Current ElectricityClass 12 Physics Part I, Chapter 3

This chapter explains how charge flows through conductors and cells. You will link current to drift velocity, use resistance and resistivity with Ohm's law, and handle power, internal resistance and combinations of cells. After revising here, you should be able to apply Kirchhoff's rules and the Wheatstone bridge condition to circuit problems.

Follows the Class 12 Physics Part I textbook, Chapter 3. Made with Classly and checked by our team.

Cheat sheet

Current Electricity on one page, for a last look before a test.

Current Electricity

Cheat sheetClass 12 · Physics

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Ohmic conductor — $V=IR$ only when resistance stays constant under the conditions.

Drift — $\vec v_d=-\dfrac{e\tau}{m}\vec E$; electron drift is opposite $\vec E$.

Cell under load — $V=\varepsilon-Ir$ while supplying current; open circuit: $V=\varepsilon$.

Kirchhoff — junction: current in = current out; closed loop: signed potential changes sum to zero.

Wheatstone null — $I_g=0 \Rightarrow \dfrac{R_1}{R_2}=\dfrac{R_3}{R_4}$.

1. Current, drift & conduction

  • Current — $I=\dfrac{\Delta Q}{\Delta t}$ for steady flow; instantaneously $I=\dfrac{dQ}{dt}$.
  • Conventional current — direction positive charge moves; in metals, electron drift is opposite.
  • Current density — $\vec j=nq\vec v_d$; for electrons, $q=-e$; uniform normal flow: $I=|j|A$.
  • Drift speed — $\vec v_d=-\dfrac{e\tau}{m}\vec E$ for electrons; $v_d=\dfrac{I}{neA}$ in magnitude.
  • Relaxation time — $\tau$, average time between successive carrier collisions.
  • Mobility — $\mu=\dfrac{|v_d|}{E}=\dfrac{e\tau}{m}$ for electrons; unit $\mathrm{m^2,V^{-1},s^{-1}}$.

Random thermal motion → average velocity zero → no net current; applied field adds a small average drift to the random motion. Collisions repeatedly interrupt acceleration, producing steady average drift.

2. Resistance, resistivity & Ohm’s law

  • Ohm’s law — $V=IR$; $V$ is proportional to $I$ only for an ohmic conductor under fixed physical conditions.
  • Resistance — $R=V/I$, unit $\Omega=\mathrm{V/A}$; depends on material, dimensions and temperature.
  • Geometry — $R=\rho\dfrac{\ell}{A}$; longer conductor → larger $R$; larger cross-section → smaller $R$.
  • Resistivity — $\rho=RA/\ell$, material property; unit $\Omega,\mathrm m$.
  • Conductivity — $\sigma=1/\rho$; $\vec j=\sigma\vec E$ or $\vec E=\rho\vec j$.
  • Drude-model link — $\sigma=\dfrac{ne^2\tau}{m}$; $\rho=\dfrac{m}{ne^2\tau}$, for the chapter’s constant-$n$, constant-$\tau$ model.
  • Material classes — metals: low $\rho$ · semiconductors: intermediate; $\rho$ generally falls as temperature rises · insulators: very high $\rho$.

Ohm’s law fails if the $V$–$I$ relation is nonlinear, changes with the sign of $V$, or gives more than one $V$ for a given $I$ (as in GaAs). $V=IR$ can still define a ratio for a device; it does not prove that the device is ohmic.

3. Temperature effects

  • Linear-range relation — $\rho_T=\rho_0[1+\alpha(T-T_0)]$; approximately valid over a limited temperature range.
  • Temperature coefficient — $\alpha$, fractional change in resistivity per unit temperature change; unit $\mathrm{K^{-1}}$ or $^\circ\mathrm C^{-1}$.
MaterialAs temperature risesRecall point
Metals$\rho$ usually increases$\alpha>0$; more collisions, nearly fixed carrier density
Semiconductors / insulators$\rho$ decreasesincreasing carrier density dominates
Nichrome, manganin, constantan$\rho$ changes littleuseful for stable resistance

For a uniform wire, $R$ changes in the same proportion as $\rho$ if its dimensions are treated as unchanged: $R_T=R_0[1+\alpha(T-T_0)]$.

