recall and use P = VI, P = I 2R and P = V 2 / R

Potential Difference & Power

What is Potential Difference?

Potential difference (voltage) is the energy per unit charge that pushes charges through a circuit. Think of it like the pressure that pushes water through a pipe.

What is Power?

Power is the rate at which energy is transferred or used. In circuits, it tells us how fast energy is being converted into heat, light, etc. It is measured in watts (W).

Key Formulas

Water‑Flow Analogy

Imagine a water tap (voltage) that pushes water (current) through a pipe (circuit). The pipe’s width (resistance) controls how much water can flow. The power is like the amount of water hitting a sprinkler per second – the more water or the higher the pressure, the more power.

Example 1: Using $P = VI$

A 9 V battery powers a light bulb that draws 0.5 A. What is the power?

1. Write the formula: $P = VI$.
2. Plug in: $P = 9\,\text{V} \times 0.5\,\text{A}$.
3. Calculate: $P = 4.5\,\text{W}$.

Answer: 4.5 W. 💡

Example 2: Using $P = I^2 R$

A resistor of 10 Ω is connected to a 12 V supply. Find the power dissipated.

1. First find the current: $I = \dfrac{V}{R} = \dfrac{12}{10} = 1.2\,\text{A}$.
2. Use $P = I^2 R$: $P = (1.2)^2 \times 10 = 1.44 \times 10 = 14.4\,\text{W}$.

Answer: 14.4 W. 🔥

Example 3: Using $P = \dfrac{V^2}{R}$

A 5 V supply is connected to a 2 Ω resistor. What is the power?

1. Use $P = \dfrac{V^2}{R}$.
2. Calculate: $P = \dfrac{5^2}{2} = \dfrac{25}{2} = 12.5\,\text{W}$.

Answer: 12.5 W. ⚡

Quick Reference Cheat Sheet

Practice Problems

1. A 15 V battery powers a 5 Ω resistor. Calculate the power using all three formulas.
2. A light bulb uses 60 W of power and is connected to a 12 V supply. What is its resistance?
3. Find the current if a 3 W lamp is connected to a 9 V battery.

Try solving them before checking the solutions below! ??

Solutions

1. Given $V = 15\,\text{V}$, $R = 5\,\Omega$.

   • $I = V/R = 15/5 = 3\,\text{A}$.
   • $P = VI = 15 \times 3 = 45\,\text{W}$.
   • $P = I^2 R = 3^2 \times 5 = 45\,\text{W}$.
   • $P = V^2 / R = 15^2 / 5 = 225/5 = 45\,\text{W}$.

2. Given $P = 60\,\text{W}$, $V = 12\,\text{V}$.

   • $I = P/V = 60/12 = 5\,\text{A}$.
   • $R = V/I = 12/5 = 2.4\,\Omega$.

3. Given $P = 3\,\text{W}$, $V = 9\,\text{V}$.

   • $I = P/V = 3/9 = 0.333\,\text{A}$.

Key Takeaways

• Power tells you how fast energy is used.
• Use $P = VI$ when you know voltage and current.
• Use $P = I^2 R$ when you know current and resistance.
• Use $P = V^2 / R$ when you know voltage and resistance.
• All three formulas are equivalent – just rearrange the variables.