Current electricity for NEET

Current electricity is a chapter where practice pays more than reading. The theory fits on two pages, but NEET questions mix it into circuits: find the current in one branch, the reading of a meter, the power in a bulb. The skill being tested is simplifying a circuit quickly and applying Kirchhoff's rules without sign errors.

It also has a few pure concept questions: why drift velocity is so small, how resistance changes when a wire is stretched, and which bulb glows brighter in series.

Current and drift velocity

Current is the rate of flow of charge: I = dQ/dt. In a metal, free electrons move randomly at high speeds, but with no field their average velocity is zero. A field gives them a small average velocity against the field, the drift velocity:

• v_d = eEτ/m, where τ is the average time between collisions (relaxation time).
• I = neAv_d, where n is the number of free electrons per unit volume and A the cross-section.
• Mobility μ = v_d/E, the drift speed per unit field.
• Current density J = I/A = σE, where σ = 1/ρ is the conductivity.

Drift speeds are tiny, of the order of a fraction of a millimetre per second. A bulb still lights instantly because the field is set up along the whole wire almost at once, and electrons everywhere start drifting together.

Ohm's law and resistivity

V = IR, with R = ρL/A. Resistivity ρ depends on the material and temperature, not on the shape. From the drift model, ρ = m/(ne²τ).

Ohm's law is not universal. Devices such as diodes have non-linear V–I graphs, and some materials conduct differently in the two directions.

Temperature dependence: ρ_T = ρ₀[1 + α(T − T₀)].

Stretching a wire. The volume stays constant, so if L becomes nL, A becomes A/n and R becomes n²R. If you are told the radius instead, halving the radius makes A a quarter, so L becomes 4 times and R becomes 16 times.

Combining resistors

• Series: R = R₁ + R₂ + … (same current).
• Parallel: 1/R = 1/R₁ + 1/R₂ + … (same voltage). The result is smaller than the smallest resistor.
• Two resistors in parallel: R = R₁R₂/(R₁ + R₂). n equal resistors R in parallel: R/n.

Cells, EMF and internal resistance

The EMF E of a cell is the potential difference across it when no current flows. A real cell has internal resistance r.

• Current: I = E/(R + r).
• Terminal voltage while supplying current: V = E − Ir. While being charged: V = E + Ir.
• Short-circuit current (R = 0): E/r.
• n identical cells in series: I = nE/(R + nr).
• m identical cells in parallel: I = E/(R + r/m).
• Two different cells in parallel: E_eq = (E₁r₂ + E₂r₁)/(r₁ + r₂), and r_eq = r₁r₂/(r₁ + r₂).

Kirchhoff's laws

1. Junction rule (conservation of charge): the total current into a junction equals the total current out.
2. Loop rule (conservation of energy): the sum of potential changes around any closed loop is zero.

For the loop rule, crossing a resistor with the assumed current is a drop of IR. Crossing a cell from − to + is a rise of E. If a current comes out negative, it simply flows the other way; do not redo the problem.

Wheatstone bridge and metre bridge

A Wheatstone bridge has four resistors P, Q, R, S with a galvanometer between the midpoints. It is balanced when

P/Q = R/S

and then no current flows through the galvanometer. You can remove that branch, or the resistor in its place, when simplifying a circuit.

The metre bridge (an experimental-skills item in the syllabus) is a Wheatstone bridge in which a 100 cm uniform wire forms two arms. With the unknown R in the left gap and a known S in the right gap, balance at length l from the left end gives

R/S = l/(100 − l)

Once R is known, the resistivity follows from ρ = RA/L for the wire under test.

Electrical power and energy

P = VI = I²R = V²/R. Energy = Pt. The commercial unit is the kilowatt-hour: 1 kWh = 3.6 × 10⁶ J.

For bulbs rated at the same voltage, R = V²/P, so a lower-wattage bulb has higher resistance.

• In series, the current is the same and P = I²R, so the lower-wattage bulb glows brighter.
• In parallel, the voltage is the same and P = V²/R, so the higher-wattage bulb glows brighter.

Worked numericals

Example 1: drift velocity

A copper wire of cross-section 1.0 mm² carries 1.36 A. Take n = 8.5 × 10²⁸ m⁻³ and e = 1.6 × 10⁻¹⁹ C.

