Home / Edexcel A Level / A Level (IAL) Physics (YPH11) / 2.33 Potential Difference- Study Notes

Edexcel A Level (IAL) Physics-2.33 Potential Difference- Study Notes- New Syllabus

Edexcel A Level (IAL) Physics -2.33 Potential Difference- Study Notes- New syllabus

Edexcel A Level (IAL) Physics -2.33 Potential Difference- Study Notes -Edexcel A level Physics – per latest Syllabus.

Key Concepts:

  •  

Edexcel A level Physics-Study Notes- All Topics

Electric Potential Difference \( V = \dfrac{W}{Q} \)

The electric potential difference (voltage) between two points in a circuit tells us how much work is done per unit charge moved between those points. It is a measure of the energy transferred to or from each coulomb of charge.

Definition of Potential Difference

The potential difference between two points is defined as:

\( V = \dfrac{W}{Q} \)

  • \( V \) = potential difference (volts, V)
  • \( W \) = work done or energy transferred (joules, J)
  • \( Q \) = charge moved (coulombs, C)

Interpretation:

  • If \( V = 1\ \mathrm{V} \), then \( 1\ \mathrm{J} \) of energy is transferred per \( 1\ \mathrm{C} \) of charge.
  • Higher voltage → more energy per coulomb.

Physical Meaning

  • Voltage is “energy per charge”.
  • A battery or power supply provides energy to charges.
  • Charges lose energy when passing through resistors, lamps, motors, etc.
  • Voltage across a component tells you how much energy each coulomb loses there.

Using \( V = \dfrac{W}{Q} \)

This formula is useful for:

  • Finding energy transferred in a circuit
  • Calculating charge moved through a component
  • Relating electrical work to other forms of energy
  • Understanding how batteries deliver energy

Rearranged forms:

\( W = VQ \)

\( Q = \dfrac{W}{V} \)

Energy Interpretation in Circuits

  • A 9 V battery gives \( 9\ \mathrm{J} \) to every coulomb of charge.
  • A lamp with \( 12\ \mathrm{V} \) across it takes \( 12\ \mathrm{J} \) of energy from every coulomb.
  • Energy conservation applies: energy supplied by the source = energy used by components.

Example (Easy)

A charge of \( 3\ \mathrm{C} \) moves through a component with a potential difference of \( 5\ \mathrm{V} \). How much energy is transferred?

▶️ Answer / Explanation

\( W = VQ = 5 \times 3 = 15\ \mathrm{J} \)

Example (Medium)

If \( 120\ \mathrm{J} \) of work is done moving \( 20\ \mathrm{C} \) of charge, calculate the potential difference.

▶️ Answer / Explanation

\( V = \dfrac{W}{Q} = \dfrac{120}{20} = 6\ \mathrm{V} \)

Example (Hard)

A device transfers \( 450\ \mathrm{J} \) of energy when a current passes for \( 15\ \mathrm{s} \) at \( 3\ \mathrm{A} \). Find the potential difference across the device.

▶️ Answer / Explanation

Step 1: Find charge.

\( Q = It = 3 \times 15 = 45\ \mathrm{C} \)

Step 2: Use \( V = \dfrac{W}{Q} \).

\( V = \dfrac{450}{45} = 10\ \mathrm{V} \)

Potential difference across device = \( 10\ \mathrm{V} \)

Scroll to Top