Video Transcript
Consider the circuit shown
below. What is the reading on the
ammeter? (A) Two 𝐼 over three, (B) 𝐼 over
two, (C) 𝐼, (D) 𝐼 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
Kirchhoff’s laws. We will begin by labeling all the
components on the circuit as follows. We’ve labeled the resistors as 𝑅
one, 𝑅 two, 𝑅 three, and 𝑅 four, and we’ve also labeled currents 𝐼 one and 𝐼
two in addition to the current 𝐼.
Now, recall that Kirchhoff’s 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 𝐼
comes into the junction highlighted in blue and the currents 𝐼 one and 𝐼 two go
out of this same junction.
So, by using Kirchhoff’s first law,
we find that 𝐼 is equal to 𝐼 one plus 𝐼 two. The reading on the ammeter will
equal the current 𝐼 one. We can make 𝐼 one the subject of
the equation by subtracting 𝐼 two from both sides. This gives us that 𝐼 one is equal
to 𝐼 minus 𝐼 two. We’ll label this as equation
one. To find the values for these
currents, we will also need Kirchhoff’s second law. Recall that Kirchhoff’s 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, we’ll consider loop
one in this circuit. We’ll label the potential
difference across resistor 𝑅 one as 𝑉 sub 𝑅 one. Similarly, we’ll label the
potential difference across resistor 𝑅 two as 𝑉 sub 𝑅 two. Then, by applying Kirchhoff’s
second law to loop one, we find that 𝑉 minus 𝑉 sub 𝑅 one minus 𝑉 sub 𝑅 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 𝑉 sub
𝑅 one and 𝑉 sub 𝑅 two by using Ohm’s law. Recall that Ohm’s law can be
written as 𝑉 is equal to 𝐼 times 𝑅, where 𝑉 is the potential difference, 𝐼 is
the current, and 𝑅 is the resistance. The current in resistor 𝑅 one is
the current 𝐼, so the potential difference across resistor 𝑅 one is 𝑉 sub 𝑅 one
is equal to 𝐼 times 𝑅 one. The current in resistor 𝑅 two is
the current 𝐼 one, so the potential difference across resistor 𝑅 two will be 𝑉
sub 𝑅 two is equal to 𝐼 one times 𝑅 two. Then, by substituting these
equations into the equation we obtained from Kirchhoff’s second law for loop one, we
have the equation 𝑉 minus 𝐼 times 𝑅 one minus 𝐼 one times 𝑅 two is equal to
zero.
We are given the resistances of
both resistors, with 𝑅 one is equal to 12 ohms and 𝑅 two is equal to nine
ohms. Substituting these values into the
equation, we get that 𝑉 minus 12𝐼 minus nine 𝐼 one is equal to zero. We’ll label this as equation
two.
Now we can consider loop two in
this circuit. We can label the potential
difference across resistor 𝑅 three as 𝑉 sub 𝑅 three and the potential difference
across resistor 𝑅 four as 𝑉 sub 𝑅 four. Applying Kirchhoff’s second law to
loop two, we find that 𝑉 minus 𝑉 sub 𝑅 one minus 𝑉 sub 𝑅 three minus 𝑉 sub 𝑅
four is equal to zero. The current in resistor one is
still the current 𝐼.
So Ohm’s law tells us that the
potential difference across resistor 𝑅 one is 𝑉 sub 𝑅 one is equal to 𝐼 times 𝑅
one. Then, the current in resistor 𝑅
three is the current 𝐼 two. Ohm’s law tells us that the
potential difference across resistor 𝑅 three will be 𝑉 sub 𝑅 three is equal to 𝐼
two times 𝑅 three. Similarly, the current in resistor
four is also equal to 𝐼 two. So, Ohm’s law tells us that the
potential difference across resistor 𝑅 four will be 𝑉 sub 𝑅 four is equal to 𝐼
two times 𝑅 four.
Substituting these expressions into
the equation we obtained from Kirchhoff’s second law for loop two, we have the
equation 𝑉 minus 𝐼 times 𝑅 one minus 𝐼 two times 𝑅 three minus 𝐼 two times 𝑅
four is equal to zero. We are given the values of these
resistances. 𝑅 one is equal to 12 ohms, 𝑅
three is equal to seven ohms, and 𝑅 four is equal to 11 ohms. Substituting those values into this
equation, we get that 𝑉 minus 12𝐼 minus seven 𝐼 two minus 11𝐼 two is equal to
zero. We can then simplify this to get 𝑉
minus 12𝐼 minus 18𝐼 two is equal to zero. We’ll 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 𝑉 minus 12𝐼 minus 18𝐼 two minus 𝑉 plus 12𝐼 plus nine 𝐼
one is equal to zero. This then simplifies to negative
18𝐼 two plus nine 𝐼 one equals zero, or nine 𝐼 one is equal to 18𝐼 two. Dividing both sides by 18, we find
that 𝐼 two is equal to 𝐼 one divided by two.
We can now substitute 𝐼 one over
two in place of 𝐼 two in equation one, which we found using Kirchhoff’s first law,
in order to find the value of 𝐼 one. This gives us that 𝐼 one is equal
to 𝐼 minus 𝐼 one over two. Adding 𝐼 one over two to both
sides of the equation, we get that three 𝐼 one over two is equal to 𝐼. Dividing both sides by three over
two, we get 𝐼 one is equal to two 𝐼 over three.
We’ve already seen that 𝐼 one is
the current that passes through the ammeter, and so the reading on the ammeter will
be equal to this current 𝐼 one. Since we’ve found that 𝐼 one is
equal to two 𝐼 over three, we know that the ammeter reading must be two 𝐼 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 𝐼 over three.
