CLASS 12 PHYSICS · CHAPTER 3 · STUDY NOTES
Current Electricity
An in-depth study of charges in motion, covering Ohm's law, the microscopic origin of resistivity through electron drift, temperature dependence, Kirchhoff's rules, and the Wheatstone bridge.
EXPLORE
Electric Current and Flow in Conductors
Electric current is the rate of flow of electric charge through a cross-section of a conductor.EXPLORE THE IDEA
Follow the carriers; count net charge
Predict: which way will electrons drift when the field points right?
Electrons move randomly among fixed positive ions. Opposite crossings cancel on average: net current is zero. Count net charge through a cross-section, not the total distance travelled by every electron.
- Electric current
- Net charge crossing a specified surface per unit time.
While static electricity deals with charges at rest, current electricity focuses on charges in motion. In nature, this is seen in lightning, where charge flow is transient. In domestic circuits, cells and batteries maintain a steady flow of charge.
The current I is defined as the net charge Q flowing across an area per unit time t. For a steady flow, I = Q/t. If the flow is not steady, we define instantaneous current as the limit of ΔQ/Δt as Δt approaches zero. The SI unit of current is the ampere (A).
In solid conductors like metals, current is carried by negatively charged electrons moving through a lattice of fixed positive ions. In the absence of an external electric field, these electrons move randomly due to thermal energy, resulting in zero net current. When an electric field is applied, electrons experience a force and begin to drift.
Let’s look a little closer.
Go deeper: Reason it through
A current is a signed rate; choose a direction before counting crossings. Random motion alone can produce many crossings with no net current.
Take a moment. What do you think?
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Question
A hypothetical wire carries 18 C in 45 s. Calculate its steady current.
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Ohm's Law and Resistance
Ohm's law establishes the relationship between the potential difference across a conductor and the current flowing through it.EXPLORE THE IDEA
Connect wire geometry to the I–V graph
Predict: at the same voltage, will a wider wire carry more or less current?
For ρ = 2 × 10⁻⁶ Ω·m, l = 1 m and A = 1 mm²: R = ρl/A = 2.00 Ω. At 6 V, I = 3.00 A. The I–V slope is 1/R. Temperature and material are held fixed; changing area is not stretching a fixed-volume wire.
- Resistivity
- A material property relating electric field to current density in a stated regime; its SI unit is Ω m.
| Factor | Change | Effect on Resistance (R) |
|---|---|---|
| Length (l) | Doubled | Doubles |
| Area (A) | Doubled | Halved |
| Resistivity (ρ) | Stated material and physical conditions | May change with temperature or material state |
For an ohmic conductor with unchanged physical conditions, current is proportional to potential difference: V = IR. The law describes this operating regime; it is not a rule obeyed by every device.
The constant R is the resistance, measured in ohms (Ω). Resistance is the opposition offered by the conductor to the flow of current. It depends on the material's nature and the conductor's dimensions.
Experimentally, resistance R is found to be directly proportional to the length (l) of the conductor and inversely proportional to its cross-sectional area (A). Combining these, R = ρ(l/A), where ρ (rho) is the resistivity of the material.
Go deeper: Reason it through
R = ρL/A assumes a uniform conductor with constant cross-section. Stretching changes both length and area; keeping area fixed is a different comparison. Temperature and material state must also be specified.
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Question
A uniform 3 Ω wire is stretched to 1.5 times its length at constant volume and resistivity. Find its new resistance.
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Drift of Electrons and Origin of Resistivity
Resistivity arises from the microscopic collisions of electrons with the ions of the conductor.EXPLORE THE IDEA
Build current from microscopic drift
Predict: will doubling the cross-sectional area change drift speed, current, or both?
For E = 1 V/m and τ = 1 × 10⁻¹⁴ s, |v_d| = eEτ/m = 1.759 mm/s, opposite E. Between collisions the electron accelerates; a collision randomises its velocity. The shown trajectory is an illustrative realisation, not a measured path or a typical thermal speed.
- Drift velocity
- The average velocity of charge carriers due to an applied electric field, superposed on their random motion.
In the relaxation-time model, an electric field gives an electron acceleration opposite to the field: a⃗ = −eE⃗/m. Scattering prevents indefinite acceleration; many trajectories average to a small drift superposed on random motion.
The drift-velocity vector is v⃗d = −eτE⃗/m, while its speed magnitude is |vd| = eτ|E|/m. Keeping this sign distinction prevents confusing electron motion with conventional current.
Current density is j⃗ = nqv⃗d. For electrons q = −e, giving j⃗ = −nev⃗d = (ne²τ/m)E⃗. Thus σ = ne²τ/m and ρ = m/(ne²τ) in the model. The electrical disturbance travels along a circuit at a finite speed set by its electromagnetic environment; local carriers need not travel from the battery to the lamp first.
