Current Electricity — Derivations & Reasoning

Class XII Physics · Chapter 3 · Card 03

This learning card builds each result from a physical model, tracks the assumptions, and explains what the equation tells us. It complements the existing Physics Part I chapter; it does not reproduce the textbook.

D01 — Derive I = neAv_d

What is being derived

The magnitude of steady current carried by mobile charge carriers in a uniform conductor.

Starting physical idea

Current is charge crossing a cross-section per unit time. Let carrier number density be n, charge magnitude e, area normal to motion A, and drift-speed magnitude v_d.

Step-by-step derivation

1. In Δt, carriers drift Δx = v_dΔt.
2. Those crossing the section occupy volume AΔx = Av_dΔt.
3. Their number is ΔN = nAv_dΔt.
4. Transported charge magnitude is ΔQ = eΔN = neAv_dΔt.
5. Thus I = ΔQ/Δt = neAv_d.

Meaning of the result

Current grows with carrier density, charge per carrier, cross-sectional area, and average drift speed. For electrons, conventional current points opposite their drift.

Common trap: Counting rapid random thermal motion as net current. Only the small average drift contributes to steady current.

D02 — Derive v_d = eEτ/m

What is being derived

The magnitude of electron drift speed in the simple relaxation-time model.

Starting physical idea

An electron accelerates in an electric field between collisions; τ is the average interval between momentum-randomising collisions.

Step-by-step derivation

1. The electron charge is −e, so F = −eE.
2. Newton’s law gives a = −eE/m.
3. Over an average free-flight interval τ, the field changes mean velocity by aτ = −eEτ/m.
4. Random thermal velocities average to zero, leaving v⃗_d = −eτE⃗/m.
5. Taking magnitudes gives v_d = eEτ/m.

Meaning of the result

Drift speed rises with field and relaxation time, and falls with carrier mass.

Common trap: Treating τ as the time since an electron was released. It is an average time between collisions; electron drift points opposite E⃗.

D03 — Derive ρ = m/(ne²τ)

What is being derived

The Drude-model resistivity in terms of carrier density and relaxation time.

Starting physical idea

Microscopic current density comes from drifting carriers; macroscopically, resistivity relates field to current density.

Step-by-step derivation

1. For electron carriers, magnitudes obey j = ne v_d.
2. Substitute D02: j = ne(eEτ/m) = (ne²τ/m)E.
3. Hence j = σE with σ = ne²τ/m.
4. Since ρ = 1/σ, ρ = m/(ne²τ).

Meaning of the result

In this model, more carriers and longer momentum-relaxation time increase conductivity; more frequent scattering increases resistivity.

Common trap: Treating this as an exact universal law. It follows from a simplified carrier and collision model.

D04 — Derive R = ρl/A

What is being derived

Resistance of a uniform ohmic conductor of length l and cross-sectional area A.

Starting physical idea

For a uniform conductor, E = ρj; potential-drop and current magnitudes follow the conductor axis.

Step-by-step derivation

1. For uniform field, V = El.
2. Current through the section is I = jA.
3. Using E = ρj: V/I = El/(jA) = ρjl/(jA).
4. Cancel j to obtain R = V/I = ρl/A.

Meaning of the result

At fixed material and temperature, a longer path raises resistance and a larger cross-section lowers it.

Common trap: Confusing resistivity (material property) with resistance (also dependent on geometry and conditions).

D05 — Derive P = VI = I²R = V²/R

What is being derived

The rate of electrical energy transfer and its resistor forms.

Starting physical idea

Moving charge q through potential difference V transfers energy W = qV; power is energy per time.

Step-by-step derivation

1. P = W/t = qV/t.
2. Since I = q/t, P = VI.
3. For an ohmic resistor V = IR, so P = I²R.
4. Using I = V/R instead gives P = V²/R.

Meaning of the result

Power is the rate of energy transfer. The forms are equivalent for a resistor; choose the one matching known quantities.

Common trap: Applying I²R or V²/R unchanged to every component; those forms use Ohm’s law for a resistor.

D06 — Derive V = ε − Ir

What is being derived

Terminal voltage of a real source delivering current.

Starting physical idea

Emf ε is source energy supplied per unit charge. Internal resistance r causes a drop while current flows.

Step-by-step derivation

1. A charge q receives source energy εq.
2. Internal heating over t is I²rt; with q = It, the lost energy per charge is Ir.
3. Energy per charge available at the terminals is therefore ε − Ir.
4. Thus V = ε − Ir while the source delivers current.

Meaning of the result

With no current V = ε; under load the internal drop makes terminal voltage smaller.

Common trap: Calling emf a force or assuming V always equals ε. Emf is energy per charge; equality is the open-circuit/ideal limit.

D07 — Derive I = ε/(R+r)

What is being derived

Current from emf ε and internal resistance r connected to external resistance R.

