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Class 12 Physics Notes

Current Electricity Class 12 Physics Notes

Complete, exam-ready notes on current electricity: electric current and drift velocity, Ohm's law and resistivity, Kirchhoff's rules, Wheatstone bridge, EMF, internal resistance and the potentiometer — written for CBSE boards, JEE and NEET revision.

Class12SubjectPhysicsCoversCBSE · JEE · NEET

Written byDeep Narayan· Science & Mathematics EducatorReviewed byPushpanjali

What is electric current in one line?

Electric current is the rate of flow of electric charge through a conductor, measured in amperes (A), defined as I = q/t for steady current or I = dq/dt in general.

Electric Current and Drift Velocity

Electric current

I=dqdtI = \frac{dq}{dt}

The rate of flow of charge. Conventional current direction is opposite to the electron flow direction. The SI unit is the ampere (1 A = 1 C/s).

Drift velocity

vd=eEmτv_d = \frac{eE}{m}\,\tau

The average velocity acquired by electrons in a conductor under an applied electric field. τ\tau is the average relaxation time between collisions. Relating it to current: I=nAevdI = nAev_d, where nn is electron density, AA the cross-section.

I=neAvd,vd=IneAI = n\,e\,A\,v_d,\quad v_d = \frac{I}{n e A}
Current in terms of drift velocity

Ohm's Law and Resistance

Ohm's law

V=IRV = IR

At constant temperature, the current through a conductor is directly proportional to the potential difference across it, where R is the resistance in ohms (Ω). Not all materials obey it — semiconductors and electrolytes do not.

  • Resistivity:\text{Resistivity:}
  • ρ=RAL\rho = \frac{RA}{L}
  • — a material property independent of dimensions.
  • Conductivity:\text{Conductivity:}
  • σ=1ρ\sigma = \frac{1}{\rho}
  • .
  • Resistance of metals increases with temperature; that of semiconductors decreases.
  • Resistors in series:
  • R=R1+R2+R = R_1 + R_2 + \dots
  • .
  • Resistors in parallel:
  • 1R=1R1+1R2+\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \dots
  • .
R=ρLA,σ=1ρR = \rho\,\frac{L}{A},\quad \sigma = \frac{1}{\rho}
Resistance and conductivity

Kirchhoff's Laws

Kirchhoff's junction rule

The algebraic sum of currents meeting at a junction is zero. Equivalent to conservation of charge: current entering a junction equals current leaving it.

Kirchhoff's loop rule

The algebraic sum of the potential differences around any closed loop is zero. Equivalent to conservation of energy.

Sign convention

Traverse a loop in a chosen direction: a rise in potential is positive, a drop is negative. A battery gives +EMF if traversed from negative to positive terminal.

Wheatstone Bridge and Meter Bridge

Wheatstone bridge condition

PQ=RS\frac{P}{Q} = \frac{R}{S}

When the four resistances P, Q, R, S are arranged in a bridge and the galvanometer shows zero deflection (balanced), the ratio condition holds — no current flows through the galvanometer.

Meter bridge

A meter bridge is a practical Wheatstone bridge. In balance: RS=l100l\frac{R}{S} = \frac{l}{100-l}, where ll is the balancing length in cm. The closer the null point to the middle, the more accurate the measurement.

EMF, Internal Resistance and Cells

EMF and terminal voltage

V=εIrV = \varepsilon - Ir

The terminal voltage of a cell is the EMF minus the drop across its internal resistance rr. II is the current drawn. For a battery of cells, connect in series to increase EMF and in parallel to reduce internal resistance.

V=εIr,ε=I(R+r)V = \varepsilon - Ir,\quad \varepsilon = I(R + r)
Cell equations

Potentiometer

A potentiometer measures EMF without drawing current from the cell, giving a more accurate result than a voltmeter. It compares EMFs because the potential gradient along a wire is uniform.

ε1ε2=l1l2\frac{\varepsilon_1}{\varepsilon_2} = \frac{l_1}{l_2}
Comparing EMFs with a potentiometer

Why not a voltmeter?

A voltmeter draws some current and measures the terminal voltage, not the true EMF. The potentiometer draws no current at balance, so it measures the true EMF.

Solved Examples

Example: A cell of EMF 2V2\,\text{V} and internal resistance 0.5Ω0.5\,\Omega is connected to a 4.5Ω4.5\,\Omega resistor. Find the current and the terminal voltage.

Solution: I=εR+r=24.5+0.5=0.4AI = \frac{\varepsilon}{R+r} = \frac{2}{4.5+0.5} = 0.4\,\text{A}. Terminal voltage V=εIr=20.4×0.5=1.8VV = \varepsilon - Ir = 2 - 0.4\times0.5 = 1.8\,\text{V}.

Revision

Key formulas at a glance

Memorise these before attempting numericals — most exam questions hinge on one of them.

Current

I=dqdtI = \frac{dq}{dt}

Drift velocity

vd=IneAv_d = \frac{I}{neA}

Ohm's law

V=IRV = IR

Resistance

R=ρLAR = \rho\frac{L}{A}

Wheatstone balance

PQ=RS\frac{P}{Q} = \frac{R}{S}

Meter bridge

RS=l100l\frac{R}{S} = \frac{l}{100-l}

Terminal voltage

V=εIrV = \varepsilon - Ir

Potentiometer

ε1ε2=l1l2\frac{\varepsilon_1}{\varepsilon_2} = \frac{l_1}{l_2}

Exam tips

How this chapter is asked

Where this topic appears in CBSE, JEE Main and NEET papers.

  • I = n e A v_d is the key link between current and drift velocity.
  • Metals: resistance rises with temperature; semiconductors: falls.
  • Wheatstone balance: P/Q = R/S.
  • Terminal voltage V = ε − Ir.
  • Potentiometer measures true EMF (draws no current).

FAQ

Common questions

What is drift velocity?

It is the small average velocity acquired by free electrons under an applied electric field, given by v_d = eEτ/m. Typical values are only a few millimetres per second.

Why is a potentiometer more accurate than a voltmeter?

At balance, a potentiometer draws no current from the cell, so it measures the true EMF. A voltmeter draws a small current and measures only the terminal voltage.

What is the Wheatstone bridge condition?

The bridge balances when P/Q = R/S, at which point no current flows through the galvanometer connecting the two ratio arms.

Does Ohm's law apply to all materials?

No. Semiconductors, electrolytes, diodes and certain substances show non-linear current-voltage behaviour and do not obey Ohm's law.

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