Go deeper: Reason it through
The relaxation-time model averages many scattering events. It does not claim that every electron has the same trajectory or that quantum electron speeds are classical thermal speeds. Signal propagation and slow carrier drift are different quantities.
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Question
Does a metal electron drift along or against the electric field, and how does this compare with conventional current?
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Mobility and Limitations of Ohm's Law
Mobility measures how easily a charge carrier moves, while limitations define where Ohm's law fails.EXPLORE THE IDEA
Compare mobility and non-Ohmic curves
Predict: which curves allow one constant resistance across all the shown conditions?
μ = |v_d|/E = eτ/m = 0.001759 m²/(V·s), for τ = 1 × 10⁻¹⁴ s. Mobility is positive; the electron’s velocity is opposite E. Increasing τ raises mobility in this simple collision model.
- Mobility
- Magnitude of drift velocity per unit electric field; its SI unit is m² V⁻¹ s⁻¹.
Mobility (μ) is defined as the magnitude of drift velocity per unit electric field: μ = |vd|/E. In SI units, it is measured in m²/Vs. For electrons, μ = eτ/m.
Ohm's law is not a fundamental law but an empirical one. It fails in several cases: (1) V ceases to be proportional to I, such as in a good conductor at high currents where the V-I graph deviates from a straight line; (2) The relation between V and I depends on the sign of V, as seen in a diode; and (3) The relation between V and I is non-unique, meaning more than one value of V exists for the same current I, as seen in Gallium Arsenide (GaAs).
Go deeper: Reason it through
The three non-ohmic plots illustrate different failures of a constant proportional relationship: nonlinearity, polarity dependence and a non-unique relationship. One straight segment is not enough to establish Ohm’s law across all operating conditions.
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Question
A hypothetical carrier has drift-speed magnitude 0.003 m/s in a field of 150 V/m. Find its mobility.
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Temperature Dependence of Resistivity
The resistivity of materials changes with temperature based on their internal structure.EXPLORE THE IDEA
Compare each material with its own reference
Predict: will heating change each material’s resistivity in the same direction?
At T = 20 °C, ρ/ρ₂₀ = 1.000 in this illustrative metal model. Metal: increased lattice scattering reduces τ. The linear approximation is limited to this moderate temperature range. The plot compares change from each material’s own reference; it does not compare their absolute resistivities.
- Temperature coefficient
- The fractional change in resistivity per unit temperature change in an approximately linear range.
For many metals over a suitable range, increased lattice scattering raises resistivity. An approximate relation is ρ(T) = ρ₀[1 + α(T − T₀)], with α and temperature units stated.
In many semiconductors, heating increases carrier density substantially, so resistivity can decrease even while scattering changes. The curve depends on material, doping and temperature range.
Alloys such as manganin and constantan can have small temperature coefficients in useful ranges, supporting stable resistors. Nichrome is useful in heating elements; low temperature sensitivity does not mean temperature has no effect.
Original linear-temperature estimate
R = R₀[1 + α(T − T₀)]For R₀ = 20 Ω at 20 °C and α = 0.004 °C⁻¹, the 50 °C rise gives R = 20(1 + 0.20) = 24 Ω. Assume the material remains in the model’s linear range.
Go deeper: Reason it through
The linear temperature equation is an approximation over a suitable range. Temperature can affect carrier density as well as scattering; alloy trends and semiconductor trends are not one universal straight-line law.
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Question
An original resistor has R₀ = 25 Ω at 10 °C and α = 0.004 °C⁻¹. Estimate R at 60 °C using the linear model.
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Electrical Energy and Power
Electrical energy is converted into heat when current flows through a resistor.EXPLORE THE IDEA
Track energy transfer and cable loss
Predict: what happens to cable heating when transmission voltage doubles at fixed power?
A 6 Ω resistor at 6 V carries 1.00 A and dissipates P = VI = I²R = V²/R = 6.0 W. In 10 s it receives 60 J. Collisions transfer electrical energy to the lattice; current is not used up.
- Electrical power
- Rate at which electrical energy is transferred; P = VI for the stated terminal quantities.
For constant terminal voltage and current, charge Q = It transfers electrical energy W = VIt. In a resistor, this energy is dissipated as thermal energy.
Power is P = VI. For an ohmic resistor, P = I²R = V²/R; identify which quantity is held fixed before predicting the effect of a resistance change.
At the same transmitted power, a higher transmission voltage allows lower current. Cable heating is I²Rcable, so lower current reduces this loss. The simple comparison omits other engineering limits.
Go deeper: Reason it through
For a fixed voltage across a resistor, increasing R reduces V²/R; for fixed current, increasing R raises I²R. These are different controlled conditions. Transmission-loss comparisons assume the same delivered power and cable resistance.
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Question
A hypothetical 12 Ω resistor carries 0.50 A. Find its power and the energy transferred in 80 s.