• neA = 8.5 × 10²⁸ × 1.6 × 10⁻¹⁹ × 1.0 × 10⁻⁶ = 1.36 × 10⁴.
• v_d = I/(neA) = 1.36/(1.36 × 10⁴) = 1.0 × 10⁻⁴ m s⁻¹, or 0.1 mm s⁻¹.

Example 2: a cell with internal resistance

A cell of EMF 12 V and internal resistance 1 Ω is connected to a 5 Ω resistor.

• I = 12/(5 + 1) = 2 A.
• Terminal voltage = 12 − 2 × 1 = 10 V.
• Power in the external resistor = I²R = 4 × 5 = 20 W; power wasted inside the cell = 4 × 1 = 4 W.

Example 3: two cells in parallel, using Kirchhoff

Cells of 6 V and 4 V, each with internal resistance 1 Ω, are connected in parallel (positive to positive) across a 2 Ω resistor. Find the current in each cell.

• Let V be the voltage across the 2 Ω resistor. The currents out of the cells are (6 − V)/1 and (4 − V)/1.
• Junction rule: (6 − V) + (4 − V) = V/2, so 10 = 2.5V and V = 4 V.
• Current in the resistor: 4/2 = 2 A. From the 6 V cell: 2 A. From the 4 V cell: zero.
• Check with the formula: E_eq = (6 + 4)/2 = 5 V, r_eq = 0.5 Ω, I = 5/2.5 = 2 A ✓.

Example 4: metre bridge

A metre bridge has an unknown resistance in the left gap and 10 Ω in the right gap. The balance point is at 60 cm.

• R/10 = 60/40, so R = 15 Ω.

Example 5: two bulbs in series

A 100 W and a 60 W bulb, both rated at 220 V, are connected in series across 220 V. Which glows brighter?

• R(100 W) = 220²/100 = 484 Ω. R(60 W) = 48,400/60 ≈ 807 Ω.
• Same current in both, so P = I²R is larger in the 60 W bulb, which glows brighter.

Practice MCQs

1. A wire is stretched to twice its length at constant volume. Its resistance becomes: (a) R/2 (b) 2R (c) 4R (d) 8R
2. Three 4 Ω resistors in parallel give: (a) 12 Ω (b) 4 Ω (c) 4/3 Ω (d) 3/4 Ω
3. A 100 W, 220 V bulb is run on 110 V. Its power is: (a) 12.5 W (b) 25 W (c) 50 W (d) 100 W
4. A 10 Ω wire is cut into 5 equal pieces, which are joined in parallel. The resistance is: (a) 0.4 Ω (b) 0.5 Ω (c) 2 Ω (d) 50 Ω
5. A cell of EMF 2 V and internal resistance 0.5 Ω is short-circuited. The current is: (a) 1 A (b) 2 A (c) 8 A (d) 4 A
6. A resistor is 10 Ω at 20 °C and has α = 0.004 °C⁻¹. Its resistance at 70 °C is: (a) 10.2 Ω (b) 12 Ω (c) 14 Ω (d) 20 Ω
7. In a metre bridge with 7 Ω in the right gap, the balance point is 30 cm from the left end. The resistance in the left gap is: (a) 3 Ω (b) 7 Ω (c) 16.3 Ω (d) 2.1 Ω
8. The same current flows through a wire whose diameter is then doubled. The drift velocity becomes: (a) 2 times (b) half (c) a quarter (d) unchanged

Answers

1. (c) R ∝ L² at constant volume.
2. (c) 4/3 Ω.
3. (b) R is fixed, P ∝ V², so power falls to a quarter.
4. (a) Each piece is 2 Ω; five in parallel give 2/5 = 0.4 Ω.
5. (d) I = E/r = 2/0.5 = 4 A.
6. (b) 10 × (1 + 0.004 × 50) = 12 Ω.
7. (a) R = 7 × 30/70 = 3 Ω.
8. (c) v_d = I/(neA), and doubling the diameter makes A four times larger.

What to do next

• Draw ten circuits from your book and reduce each to one resistor, saying out loud whether each step is series, parallel or a balanced bridge.
• Solve five two-loop Kirchhoff problems, writing the sign of every term before adding.
• Revise the metre bridge method as both a numerical and an experimental-skills question.
• Revise potential and capacitors in electrostatics for NEET, since mixed RC-circuit questions draw on both.

Next in the syllabus: moving charges and magnetism.