NEET UG · Physics · chapter
Current Electricity for NEET: circuits, bridges and the marks in between
Current Electricity questions in NEET come to about three to five of the forty-five Physics questions, and nearly all of them are procedural: reduce the network, apply a rule, substitute. This guide gives the formulas worth memorising, the bridge and potentiometer conditions, the five mistakes that cost the most marks, a free quiz and worked examples.
Quick answer
- Current Electricity carries about 3–5 of the 45 NEET Physics questions in recent papers — our estimate from past papers, not an official figure.
- The repeat areas are series and parallel reduction, cells with internal resistance, Kirchhoff’s rules, the Wheatstone bridge, the metre bridge and the potentiometer.
- It is a Class 12 chapter, but it assumes you are fluent with algebra and ratios — nothing here needs calculus.
- The commonest loss is reducing a network by habit without checking which points are at the same potential.
Current Electricity at a glance
Questions per paper
3–5
Of 45 Physics questions · BrainStride analysis of past papers
Question style
Circuit reduction
Numerical, one or two steps each
Class
12
Builds on electrostatics and basic algebra
Marks at stake
Up to 20
At +4 each, before negative marking
Questions per year
- 20194 Q
- 20203 Q
- 20214 Q
- 20225 Q
- 20234 Q
- 20244 Q
- 20255 Q
- 20264 Q
BrainStride analysis of past papers
Key concepts
Current, drift velocity and resistivity
Current as charge per unit time, and the microscopic picture behind it: electrons drifting slowly against a lattice. Resistivity depends on the material and its temperature, never on the length or the cross-section — that distinction is worth marks on its own.
Ohm's law and combinations
Series adds resistances, parallel adds conductances. Most NEET questions are a reduction problem with a one-line calculation at the end, so the marks go to whoever reduces the network correctly.
Cells, emf and internal resistance
Terminal voltage falls below the emf as soon as current flows. Cells in series add emfs, and identical cells in parallel keep the emf but divide the internal resistance.
Kirchhoff's rules
The junction rule is conservation of charge; the loop rule is conservation of energy. Fix a sign convention before writing any loop equation and keep it for the whole problem.
Wheatstone bridge and metre bridge
At balance no current flows through the galvanometer and the ratio condition holds. The metre bridge is the same condition written with lengths, which is why a balance point at 40 cm tells you a resistance without measuring any current.
Potentiometer
The one instrument that measures emf rather than terminal voltage, because it draws no current from the cell at balance. Questions test that idea more often than the arithmetic around it.
Power, heating and meters
Power as VI, I²R or V²/R — choose the form whose quantities you already know. Converting a galvanometer into an ammeter needs a shunt in parallel; into a voltmeter, a high resistance in series.
Formulas to know cold
Ohm's law
V = IR
Valid for ohmic conductors at constant temperature.
Resistance of a conductor
R = ρL/A
Stretching a wire keeps its volume constant, so R ∝ L² — a favourite trick.
Drift velocity
I = nAevᵈ · vᵈ = eEτ/m
Drift speeds are of the order of 10⁻⁴ m/s, far slower than the signal itself.
Current density
J = I/A = σE
The vector form of Ohm’s law, with σ the conductivity.
Combinations
Series: R = R₁ + R₂ + … · Parallel: 1/R = 1/R₁ + 1/R₂ + …
n identical resistors R in parallel give R/n.
Terminal voltage
V = ε − Ir
Equal to the emf only when no current is drawn, which is why a potentiometer measures emf.
Power
P = VI = I²R = V²/R
In series the largest resistance dissipates most power; in parallel, the smallest.
Wheatstone bridge
P/Q = R/S at balance
At balance the galvanometer carries no current, so it can be ignored while reducing the rest.
Metre bridge
R/S = l/(100 − l)
l is the balancing length in centimetres measured from the left end.
Potentiometer comparison
ε₁/ε₂ = l₁/l₂
Only valid while the driver cell’s emf exceeds both cells being compared.
Temperature dependence
ρ = ρ₀[1 + α(T − T₀)]
Positive α for metals; semiconductors behave the other way.
Galvanometer conversion
Ammeter: S = IᶢG/(I − Iᶢ) · Voltmeter: R = V/Iᶢ − G
A shunt in parallel for an ammeter, a high resistance in series for a voltmeter.