Starting physical idea

The same steady current passes through the source resistance and a single series load.

Step-by-step derivation

1. Load voltage is V = IR.
2. Source terminal voltage is V = ε − Ir.
3. Equate: IR = ε − Ir.
4. Rearrange: I(R+r) = ε, so I = ε/(R+r).

Meaning of the result

Internal resistance acts in series and lowers current below ideal ε/R.

Common trap: Omitting r for a real cell, or treating a non-series network as a single R+r loop.

D08 — Equivalent emf and resistance of cells

What is being derived

Equivalent parameters for series cells and parallel cells.

Starting physical idea

Model each cell as ideal emf in series with internal resistance. Give each emf a sign from its polarity along a chosen direction.

Step-by-step derivation

1. Series: Same current crosses each internal resistance, so drops add: r_eq = Σr_i. Signed source rises add: ε_eq = Σε_i; a reversed cell contributes negative emf.
2. Parallel: All branches share terminal voltage V. The signed current supplied by cell i is (ε_i−V)/r_i.
3. At open output, junction balance gives Σ(ε_i−V)/r_i = 0.
4. Solving: ε_eq = [Σ(ε_i/r_i)]/[Σ(1/r_i)], and r_eq = 1/[Σ(1/r_i)].
5. For n identical aligned cells in parallel: ε_eq = ε and r_eq = r/n.

Meaning of the result

Series adds signed source rises and internal resistances; parallel combines conductances and gives a resistance-weighted signed emf.

Common trap: Adding parallel emfs directly or ignoring polarity. Unequal parallel cells may drive circulating current.

D09 — Kirchhoff junction rule

What is being derived

Algebraic current balance at a circuit junction.

Starting physical idea

Charge is conserved: a junction cannot steadily accumulate or destroy net charge.

Step-by-step derivation

1. Choose a junction and mark currents entering or leaving.
2. Over Δt, incoming charge is ΣI_inΔt and outgoing charge is ΣI_outΔt.
3. Steady junction charge means these are equal.
4. Cancel Δt: ΣI_in = ΣI_out, or ΣI = 0 with consistent signs.

Meaning of the result

At a steady ordinary node, total current entering equals total current leaving.

Common trap: Mixing sign conventions. Choose entering/leaving signs and use them consistently.

D10 — Kirchhoff loop rule

What is being derived

The zero algebraic sum of potential changes around a closed loop.

Starting physical idea

Energy is conserved. A charge returning to its starting point has no net change in potential energy.

Step-by-step derivation

1. Choose a traversal direction and track a test charge around the loop.
2. Across a source from negative to positive terminal, energy per charge rises by ε; reverse traversal gives −ε.
3. Across a resistor along current, energy per charge drops by IR; opposite traversal gives a rise IR.
4. After the loop, the charge returns to its starting potential, so total energy change per charge is zero.
5. Thus signed source rises and resistor drops balance: Σε − ΣIR = 0, with signs set by traversal.

Meaning of the result

Every closed path balances source energy gains against drops and other signed changes.

Common trap: Assigning signs by rote instead of the chosen traversal and current directions.

D11 — Wheatstone bridge balance

What is being derived

The null condition for a four-arm bridge whose two intermediate junctions are connected by a galvanometer.

Starting physical idea

At balance, galvanometer current is zero. The joined junctions have equal potential; each side carries its own branch current.

Step-by-step derivation

1. Label upper-left and lower-left arms P,Q; upper-right and lower-right arms R,S.
2. Let branch currents be I_L and I_R. At null, no current passes through the galvanometer.
3. Equal junction potentials require equal top-to-junction drops: I_LP = I_RR.
4. They also require equal junction-to-bottom drops: I_LQ = I_RS.
5. Divide to eliminate branch currents: P/Q = R/S, equivalently PS = QR.

Meaning of the result

At balance the galvanometer reads null; an unknown arm follows from the other-arm ratio. The exact null condition does not depend on galvanometer resistance.

Common trap: Pairing the wrong arms. Keep top-to-bottom order consistent on both sides.

Reasoning prompts

• Why is electron drift opposite to the electric field? The force is qE; for q = −e, acceleration and mean drift oppose E⃗.
• Why can a tiny drift speed produce a large current? I = neAv_d: a macroscopic cross-section contains an enormous number of mobile carriers.
• Why does resistance increase with length? At fixed material and area, a longer path gives a proportionally larger drop for the same current: R = ρl/A.
• Why can a negative current from Kirchhoff equations be physically valid? It means the actual current runs opposite to the trial direction.
• Why is emf a potential difference rather than a force? Emf is work/energy supplied per charge, measured in joules per coulomb (volts), not newtons.
• Why does high-voltage transmission reduce line losses? For fixed P = VI, higher V means lower I; line heating I²R falls.

Model note: Microscopic relations use the simple relaxation-time model. Circuit reference directions are arbitrary; consistent signs preserve the measurable result.