The potential difference (voltage) across a component is the amount of energy transferred to each unit of charge that passes through it.
\$V=\frac{W}{Q}\$
where \$W\$ is the work (energy) in joules and \$Q\$ is the charge in coulombs.
Water‑flow analogy: Think of a water pipe. The pressure difference between two points pushes water through. Similarly, a voltage difference pushes electric charge through a circuit.
⚡️ Tip for exams: When you see a question about energy per unit charge, remember the formula \$V=W/Q\$ and that the units are volts (V).
If a charge \$Q\$ moves through a potential difference \$V\$, the energy transferred is
\$W = Q \times V\$
For example, moving 0.02 C through a 12 V battery transfers
\$W = 0.02 \times 12 = 0.24\;\text{J}\$
Practical example: A 9 V battery powering a 3 W LED. The LED draws
\$I = \frac{P}{V} = \frac{3}{9} = 0.33\;\text{A}\$
and each coulomb of charge carries 9 J of energy.
Power is the rate at which energy is used or transferred. In terms of voltage and current:
\$P = V \times I\$
where \$P\$ is power in watts, \$V\$ is voltage, and \$I\$ is current.
Water‑flow analogy: Power is like the volume of water flowing per second through a pipe. More pressure (voltage) or more flow (current) gives higher power.
📝 Exam tip: If a problem gives power and voltage, find current with \$I=P/V\$. If it gives current and voltage, find power with \$P=VI\$.
| Quantity | Symbol | SI Unit | Example |
|---|---|---|---|
| Energy | \$W\$ | Joule (J) | 0.24 J (example above) |
| Charge | \$Q\$ | Coulomb (C) | 0.02 C (example above) |
| Voltage | \$V\$ | Volt (V) | 12 V (example above) |
| Current | \$I\$ | Ampere (A) | 0.33 A (example above) |
| Power | \$P\$ | Watt (W) | 3 W (example above) |