CIE AS/A Level Physics 9.1 Electric current Study Notes- 2025-2027 Syllabus
CIE AS/A Level Physics 9.1 Electric current Study Notes – New Syllabus
CIE AS/A Level Physics 9.1 Electric current Study Notes at IITian Academy focus on specific topic and type of questions asked in actual exam. Study Notes focus on AS/A Level Physics latest syllabus with Candidates should be able to:
- understand that an electric current is a flow of charge carriers
- understand that the charge on charge carriers is quantised
- recall and use Q = It
- use, for a current-carrying conductor, the expression I = Anvq, where n is the number density of charge carriers
Electric Current as a Flow of Charge Carriers
An electric current is the rate of flow of electric charge through a conductor or any medium. It results from the movement of charge carriers such as electrons, ions, or holes, depending on the material.
\( \mathrm{I = \dfrac{Q}{t}} \)
Explanation:
- In a metallic conductor, the charge carriers are free electrons moving through a lattice of positive ions.
- In an electrolyte, the charge carriers are positive and negative ions moving in opposite directions.
- In a semiconductor, charge carriers may be both electrons and “holes”.
Conventional vs Electron Flow:
- Conventional current flows from the positive to the negative terminal of a source.
- Electron flow is opposite — from the negative to the positive terminal.
Electric current is caused by the motion of charge carriers. The size of the current depends on how much charge passes a given point per unit time.
Example
A current of \( \mathrm{2.0 \ A} \) flows through a wire. How much charge passes through the cross-section of the wire in \( \mathrm{30 \ s} \)?
▶️ Answer / Explanation
Using \( \mathrm{Q = It} \):
\( \mathrm{Q = 2.0 \times 30 = 60 \ C.} \)
Therefore, 60 coulombs of charge pass through the wire in 30 seconds.
Example
If a current of \( \mathrm{5.0 \ A} \) flows for \( \mathrm{2 \ minutes} \), calculate the total charge transferred.
▶️ Answer / Explanation
\( \mathrm{t = 2 \ minutes = 120 \ s.} \)
Using \( \mathrm{Q = It} \):
\( \mathrm{Q = 5.0 \times 120 = 600 \ C.} \)
Hence, 600 coulombs of charge pass through the conductor.
Quantisation of Charge
The charge of matter is quantised, meaning it exists only in discrete units that are integer multiples of the elementary charge \( \mathrm{e} \).
\( \mathrm{Q = n e} \)
Where:
- \( \mathrm{Q} \): total charge (C)
- \( \mathrm{n} \): number of charge carriers (integer)
- \( \mathrm{e = 1.6 \times 10^{-19} \ C} \): charge of one electron or proton
Explanation: All free charge in nature is carried by particles with a charge that is an integer multiple of \( \mathrm{e} \). For example:
- Electron: \( \mathrm{-1.6 \times 10^{-19} \ C} \)
- Proton: \( \mathrm{+1.6 \times 10^{-19} \ C} \)
- Ion with charge \( \mathrm{+2e} \): \( \mathrm{+3.2 \times 10^{-19} \ C} \)
All observable electric charge is made up of whole multiples of the elementary charge \( \mathrm{e} \). Fractional charges do not exist in isolation.
Example
Find the total charge of \( \mathrm{5 \times 10^{18}} \) electrons.
▶️ Answer / Explanation
Using \( \mathrm{Q = n e} \):
\( \mathrm{Q = 5 \times 10^{18} \times 1.6 \times 10^{-19} = 0.8 \ C.} \)
Since electrons are negatively charged, \( \mathrm{Q = -0.8 \ C.} \)
Therefore, the total charge is \( \mathrm{-0.8 \ C.} \)
Example
How many electrons flow through a circuit if a charge of \( \mathrm{2.4 \ C} \) passes?
▶️ Answer / Explanation
Using \( \mathrm{n = \dfrac{Q}{e}} \):
\( \mathrm{n = \dfrac{2.4}{1.6 \times 10^{-19}} = 1.5 \times 10^{19} \ electrons.} \)
Therefore, approximately \( \mathrm{1.5 \times 10^{19}} \) electrons flow through the circuit.
