Question Video: Calculating the Current in a Circuit Containing a Cell and a Resistor | Nagwa Question Video: Calculating the Current in a Circuit Containing a Cell and a Resistor | Nagwa

# Question Video: Calculating the Current in a Circuit Containing a Cell and a Resistor Science • Third Year of Preparatory School

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The diagram shows a circuit consisting of a cell and a resistor. The cell provides a potential difference of 6 volts, and the resistor has a resistance of 3 ohms. What is the current at point π in the circuit?

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### 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 Ohmβs law. Ohmβs 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.

Ohmβs law can be written as an equation, where π stands for the potential difference across the resistor, πΌ 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 πΌ 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 πΌ is equal to the potential difference π divided by the resistance π.

Before we begin putting in the given values, letβs 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.

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