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Potential difference and electromotive force (EMF) are two of the most important concepts in electricity. Though both are measured in volts, they describe fundamentally different things. Understanding the distinction is essential for circuit analysis.
Potential difference (p.d.) between two points is the energy transferred per unit charge as charge moves between those two points.
V=QW
where:
One volt equals one joule per coulomb (1 V = 1 J C⁻¹).
When charge flows through a component (such as a resistor or lamp), electrical energy is transferred to other forms (heat, light, etc.). The p.d. across the component tells us how much energy is transferred per coulomb of charge.
Key Definition: The potential difference across a component is the work done (energy transferred) per unit charge as charge passes through the component.
A 12 V battery drives a charge of 50 C through a circuit. Calculate the total energy transferred.
Solution:
V = W/Q, so W = QV
W = 50 × 12 = 600 J
A heater operates at 230 V and transfers 46 kJ of energy. How much charge flows through it?
Solution:
W = 46 kJ = 46 000 J
Q = W/V = 46 000 / 230 = 200 C
EMF (ε) is the energy transferred per unit charge by a source (battery, cell, generator, solar cell) in driving charge around a complete circuit.
ε=QW
where:
EMF is also measured in volts (V) because it has the same dimensions as potential difference (energy per unit charge).
| EMF (ε) | Potential Difference (V) | |
|---|---|---|
| What it measures | Energy supplied per unit charge by a source | Energy transferred per unit charge by a component |
| Where it applies | Sources (batteries, cells, generators) | Components (resistors, lamps, motors) |
| Energy conversion | Other forms → electrical energy | Electrical energy → other forms |
Common Misconception: EMF is not a "force" despite the name. It is an energy per unit charge (measured in volts), not a force (which would be measured in newtons). The name is historical and unfortunately misleading.
When current I flows through a component with resistance R, the potential difference across it is:
V=IR
This is one of the most frequently used equations in electricity. It can be rearranged to:
A 470 Ω resistor carries a current of 25 mA. Calculate the potential difference across it.
Solution:
I = 25 mA = 25 × 10⁻³ = 0.025 A
V = IR = 0.025 × 470 = 11.75 V
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