4. Electrical energy, power & transmission

  • Electrical energy transferred — $W=IV\Delta t$; in a resistor, dissipated as heat.
  • Power — $P=IV=I^2R=\dfrac{V^2}{R}$; choose the form matching known quantities.
  • Transmission loss — $P_c=I^2R_c=\dfrac{P^2R_c}{V^2}$ for delivered power $P=VI$; high transmission voltage lowers cable loss.

5. Cells & equivalent sources

  • EMF — $\varepsilon$, work supplied per unit charge; open-circuit terminal voltage, not a force.
  • Internal resistance — $r$, resistance inside the cell; while discharging, $V=\varepsilon-Ir$ and $I=\dfrac{\varepsilon}{R+r}$ for external resistance $R$.
  • Maximum theoretical current — $I_{\max}=\varepsilon/r$ at $R=0$; practical cells may have a lower safe limit.
Cell combinationEquivalent emfEquivalent internal resistance
Series, aiding$\varepsilon_{eq}=\sum_i\varepsilon_i$$r_{eq}=\sum_i r_i$
Series, opposingalgebraic sum; opposing emf is negative$r_{eq}=\sum_i r_i$
Parallel$\varepsilon_{eq}=\dfrac{\sum_i(\varepsilon_i/r_i)}{\sum_i(1/r_i)}$$\dfrac{1}{r_{eq}}=\sum_i\dfrac{1}{r_i}$

Parallel-cell formula uses consistent polarity signs; for two cells, $\varepsilon_{eq}=\dfrac{\varepsilon_1r_2+\varepsilon_2r_1}{r_1+r_2}$.

6. Circuit rules & bridge

  • Junction rule — $\sum I_{in}=\sum I_{out}$; charge does not build up at a steady-current junction.
  • Loop rule — $\sum \Delta V=0$ around any closed loop; potential returns to its starting value.
  • Current sign — choose branch directions first; a negative result means actual flow is opposite the chosen arrow.
  • Cell sign in a loop — crossing $-\to+$ gives $+\varepsilon$; crossing $+\to-$ gives $-\varepsilon$.
  • Resistor sign in traversal — along current: $-IR$; against current: $+IR$.
  1. Mark a trial direction for each branch current.
  2. Use junction equations to relate currents.
  3. Choose independent closed loops; add signed voltage changes.
  4. Solve simultaneous equations; reverse any negative-current arrow.

Wheatstone bridge — four arms $R_1,R_2,R_3,R_4$; galvanometer joins the other pair of vertices.

  • Balanced bridge: $I_g=0$ → $\dfrac{R_1}{R_2}=\dfrac{R_3}{R_4}$ → $R_1R_4=R_2R_3$.
  • Unknown arm, when $R_4$ is unknown: $R_4=\dfrac{R_2R_3}{R_1}$.
  • Meter bridge — practical resistance measurement using the balanced-bridge principle.

7. Traps

  • Current is a vector → current is scalar; its arrow marks direction, not vector addition.
  • All electrons drift together → random motion remains; drift is only their small average motion.
  • $V=IR$ guarantees Ohm’s law → it defines resistance; Ohm’s law requires a linear $V$–$I$ relation under fixed conditions.
  • EMF is a force → it is energy supplied per unit charge, measured in volts.
  • Terminal voltage equals $\varepsilon$ under load → while discharging, internal resistance gives a drop $Ir$.
  • Use a guessed current direction as certain → a negative solved value means the actual direction is reversed.

Flashcards

10 cards on Current Electricity. Flip them here, or read them all below.

Current Electricity flashcards1 of 10

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All 10 cards

  1. 1

    Write Ohm’s law relating potential difference, current and resistance.

    $V=IR$, with $R$ constant (an ohmic conductor)

  2. 2

    What formula gives a uniform conductor’s resistance in terms of its resistivity, length and cross-sectional area?

    $R=\rho\frac{l}{A}$

  3. 3

    What is the relation between electron drift velocity and electric field in the model with mean collision time $\tau$?

    $\vec{v}_d=-\frac{e\tau}{m}\vec{E}$

  4. 4

    Write the approximate temperature relation for a metal’s resistivity near reference temperature $T_0$.

    $\rho_T=\rho_0[1+\alpha(T-T_0)]$

  5. 5

    Give the equivalent expressions for power dissipated in a resistor.

