### Video Transcript

Consider the given circuit. What is the value of 𝐼? (A) Nine amps, (B) six amps, (C)
4.5 amps, (D) three amps.

In this question, we are presented
with the following circuit, and we want to calculate the value of the current 𝐼
labeled in the circuit. To answer this question, we will
use Kirchhoff’s laws. Recall that Kirchhoff’s first law
states that the sum of the currents into a junction, or node, in the circuit must be
the same as the sum of the currents out of the junction, or node.

Let’s label the circuit diagram
with the currents in the circuit. We can see that the current 𝐼
comes into the node highlighted in blue, and the currents 𝐼 one and 𝐼 two go out
of the node highlighted in blue. Therefore, by using Kirchhoff’s
first law at this node, we find that 𝐼 equals 𝐼 one plus 𝐼 two. We are given the value of 𝐼
one. So if we can find out the value of
the current 𝐼 two, then we will be able to calculate the value of 𝐼.

To find the value of 𝐼 two, we
will use Kirchhoff’s second law. Recall that Kirchhoff’s second law
states that the sum of the potential difference across each component in a loop is
equal to zero. Let’s consider loop one, which
we’ve highlighted in the circuit. We can label the potential
difference across the five-ohm resistor as 𝑉 𝑅 one, and we can label the potential
difference across the 10-ohm resistor as 𝑉 𝑅 two. Then, by applying Kirchhoff’s
second law to loop one, we find that 𝑉 𝑅 one plus 𝑉 𝑅 two equals zero.

We can get expressions for 𝑉 𝑅
one and 𝑉 𝑅 two by using Ohm’s law. Recall that Ohm’s law can be
written as 𝑉 equals 𝐼𝑅, where 𝑉 is the potential difference, 𝐼 is the current,
and 𝑅 is the resistance. We also need to use some sign
conventions when using Kirchhoff’s second law. When the loop passes through a
resistor in the same direction as the current, the 𝐼𝑅 term is negative because the
current goes in the direction of decreasing potential. When the loop passes through a
resistor in the direction opposite to the current, the 𝐼𝑅 term is positive because
this represents an increase in potential.

Now we can apply Ohm’s law. Let’s start with the five-ohm
resistor. The current through the five-ohm
resistor is the current 𝐼 one, which we’re told has a value of three amps. We also note that the loop passes
through the resistor in the same direction as the current, so our 𝐼𝑅 term must be
negative. When we apply Ohm’s law, we find
that the potential difference across the five-ohm resistor 𝑉 𝑅 one is equal to
minus three amps times five ohms equals minus 15 volts.

Next, let’s look at the 10-ohm
resistor. The current through the 10-ohm
resistor is the current 𝐼 two. We note this time that the loop
passes through the resistor in the direction opposite to the current, so our 𝐼𝑅
term is positive. The potential difference across the
10-ohm resistor 𝑉 𝑅 two will be equal to 𝐼 two times 10 ohms. So by substituting the values of 𝑉
𝑅 one and 𝑉 𝑅 two into the equation we obtained from Kirchhoff’s second law for
loop one, we have the equation minus 15 volts plus 10 ohms times 𝐼 two equals
zero. Adding 15 volts to both sides of
the equation, we get 10 ohms times 𝐼 two equals 15 volts. And by dividing both sides by 10
ohms, we find that 𝐼 two equals 15 volts divided by 10 ohms equals 1.5 amps.

Now that we have values for both 𝐼
one and 𝐼 two, we can calculate the value of 𝐼 using Kirchhoff’s first law. We find that 𝐼 equals 𝐼 one plus
𝐼 two equals three amps plus 1.5 amps equals 4.5 amps. This value corresponds with option
(C), so the value of 𝐼 must equal 4.5 amps. Any other value of 𝐼 would violate
the conservation of charge and the conservation of energy in a circuit. So the correct answer is option
(C).