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:
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:
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\( V = \dfrac{W}{Q} \)
- \( V \) = 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:
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- 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 = \( 10\ \mathrm{V} \)
