Is Current Electricity part of the EAPCET Physics syllabus?

Yes. Current Electricity is taught in Intermediate second year Physics, and the EAPCET Engineering syllabus is based on the Intermediate first- and second-year syllabus. Check the official notification each year for any changes.

How many Physics questions are there in EAPCET?

40 of the 160 questions in the Engineering paper are Physics, spread across 30 chapters.

What mistakes do students make in Current Electricity?

Assuming resistance doubles when a wire is stretched to twice its length — the volume stays fixed, so the area halves and R becomes four times larger. Treating resistivity as dependent on the length or thickness of a wire — it depends only on the material and temperature. Taking the terminal voltage equal to the emf while current flows — subtract the drop Ir across the internal resistance. Mixing up signs in Kirchhoff’s loop rule — choose a direction round the loop and keep the sign of each IR and emf consistent with it.

EAPCET Physics · Intermediate second year

Current Electricity

Current Electricity covers current, resistance, Ohm’s law, cells, Kirchhoff’s laws and measuring instruments such as the Wheatstone bridge and potentiometer.

Subject: Physics
Taught in: Intermediate second year
Physics in the paper: 40 of 160 questions

Key formulas and results

$V = IR$ and $R = \frac{\rho l}{A}$
$I = neAv_{d}$
Series: $R = \sum R_{i}$ ; parallel: $\frac{1}{R} = \sum \frac{1}{R_{i}}$
Terminal voltage $V = E - Ir$ ; maximum power transfer when $R = r$
Balanced Wheatstone bridge: $\frac{P}{Q} = \frac{R}{S}$
Power $P = VI = I^{2}R = \frac{V^{2}}{R}$
Current density: $J = I/A = nev_{d} = \sigma E$
Drift velocity $v_{d} = \frac{eE\tau}{m}$ ; mobility $\mu = v_{d}/E$ ; conductivity $\sigma = \frac{1}{\rho} = \frac{ne^{2}\tau}{m} = ne\mu$
Resistance of a truncated cone of length $l$ with end radii $a$ and $b$ : $R = \frac{\rho l}{\pi ab}$
Terminal voltage of a cell being charged (current entering the positive terminal): $V = E + Ir$
Cells in series (some reversed) with external $R$ : $I = \dfrac{\sum \text{signed emfs}}{R + \sum r}$
Cells in parallel feeding an external $R$ : $I = \dfrac{\sum E_{i}/r_{i}}{1 + R\sum 1/r_{i}}$
Shunt for an ammeter of range $I$ : $S = \frac{I_{g}G}{I - I_{g}}$ ; series resistance for a voltmeter of range $V$ : $R = \frac{V}{I_{g}} - G$
Heat from a resistor used to warm a body: $Pt = ms\,\Delta T$

What this chapter covers

Current and drift velocity
Ohm’s law and resistivity
Combinations of resistors
Cells, emf and internal resistance
Kirchhoff’s laws
Wheatstone bridge, meter bridge and potentiometer

How it is usually asked

Equivalent resistance of networks
Kirchhoff’s law circuits
Meter bridge and potentiometer
Power dissipated

Understanding Current Electricity

Electric current is the rate of flow of charge. In a metal, free electrons drift slowly against the field, and the current I = neAv_d depends on their number density and drift speed. Ohm’s law, V = IR, holds for metallic conductors at constant temperature. Resistance R = ρl/A depends on the length and area, while resistivity ρ is a property of the material that changes with temperature.

A real cell has an emf E and an internal resistance r, so its terminal voltage V = E − Ir drops as more current is drawn. For complicated networks, Kirchhoff’s junction rule (charge conservation) and loop rule (energy conservation) give enough equations to solve for every current. A balanced Wheatstone bridge, the meter bridge and the potentiometer all compare resistances or emfs using a null point.

EAPCET usually asks for equivalent resistance, terminal voltage, Kirchhoff networks, bridge balance and power dissipated. Look for symmetry and balanced bridges before writing equations: if a bridge is balanced, the middle resistor carries no current and can be removed. For power, use P = V²/R when the voltage is common (bulbs in parallel) and P = I²R when the current is common (bulbs in series).

Worked example

EAPCET-style question

A cell of emf 2 V and internal resistance 0.5 Ω is connected across an external resistor of 3.5 Ω. The terminal voltage of the cell is

(1)1.75 V
(2)2 V
(3)0.25 V
(4)1.5 V

Solution

Current $I = \frac{E}{R + r} = \frac{2}{3.5 + 0.5} = 0.5\ \text{A}$ .
Voltage lost inside the cell $= Ir = 0.5 \times 0.5 = 0.25\ \text{V}$ .
Terminal voltage $V = E - Ir = 2 - 0.25 = 1.75\ \text{V}$ .
Check: $V = IR = 0.5 \times 3.5 = 1.75\ \text{V}$ .

Answer: 1.75 V

Shortcuts and results to remember

In a wire of non-uniform cross-section the current is the same everywhere, but $J$ , $v_{d}$ and $E$ are all larger where the area is smaller.
$n$ identical cells in series with $m$ of them reversed give a net emf $(n - 2m)E$ , while the internal resistance stays $nr$ .
For a fixed potential difference across a wire, drift velocity $v_{d} = \mu V/l$ does not depend on the wire’s diameter.
Resistivity depends only on the material and temperature, not on the shape or size of the body.
In series the current is the same and voltages differ; in parallel the voltage is the same and currents differ; points joined by a plain wire share one potential.

Common mistakes to avoid

Assuming resistance doubles when a wire is stretched to twice its length — the volume stays fixed, so the area halves and R becomes four times larger.
Treating resistivity as dependent on the length or thickness of a wire — it depends only on the material and temperature.
Taking the terminal voltage equal to the emf while current flows — subtract the drop Ir across the internal resistance.
Mixing up signs in Kirchhoff’s loop rule — choose a direction round the loop and keep the sign of each IR and emf consistent with it.

How to prepare it

Stretching a wire to n times its length multiplies its resistance by n² — a frequent one-line question.

Once the formulas feel familiar, test them the way the exam does: practise Current Electricity questions chapter-wise, then attempt a full previous year paper to see whether it holds up under a 3-hour timer.

Related Physics chapters

See every chapter in the EAPCET Physics syllabus.

Practise Current Electricity chapter-wise

Chapter-wise quizzes show exactly which topics in this chapter you still get wrong — before a full paper does.

Frequently asked questions

Is Current Electricity part of the EAPCET Physics syllabus?: Yes. Current Electricity is taught in Intermediate second year Physics, and the EAPCET Engineering syllabus is based on the Intermediate first- and second-year syllabus. Check the official notification each year for any changes.
How should I prepare Current Electricity for EAPCET?: Stretching a wire to n times its length multiplies its resistance by n² — a frequent one-line question.
How many Physics questions are there in EAPCET?: 40 of the 160 questions in the Engineering paper are Physics, spread across 30 chapters.
What mistakes do students make in Current Electricity?: Assuming resistance doubles when a wire is stretched to twice its length — the volume stays fixed, so the area halves and R becomes four times larger. Treating resistivity as dependent on the length or thickness of a wire — it depends only on the material and temperature. Taking the terminal voltage equal to the emf while current flows — subtract the drop Ir across the internal resistance. Mixing up signs in Kirchhoff’s loop rule — choose a direction round the loop and keep the sign of each IR and emf consistent with it.