Relationship Between Charge, Current, and Time
Formula:
\( \mathrm{Q = It} \)
Meaning: The total charge \( \mathrm{Q} \) that flows through a conductor is equal to the product of the current \( \mathrm{I} \) and the time \( \mathrm{t} \) for which the current flows.
This equation is derived from the definition of current as the rate of flow of charge: \( \mathrm{I = \dfrac{Q}{t}}. \)
Example
A current of \( \mathrm{0.5 \ A} \) flows through a circuit for \( \mathrm{10 \ minutes.} \) Find the charge transferred.
▶️ Answer / Explanation
\( \mathrm{t = 10 \ minutes = 600 \ s.} \)
Using \( \mathrm{Q = It} \):
\( \mathrm{Q = 0.5 \times 600 = 300 \ C.} \)
Therefore, 300 coulombs of charge pass through the circuit.
Example
How long will a current of \( \mathrm{4.0 \ A} \) take to transfer \( \mathrm{800 \ C} \) of charge?
▶️ Answer / Explanation
Using \( \mathrm{t = \dfrac{Q}{I}} \):
\( \mathrm{t = \dfrac{800}{4.0} = 200 \ s.} \)
Hence, the charge is transferred in 200 seconds (3 minutes 20 seconds).
Current in a Conductor — \( \mathrm{I = n A v q} \)
For a conductor carrying current, the current can be expressed in terms of the properties of the charge carriers using:
\( \mathrm{I = n A v q} \)
Where:
- \( \mathrm{I} \): current (A)
- \( \mathrm{n} \): number density of charge carriers (m\(^{-3}\))
- \( \mathrm{A} \): cross-sectional area of the conductor (m\(^2\))
- \( \mathrm{v} \): drift velocity of the charge carriers (m/s)
- \( \mathrm{q} \): charge on each carrier (C)
Explanation:
- Each charge carrier moves with a slow average drift velocity \( \mathrm{v} \) under the influence of an electric field.
- In time \( \mathrm{t} \), each carrier moves a distance \( \mathrm{v t} \), and the total charge passing through the cross-section is \( \mathrm{n A v t q.} \)
- Thus, the rate of charge flow (current) is \( \mathrm{I = n A v q.} \)
This formula connects microscopic particle motion (drift of charge carriers) to the macroscopic quantity of electric current.
Example
A copper wire has a cross-sectional area \( \mathrm{A = 1.0 \times 10^{-6} \ m^2} \), an electron number density \( \mathrm{n = 8.5 \times 10^{28} \ m^{-3}} \), and each electron carries a charge \( \mathrm{q = 1.6 \times 10^{-19} \ C.} \) If the drift velocity is \( \mathrm{2.5 \times 10^{-4} \ m/s,} \) find the current in the wire.
▶️ Answer / Explanation
Using \( \mathrm{I = n A v q} \):
\( \mathrm{I = 8.5 \times 10^{28} \times 1.0 \times 10^{-6} \times 2.5 \times 10^{-4} \times 1.6 \times 10^{-19}} \)
\( \mathrm{I = 3.4 \ A.} \)
Therefore, the current in the wire is approximately 3.4 A.
Example
The current in a copper wire is \( \mathrm{5.0 \ A.} \) If the electron number density is \( \mathrm{8.5 \times 10^{28} \ m^{-3}} \) and the wire’s cross-sectional area is \( \mathrm{2.0 \times 10^{-6} \ m^2,} \) calculate the drift velocity of the electrons.
▶️ Answer / Explanation
Using \( \mathrm{I = n A v q} \Rightarrow \mathrm{v = \dfrac{I}{n A q}} \):
\( \mathrm{v = \dfrac{5.0}{(8.5 \times 10^{28})(2.0 \times 10^{-6})(1.6 \times 10^{-19})}} \)
\( \mathrm{v = 1.84 \times 10^{-4} \ m/s.} \)
Hence, the drift velocity of electrons is \( \mathrm{1.8 \times 10^{-4} \ m/s.} \)