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The circuit diagram provided shows a simple series circuit with a resistor of \(50 \, \Omega\) and a current of \(0.3 \, A\) flowing through it. The goal is to find the potential difference (voltage) provided by the battery.

Using **Ohm's Law**:

\[
V = I \times R
\]

Where:
- \( V \) is the potential difference (voltage),
- \( I \) is the current,
- \( R \) is the resistance.

Substitute the values:

\[
V = 0.3 \, A \times 50 \, \Omega
\]

Now, let's calculate it.The potential difference provided by the battery is \( 15 \, V \).

Would you like further clarification on this, or have any additional questions? Here are some related questions for deeper understanding:

1. What happens to the voltage if the resistance is increased but the current remains the same?
2. How would the current change if the resistance were decreased while maintaining the same voltage?
3. What is the formula for Ohm's Law, and how can it be rearranged to find current or resistance?
4. How does the power dissipated in the resistor relate to the voltage and current?
5. What would the total resistance be in a circuit with two resistors connected in series?

**Tip:** In a series circuit, the current remains the same throughout, while the voltage is divided among components.

The diagram shows a circuit with a battery that provides an unknown potential difference. Work out the potential difference provided by the battery. The circuit includes a 50-ohm resistor and a current of 0.3 A.

Solve the physics problem of calculating the potential difference in a circuit that includes a 50-ohm resistor and a current of 0.3 A using Ohm's Law. Learn how to apply V = I × R to find that the voltage provided by the battery is 15 V. Suitable for students in Grades 9-10.

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