ one. Since weโve found that ๐ผ one is
equal to two ๐ผ over three, we know that the ammeter reading must be two ๐ผ over
three. Comparing this answer with the
available answer options, we find that it matches option (A). The reading on the ammeter is equal
to two ๐ผ over three.