    $P=IV=I^2R=\frac{V^2}{R}$

  6. 6

    A cell’s emf is the open-circuitbetween its terminals.

    The potential difference between its terminals when no current is drawn

  7. 7

    What is the terminal voltage of a cell supplying current $I$ when its emf is $\varepsilon$ and internal resistance is $r$?

    $V=\varepsilon-Ir$

  8. 8

    For two cells connected in series aiding one another, what are the equivalent emf and internal resistance?

    $\varepsilon_{eq}=\varepsilon_1+\varepsilon_2,\quad r_{eq}=r_1+r_2$

  9. 9

    State Kirchhoff’s junction rule.

    At a junction, total current entering equals total current leaving

  10. 10

    What is the balance condition for a Wheatstone bridge when the galvanometer current is zero?

    $\frac{R_1}{R_2}=\frac{R_3}{R_4}$

Quiz

5 questions, each with the reason behind its answer.

1 of 5

A uniform wire has resistance $R$. It is stretched until its length is twice its original length, with its volume unchanged and resistivity constant. What is its new resistance?

Answer key
  1. 1

    A uniform wire has resistance $R$. It is stretched until its length is twice its original length, with its volume unchanged and resistivity constant. What is its new resistance?

    Answer: C. 4R4R

    With volume fixed, doubling the length halves the cross-sectional area. Since resistance is proportional to length divided by area, the new resistance is $\frac{2}{1/2}R=4R$.

  2. 2

    A metallic wire carries a steady current. If the current is doubled while the material, cross-sectional area, and charge-carrier density remain unchanged, what happens to the magnitude of the electrons’ drift velocity?

    Answer: B. It is doubled.

    For a uniform conductor, the current magnitude is $I=neAv_d$. With $n$, $e$, and $A$ unchanged, doubling $I$ doubles the drift speed.

  3. 3

    A device gives different current magnitudes for equal-magnitude voltages applied in opposite directions. Which conclusion best follows?

    Answer: D. The device does not obey Ohm’s law in that range.

    For an ohmic device, reversing the voltage should reverse the current without changing its magnitude. The described response depends on voltage polarity, so the device does not obey Ohm’s law in that range.

  4. 4

    A cell of emf $12,\mathrm{V}$ and internal resistance $1,\Omega$ is connected to an external resistor of $5,\Omega$. What is the potential difference across the external resistor?

    Answer: A. $10,\mathrm{V}$

    The circuit current is $I=\frac{12}{5+1}=2,\mathrm{A}$. The voltage across the external resistor is $IR=2\times5=10,\mathrm{V}$ (equivalently, the emf less the internal drop).

  5. 5

    A Wheatstone bridge is balanced when its arm resistances satisfy $\frac{R_2}{R_1}=\frac{R_4}{R_3}$. If $R_1=2,\Omega$, $R_2=4,\Omega$, and $R_3=3,\Omega$, what must $R_4$ be?

    Answer: C. $6,\Omega$

    Substituting the known values gives $\frac{4}{2}=\frac{R_4}{3}$. Thus $R_4=6,\Omega$, which makes the bridge balanced.

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Questions about Current Electricity

What are the most important topics in Class 12 Current Electricity?

Focus on drift velocity and current density, Ohm's law with resistance and resistivity, and how resistivity changes with temperature. Then practise power, emf versus terminal voltage, cells in series and parallel, Kirchhoff's rules and the Wheatstone bridge.

How do I apply Kirchhoff's rules to a circuit?

Mark a trial current direction in every branch, then write the junction rule so that current in equals current out. Pick independent closed loops and add the signed potential changes, taking a drop of $IR$ along the current and a rise against it. Solve the equations together; a negative answer only means the real current flows opposite to your arrow.

What is the difference between resistance and resistivity?

Resistance belongs to a particular conductor and depends on its material, length, cross-section and temperature. Resistivity belongs to the material alone and is linked to resistance by $R=\rho \ell/A$. Two wires of the same metal can have different resistances but the same resistivity.

Why is the terminal voltage of a cell less than its emf?

When a cell supplies current, part of its emf is used up across its own internal resistance, so the terminal voltage is $V=\varepsilon-Ir$. Only when no current flows does the terminal voltage equal the emf. You can also make your own flashcards or quiz for any chapter in Classly.