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Cells, EMF, and Internal Resistance
Separate the source’s emf from its loaded terminal voltage, then combine sources with their polarity.EXPLORE THE IDEA
Follow source polarity and terminal voltage
Predict: will terminal voltage rise or fall when discharge current increases?
With ε = 6 V, r = 1 Ω and external R = 5 Ω: I = ε/(R+r) = 1.00 A. Terminal V = ε−Ir = 5.00 V. EMF is energy supplied per charge, measured in volts; it is not a force. Open circuit gives I = 0 and V = ε.
- Electromotive force (emf)
- Energy supplied by a source per unit charge; measured in volts, despite the word force.
Emf ε is energy supplied per unit charge, measured in volts. For a source at open circuit in this model, terminal voltage equals ε. A real cell with internal resistance does not maintain one constant terminal voltage at every load.
During discharge, V = ε − Ir and I = ε/(R + r). During charging, current enters the positive terminal and the terminal voltage is ε + Ir in the same resistance model. The short-circuit expression ε/r is a model limit, not a safe operating procedure.
For series cells, add emfs with signs set by polarity and add internal resistances. Two identical aiding cells give 2ε and 2r; reversing one changes the net emf. For n identical, same-polarity parallel cells, εeq = ε and req = r/n. Unequal sources need a fuller circuit calculation.
Go deeper: Reason it through
Series cells add signed emfs and internal resistances. Identical parallel cells have the same equivalent emf and a smaller equivalent internal resistance. Unequal parallel cells can drive circulating currents, so the identical-cell formula cannot simply be reused.
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Question
Two identical cells each have emf 1.5 V and internal resistance 0.20 Ω. Find equivalent source quantities in series-aiding and in parallel.
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Kirchhoff's Rules
Kirchhoff's rules are essential for analyzing complex circuits that cannot be simplified by series-parallel rules.EXPLORE THE IDEA
Balance charge at junctions and energy in loops
Predict: does reversing your assumed arrow change the actual circuit?
A 6 V source feeds parallel 3 Ω and 6 Ω branches: 2 A + 1 A = 3 A at the junction. Each loop gives +6 V−6 V = 0. Junction balance conserves charge; the closed-loop sum conserves energy. Both branches have the same terminal voltage.
- Kirchhoff’s junction rule
- In a steady circuit, total current entering a junction equals total current leaving it.
Junction Rule: At any junction, the sum of currents entering must equal the sum of currents leaving. This is a statement of the Law of Conservation of Charge.
Loop Rule: The algebraic sum of changes in potential around any closed loop is zero. This follows from the Law of Conservation of Energy, as the net work done in moving a charge around a closed path in an electrostatic field is zero.
When applying the loop rule: (1) Potential falls by IR when moving in the direction of current. (2) Potential rises when moving from the negative to the positive terminal of a cell.
Original loop calculation
9 − 3I − 6I = 0An ideal 9 V source and 3 Ω and 6 Ω series resistors carry I = 9/(3 + 6) = 1 A. In the direction of current, resistor potential changes are −3 V and −6 V.
Go deeper: Reason it through
Choose current directions first; a negative solved current means the real direction is opposite to that choice. The usual loop rule assumes a lumped circuit without an unaccounted changing magnetic flux through the loop.
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Question
A hypothetical junction receives 2.4 A and 0.8 A. One outgoing branch carries 1.1 A. Find the other outgoing current.
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The Wheatstone Bridge
The Wheatstone bridge is a specific circuit arrangement used to measure an unknown resistance accurately.EXPLORE THE IDEA
Find equal detector potentials
Predict: how will the detector current change as R₄ passes through balance?
R₁ = R₂ = R₃ = 100 Ω; R₄ = 100 Ω; R_g = 100 Ω; supply = 6 V. V_B = 3.000 V and V_D = 3.000 V. I_g (B→D) = 0.000 mA. Balanced: V_B = V_D, so I_g = 0. R₂/R₁ = R₄/R₃.
- Balanced Wheatstone bridge
- A bridge in which the two detector junctions have equal potential, so detector current is zero.
It consists of four resistors R1, R2, R3, and R4 arranged in a quadrilateral. A galvanometer is connected between one pair of opposite junctions, and a cell is connected across the other pair.
The bridge is said to be 'balanced' when no current flows through the galvanometer (Ig = 0). Under this condition, the potential at the two galvanometer terminals is equal.
The balance condition is R2/R1 = R4/R3. If three resistances are known, the fourth can be calculated as R4 = R3 * (R2/R1). This principle is used in the Meter Bridge.
Go deeper: Reason it through
The resistance-ratio balance follows from the named arms and their connections. Away from balance, detector loading changes both junction potentials; independent unloaded divider voltages cannot give the exact detector current.
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Question
For the stated balance condition R₂/R₁ = R₄/R₃, choose R₁ = 12 Ω, R₂ = 18 Ω and R₃ = 8 Ω. Find R₄.
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