1. Chapter at a Glance
Current electricity deals with charges in motion. The chapter explains electric current, drift of electrons, Ohm's law, resistivity and its temperature dependence, cells with internal resistance, combinations of resistors and cells, Kirchhoff's rules, the Wheatstone bridge and the metre bridge. It is one of the highest-scoring chapters for NEET: two or three questions come from it most years.
- Drift velocity and current density.
- Change of resistance when a wire is stretched or its temperature changes.
- Equivalent resistance of networks; current and voltage division.
- EMF, internal resistance, terminal voltage; cells in series and parallel.
- Kirchhoff's rules, Wheatstone bridge and metre bridge; electric power.
2. Current, Drift Velocity and Ohm's Law
- The direction of conventional current is the direction of flow of positive charge, opposite to electron flow.
- In a metal, free electrons move randomly at high speed (~10⁵ m s⁻¹), but in a field they gain a small average drift velocity (~10⁻⁴ m s⁻¹) opposite to E.
- vd = eEτ/m, where τ is the average time between collisions (relaxation time). Mobility μ = vd/E = eτ/m.
- A bulb glows instantly because the electric field travels through the wire at nearly the speed of light, not because electrons move fast.
- Resistance depends on length, area, material and temperature; resistivity depends only on material and temperature.
- Ohm's law fails for non-ohmic devices such as diodes and electrolytes.
3. Resistivity and Temperature
| Material | Behaviour with temperature | NEET Note |
|---|---|---|
| Metals (Cu, Al) | ρ increases (α positive) | More collisions, τ falls |
| Alloys (nichrome, manganin, constantan) | High ρ, very small α | Used in standard resistors and heaters |
| Semiconductors (Si, Ge) | ρ decreases (α negative) | More charge carriers, n rises |
When a wire is stretched to n times its length, its volume stays the same, so the area falls to A/n. Then R ∝ l/A becomes n²R. If the radius is reduced to 1/n (by drawing it thinner), R becomes n⁴R.
4. Combination of Resistors
| Property | Series | Parallel |
|---|---|---|
| Equivalent resistance | R = R₁ + R₂ + … | 1/R = 1/R₁ + 1/R₂ + … |
| Same through each | Current | Voltage |
| Division rule | V ∝ R | I ∝ 1/R |
| NEET Note | Req > largest R | Req < smallest R |
Two resistors in parallel: R = R₁R₂/(R₁ + R₂); the current in R₁ is I × R₂/(R₁ + R₂).
5. EMF, Internal Resistance and Cells
- EMF is the potential difference across the cell terminals when no current is drawn.
- Internal resistance r causes a loss of voltage Ir inside the cell.
- Series (n cells): Eeq = nE, req = nr.
- Parallel (two cells): Eeq = (E₁r₂ + E₂r₁)/(r₁ + r₂), 1/req = 1/r₁ + 1/r₂.
- Maximum power is delivered to the external resistor when R = r; Pmax = E²/4r.
6. Kirchhoff's Rules and Bridges
- Junction rule (conservation of charge): the currents entering a junction equal the currents leaving it.
- Loop rule (conservation of energy): around any closed loop, the algebraic sum of the EMFs equals the sum of the IR drops.
- Wheatstone bridge: balanced when P/Q = R/S; then no current flows through the galvanometer, and that arm can be removed.
- Metre bridge: a 1 m uniform wire forms two arms. At the balance point l (from the left end), R/S = l/(100 − l).
7. Electrical Energy and Power
- A bulb rated P watts at V volts has resistance R = V²/P.
- Bulbs in series: the one with the higher resistance (lower wattage rating) glows brighter. In parallel, the higher-wattage bulb glows brighter.
- Power is transmitted at high voltage and low current to reduce I²R loss in the lines.
8. Formula Master Table
| Quantity | Formula |
|---|---|
| Current | I = Q/t = neAvd |
| Drift velocity | vd = eEτ/m |
| Mobility | μ = vd/E |
| Resistance | R = ρl/A |
| Resistivity | ρ = m/(ne²τ) |
| Temperature dependence | RT = R₀[1 + αΔT] |
| Cell current | I = E/(R + r) |
| Terminal voltage | V = E − Ir |
| Maximum power | Pmax = E²/4r at R = r |
| Wheatstone balance | P/Q = R/S |
| Metre bridge | R/S = l/(100 − l) |
| Power | P = VI = I²R = V²/R |
9. Master Summary Table
| Concept | Key idea | NEET Note |
|---|---|---|
| Drift velocity | ~10⁻⁴ m s⁻¹ | vd ∝ I/A |
| Resistivity | Material property | Not changed by stretching |
| Stretched wire | R → n²R | Volume constant |
| Metals vs semiconductors | α positive vs negative | Classic 1-mark question |
| Internal resistance | V = E − Ir | V = E only when I = 0 |
| Max power transfer | R = r | Efficiency then 50% |
| Kirchhoff's rules | Charge and energy conservation | Use for multi-loop circuits |
| Balanced bridge | No galvanometer current | Middle arm can be removed |
10. Mind Map
- I = neAvd
- Mobility
- Ohm's law
- ρl/A
- Temperature
- Series / parallel
- EMF, r
- V = E − Ir
- Combinations
- Kirchhoff's rules
- Wheatstone
- Metre bridge
11. Tips and Tricks
- n equal resistors: series nR, parallel R/n; ratio of series to parallel = n².