Where marks leak in this chapter
Networks reduced by habit
Why it happens
Resistors that look adjacent in the drawing are assumed to be in series or parallel without checking the nodes.The fix
Label every node before reducing. Two resistors are in parallel only when both ends sit at the same pair of nodes; redraw the circuit with the nodes spread out if the figure hides it.Internal resistance dropped
Why it happens
The cell is treated as an ideal source once the external network looks complicated.The fix
Write the total resistance as external plus internal before finding the current, then get the terminal voltage from V = ε − Ir.Resistance and resistivity confused
Why it happens
Both are called "resistance" in speech, so a question about stretching a wire gets answered as though ρ changed.The fix
Remember what each depends on: ρ on material and temperature, R also on length and area. In a stretching problem, volume stays constant and R ∝ L².Sign convention abandoned mid-loop
Why it happens
Kirchhoff's loop equations are written while reading the figure, so the sign flips when the direction of travel changes.The fix
Mark the assumed current direction and the traversal direction on the figure first, then apply the same rule to every element. A negative answer simply means the current runs the other way.Potentiometer balance misunderstood
Why it happens
It is revised as another way to measure voltage, so the no-current condition gets forgotten.The fix
State the condition every time: at balance the cell drives no current, so the reading is the emf, and the driver cell must have the larger emf for a balance point to exist at all.Check yourself
6 questions
Question 1
A wire of resistance R is stretched to twice its original length, with its volume unchanged. Its new resistance is:
Question 2
Three resistors of 6 Ω each are connected in parallel. The equivalent resistance is:
Question 3
A cell of emf 2 V and internal resistance 0.5 Ω is connected across a 3.5 Ω resistor. The current in the circuit is:
Question 4
At the balance point of a Wheatstone bridge:
Question 5
If the same current flows through a conductor whose cross-sectional area is doubled, the drift velocity of the electrons:
Question 6
The resistivity of a metallic conductor depends on:
Solved examples
Single-step numerical · metre bridge
In a metre bridge, the balance point is found at 40 cm from the left end when a 6 Ω resistance is connected in the right gap. What is the unknown resistance in the left gap?
- Write the balance condition: R/S = l/(100 − l), with l measured from the left end.
- Substitute l = 40 cm and S = 6 Ω: R/6 = 40/60.
- Simplify: R = 6 × (2/3).
Answer: The unknown resistance is 4 Ω.
Two-step numerical · cell with internal resistance
A cell of emf 2 V and internal resistance 0.5 Ω is connected to an external resistance of 3.5 Ω. Find the current, the terminal voltage and the power delivered to the external resistor.
- Total resistance = R + r = 3.5 + 0.5 = 4 Ω, so I = ε/(R + r) = 2/4 = 0.5 A.
- Terminal voltage V = ε − Ir = 2 − (0.5 × 0.5) = 1.75 V.
- Power in the external resistor P = I²R = (0.5)² × 3.5 = 0.875 W.
Answer: The current is 0.5 A, the terminal voltage 1.75 V and the external power 0.875 W.
Single-step numerical · drift velocity
A copper wire of cross-sectional area 1 × 10⁻⁶ m² carries a current of 1.6 A. If the free electron density is 8 × 10²⁸ m⁻³, find the drift velocity. Take e = 1.6 × 10⁻¹⁹ C.
- Start from I = nAevᵈ, so vᵈ = I/(nAe).
- Compute the denominator: nAe = (8 × 10²⁸)(1 × 10⁻⁶)(1.6 × 10⁻¹⁹) = 1.28 × 10⁴.
- Divide: vᵈ = 1.6 / (1.28 × 10⁴).
Answer: The drift velocity is 1.25 × 10⁻⁴ m/s — a fraction of a millimetre per second, which is why the current starts long before any electron crosses the wire.
Current Electricity: common questions
How many questions come from Current Electricity in NEET?
In recent NEET papers, Current Electricity has carried about 3 to 5 of the 45 Physics questions. That is BrainStride’s analysis of past papers rather than an official figure, since NTA publishes no chapter-wise weightage. It is consistently one of the heavier Physics chapters, alongside electrostatics and modern physics.
Is Current Electricity a scoring chapter for NEET?
Yes, more than most of Physics. The questions are procedural rather than conceptual traps: reduce the network, apply Ohm’s law or Kirchhoff’s rules, substitute. Students who practise circuit reduction deliberately tend to convert almost every question here, which is not true of mechanics.
What is the difference between emf and terminal voltage?
Emf is the potential difference across a cell when it drives no current; terminal voltage is what you measure while current flows, given by V = ε − Ir. The two are equal only in an open circuit, which is exactly why a potentiometer, which draws no current at balance, measures emf directly.
How do I get better at circuit reduction?
Label the nodes before touching the resistors. Two resistors are in parallel only if both their ends meet at the same two nodes, which is often hidden by how the figure is drawn. Redrawing the circuit with the nodes spread apart takes twenty seconds and removes the most common error in the chapter.
Which formulas from Current Electricity are asked most often?
Ohm’s law with series and parallel combinations, R = ρL/A with the stretched-wire result R ∝ L², terminal voltage V = ε − Ir, the three power forms, the Wheatstone balance condition and the metre bridge length ratio. Drift velocity appears as a single-step substitution.
Do I need Class 11 Physics for this chapter?
Only lightly. Current Electricity assumes comfort with electrostatics ideas such as potential difference, and with ratios and simultaneous equations. It needs no calculus, which makes it one of the first Class 12 chapters a dropper can rebuild confidence on.
Circuit questions reward the student who labels nodes, not the one who recognises shapes.
The 20-minute diagnostic shows whether your Physics marks leak here or in mechanics, then writes the practice for it. Free, no card.