Video Transcript
What is the current 𝐼 in the
circuit shown? Give your answer to two decimal
places.
In this question, we’ve been given
a circuit diagram consisting of two loops. Each loop contains a cell and a
resistor. We want to find the current 𝐼 in
the central branch of the circuit.
Now, because of the way the circuit
has been put together, the resistors are neither connected in series nor in parallel
with each other. This means it’s not possible to
work out the total resistance, or equivalent resistance, of the 12-ohm resistor and
the 15-ohm resistor together. Instead, we’ll need to use
Kirchhoff’s laws to calculate the current 𝐼 in the circuit.
We can start by labeling some of
the quantities in our circuit diagram. Let’s start with the 12-ohm
resistor. We’ll say that its resistance is 𝑅
one. Similarly, we’ll say that the
resistance of the 15-ohm resistor is 𝑅 two. We’ll also say that the potential
difference of the cell on the left is called 𝑉 one and the potential difference of
the cell on the right is 𝑉 two.
Next, let’s label the currents in
each branch of the circuit. Here, the orientation of the cells
tells us which way the conventional current goes in each loop. On the left, the positive terminal
of the cell is on the bottom and the negative terminal is on top. This means that the conventional
current, which always goes from positive to negative, goes in this direction in the
loop on the left. We’ll label this current 𝐼
one. On the right of the circuit
diagram, the cell is oriented with the positive terminal on top. This means we have conventional
current going in this direction. Let’s label this current 𝐼
two.
Okay, so we can see we have two
loops in the circuit, both with current going anticlockwise. Let’s call these loop one and loop
two. We’re going to start by considering
loop one on the left.
Recall that Kirchhoff’s second law,
also known as Kirchhoff’s loop rule, states that the sum of the potential difference
across each component in a loop is equal to zero. We can label the potential
difference across this resistor 𝑉 𝑅 one. Then, by applying Kirchhoff’s
second law to loop one, we find that 𝑉 one minus 𝑉 𝑅 one must be equal to
zero.
Notice that in this equation, the
potential difference 𝑉 one provided by the cell is positive, whereas the potential
difference 𝑉 𝑅 one across the resistor is negative. This is because cells provide an
increase in potential, whereas there is a decrease in potential across the
resistor.
Looking at this expression, we can
rearrange it to find that 𝑉 𝑅 one is equal to 𝑉 one. And from the diagram, we know that
the cell in loop one provides a potential difference of 15 volts. So 𝑉 one is equal to 15 volts. Therefore, we find that 𝑉 𝑅 one,
the voltage across the resistor, is also 15 volts.
Now we know this, we can use Ohm’s
law to calculate the current through the resistor. Recall that Ohm’s law can be
written as 𝑉 equals 𝐼𝑅, where 𝑉 is the voltage across a resistor, 𝐼 is the
current through the resistor, and 𝑅 is the resistance of that resistor. Now, here, we want to calculate a
current. So we need to rearrange Ohm’s law
to make 𝐼 the subject. To do this, we divide both sides of
the equation by 𝑅 to give us 𝑉 over 𝑅 equals 𝐼. Then, swapping the left and right
sides around, we get 𝐼 equals 𝑉 over 𝑅.
In this specific case, we have 𝐼
one, that’s the current through the 12-ohm resistor, is equal to 𝑉 𝑅 one, which is
the potential difference across the 12-ohm resistor, divided by 𝑅 one, which is its
resistance. Substituting in the values of 𝑉 𝑅
one and 𝑅 one, we find that 𝐼 one is equal to 15 volts divided by 12 ohms, which
is equal to 1.25 amps. So we’ve now found the current 𝐼
one. Let’s label this on our diagram,
and now we can apply the same process to loop two.
So, first, we’ll label the
potential difference across the 15-ohm resistor as 𝑉 𝑅 two. And then by applying Kirchhoff’s
second law to loop two, we find that 𝑉 two minus 𝑉 𝑅 two must be equal to
zero. Again, we’ve expressed the
potential difference of the cell as a positive value and the potential difference of
the resistor as a negative value. We can rearrange this expression to
find that 𝑉 two is equal to 𝑉 𝑅 two.
Now, since we know the potential
difference of the cell, 𝑉 two, is 20 volts, this means that the potential
difference across the resistor is also 20 volts. And we can now use Ohm’s law to
calculate the current 𝐼 two through the 15-ohm resistor. In this case, we can say that the
current in the 15-ohm resistor, that’s 𝐼 two, is equal to the potential difference
𝑉 𝑅 two across this resistor divided by the resistance of the resistor 𝑅 two. Substituting in values for 𝑉 𝑅
two and 𝑅 two, we have 20 volts over 15 ohms, which is equal to 1.3 recurring
amps. Let’s label this on our diagram
too.
Okay, now that we’ve worked out
values for the currents 𝐼 one and 𝐼 two, we can use Kirchhoff’s first law to find
the current 𝐼. Recall that Kirchhoff’s first law
states that the sum of the currents going into a junction or node in a circuit must
be equal to the sum of the currents coming out of that junction.
Let’s apply this law to this
junction at the bottom of our diagram. We can see that we have two
currents going into this junction: 𝐼 and 𝐼 one. We also have one current coming
out; that’s 𝐼 two. Kirchhoff’s first law tells us that
the sum of the currents going into this junction, that’s 𝐼 plus 𝐼 one, is equal to
the sum of the currents coming out, that’s 𝐼 two.
Now, since we want to find the
current 𝐼, we can rearrange this to make 𝐼 the subject. This gives us 𝐼 equals 𝐼 two
minus 𝐼 one. Substituting in the calculated
values of 𝐼 one and 𝐼 two, we find that 𝐼 is equal to 1.3 recurring amps minus
1.25 amps, which is equal to 0.083 recurring amps.
The final bit we need to complete
this question is to round our answer to two decimal places. And 0.083 recurring to two decimal
places is just 0.08. So our final value for the current
𝐼 in the circuit shown is 0.08 amps.