- In a symmetric network, points at the same potential can be joined or separated freely.
- For a metre bridge, when the resistances are interchanged, the balance point moves from l to 100 − l.
- For household bills, units = kWh = (watts × hours)/1000.
- Thinking resistivity changes when a wire is stretched (only resistance changes).
- Using terminal voltage and EMF interchangeably.
- Forgetting that a lower-wattage bulb has a higher resistance.
- Applying Ohm's law to a diode.
12. Important One-liners
- Current is a scalar, even though it has a direction.
- The drift velocity of electrons is of the order of 10⁻⁴ m s⁻¹.
- The resistivity of metals increases with temperature; that of semiconductors decreases.
- Manganin and constantan are used for standard resistors because their α is very small.
- Kirchhoff's junction rule is based on conservation of charge; the loop rule on conservation of energy.
- A metre bridge is most sensitive when the balance point is near the middle of the wire.
- 1 kWh = 3.6 MJ.
- A cell gives maximum power when the external resistance equals its internal resistance.
13. Solved Numericals: Average to Advanced
Five average and five advanced numericals. Take e = 1.6 × 10⁻¹⁹ C. Click to reveal each answer and working.
Average level
A current of 0.5 A flows through a wire for 2 minutes. How much charge passes, and how many electrons is that?
Show answer & solution
Concept: Q = It; n = Q/e.
- t = 2 × 60 = 120 s; Q = 0.5 × 120 = 60 C.
- n = 60/1.6 × 10⁻¹⁹ = 3.75 × 10²⁰.
Answer: 60 C; 3.75 × 10²⁰ electrons
A copper wire of cross-section 2 mm² carries 4 A. The free-electron density of copper is 8.5 × 10²⁸ m⁻³. Find the drift velocity.
Show answer & solution
Concept: vd = I/(neA).
- neA = 8.5 × 10²⁸ × 1.6 × 10⁻¹⁹ × 2 × 10⁻⁶ = 2.72 × 10⁴.
- vd = 4/2.72 × 10⁴ ≈ 1.47 × 10⁻⁴ m s⁻¹.
Answer: about 1.5 × 10⁻⁴ m s⁻¹
Find the resistance of a copper wire 50 m long with cross-section 1 mm² (ρ = 1.7 × 10⁻⁸ Ω m).
Show answer & solution
Concept: R = ρl/A.
- R = 1.7 × 10⁻⁸ × 50 / 10⁻⁶.
- R = 0.85 Ω.
Answer: 0.85 Ω
A 5 Ω wire is stretched uniformly to three times its original length. Find its new resistance. Does its resistivity change?
Show answer & solution
Concept: volume constant ⇒ A becomes A/3; R ∝ l/A.
- R′ = ρ(3l)/(A/3) = 9ρl/A = 9R.
- R′ = 9 × 5 = 45 Ω.
- Resistivity is a property of the material, so it does not change.
Answer: 45 Ω; resistivity unchanged
A 3 Ω and a 6 Ω resistor are in parallel, and this pair is in series with a 4 Ω resistor across 12 V. Find the total current and the current in each parallel resistor.
Show answer & solution
Concept: reduce the network, then divide current inversely with resistance.
- Parallel: 3 × 6/9 = 2 Ω; total R = 2 + 4 = 6 Ω.
- I = 12/6 = 2 A.
- In 3 Ω: 2 × 6/9 = 4/3 A ≈ 1.33 A; in 6 Ω: 2 × 3/9 = 2/3 A ≈ 0.67 A.
Answer: 2 A; 1.33 A (3 Ω) and 0.67 A (6 Ω)
Advanced level
A cell of EMF 9 V and internal resistance 0.5 Ω is connected to a 4 Ω resistor. Find the current, the terminal voltage and the power used in the resistor.
Show answer & solution
Concept: I = E/(R + r); V = E − Ir.
- I = 9/(4 + 0.5) = 2 A.
- V = 9 − 2 × 0.5 = 8 V (also = IR = 2 × 4).
