### Video Transcript

The diagram shows a circuit
consisting of a cell and a resistor. The cell provides a potential
difference of six volts, and the resistor has a resistance of three ohms. What is the current at point π in
the circuit?

In this question, we are given a
diagram of an electric circuit, and we are asked to find the current at a specific
point π in the circuit. The circuit consists of a single
loop, so the current will be the same at all points around the loop. In the question, we are given the
potential difference that is provided by the cell, which is six volts, and the
resistance of the resistor, which is three ohms. In order to solve this problem, we
will need to use Ohmβs law. Ohmβs law states, for two points in
a circuit, the potential difference across the points equals the current between the
points multiplied by the resistance of the object between the points.

Ohmβs law can be written as an
equation, where π stands for the potential difference across the resistor, πΌ
stands for the current in the resistor, and π
stands for the resistance of the
resistor. To find the current, we must make
the current πΌ the subject of the equation. We can do this by dividing both
sides of the equation by the resistance. This gives us the equation current
πΌ is equal to the potential difference π divided by the resistance π
.

Before we begin putting in the
given values, letβs take a look at the units in this equation. We have the unit of volts divided
by the unit of ohms on the right-hand side and the unit of amperes, the unit of
electric current, on the left-hand side. Since our units are correct, we can
substitute our values for π and π
to find the value of current. The current through the resistor is
equal to six volts divided by three ohms, which gives us a value of two amperes. Since we said that the current will
be the same at every point in the circuit, we know that this current through the
resistor will also be the current at point π. So our final answer is the current
at point π in the circuit is two amperes.