- P = I²R = 4 × 4 = 16 W.
Answer: 2 A; 8 V; 16 W
A metal coil has resistance 40 Ω at 25 °C and 48 Ω at 125 °C. Find its temperature coefficient of resistance.
Show answer & solution
Concept: RT = R₀[1 + α(T − T₀)] with T₀ = 25 °C.
- 48 = 40[1 + α × 100].
- 1 + 100α = 1.2 ⇒ α = 0.2/100.
- α = 2 × 10⁻³ °C⁻¹ (positive, as expected for a metal).
Answer: 2 × 10⁻³ °C⁻¹
(a) In a Wheatstone bridge, P = 4 Ω, Q = 6 Ω and R = 10 Ω. Find S for balance. (b) In a metre bridge, 12 Ω is in the left gap and an unknown X in the right gap; the balance point is 60 cm from the left end. Find X.
Show answer & solution
Concept: P/Q = R/S; metre bridge R/X = l/(100 − l).
- (a) S = RQ/P = 10 × 6/4 = 15 Ω.
- (b) 12/X = 60/40 ⇒ X = 12 × 40/60 = 8 Ω.
Answer: (a) 15 Ω; (b) 8 Ω
Two cells, 8 V with internal resistance 2 Ω and 4 V with internal resistance 2 Ω, are joined in parallel (positive to positive) across a 3 Ω resistor. Find the current through the resistor.
Show answer & solution
Concept: Eeq = (E₁/r₁ + E₂/r₂)/(1/r₁ + 1/r₂); req from parallel rule.
- Eeq = (8/2 + 4/2)/(1/2 + 1/2) = 6/1 = 6 V.
- req = 2 × 2/(2 + 2) = 1 Ω.
- I = 6/(3 + 1) = 1.5 A.
- Check with Kirchhoff: terminal V = 4.5 V; the 8 V cell gives (8 − 4.5)/2 = 1.75 A; the 4 V cell takes (4.5 − 4)/2 = 0.25 A in reverse; 1.75 − 0.25 = 1.5 A ✓.
Answer: 1.5 A
A 10 V battery has an internal resistance of 2 Ω. For what external resistance is the power delivered to it maximum? Find this maximum power and the efficiency at that point.
Show answer & solution
Concept: maximum power transfer when R = r.
- R = r = 2 Ω; I = 10/(2 + 2) = 2.5 A.
- Pmax = I²R = 6.25 × 2 = 12.5 W (= E²/4r = 100/8).
- Total power from the cell = EI = 25 W; efficiency = 12.5/25 = 50%.
Answer: 2 Ω; 12.5 W; 50%
14. NEET-pattern Questions with Detailed Answers
These questions are original and based on concepts that NEET UG asks again and again.
The resistivity of a wire depends on:
- Its length
- Its cross-sectional area
- Its material and temperature
- The current through it
Show solution
Resistivity is a material property that changes with temperature. Length and area change the resistance, not the resistivity. Current has no effect on either.
Answer: (C)
Kirchhoff's junction rule is a consequence of the conservation of:
- Energy
- Charge
- Momentum
- Mass
Show solution
Charge cannot pile up at a junction, so what flows in must flow out. The loop rule, not the junction rule, comes from energy conservation.
Answer: (B)
A wire of resistance R is cut into four equal parts, which are then joined in parallel. The new resistance is:
- R/4
- R/16
- 4R
- R
Show solution
Each part has R/4. Four equal resistors in parallel give (R/4)/4 = R/16. (A) forgets the parallel step.
Answer: (B)
A 60 W and a 100 W bulb (both rated for 220 V) are connected in series to 220 V. Which glows brighter?
- 60 W bulb
- 100 W bulb
- Both equally
- Neither glows
Show solution
R = V²/P, so the 60 W bulb has the larger resistance. In series the current is the same, so P = I²R is larger for the 60 W bulb, and it glows brighter.
Answer: (A)
When the temperature of a semiconductor increases, its resistance:
- Increases
- Decreases
- Stays the same
- First increases, then decreases
Show solution
Heating breaks more covalent bonds and releases more charge carriers. So resistance falls (α is negative). Metals behave the opposite way.
Answer: (B)
A cell of EMF E and internal resistance r gives a terminal voltage V while supplying current I. Then:
- V = E + Ir
- V = E − Ir
- V = E
- V = Ir
Show solution
While discharging, some voltage is lost inside the cell: V = E − Ir. (A) is true only while the cell is being charged, and (C) only when I = 0.
Answer: (B)
The drift velocity of electrons in a wire is v. If the wire is replaced by one of the same material with double the radius, carrying the same current, the drift velocity becomes:
- 2v
- v/2
- v/4
- 4v
Show solution
vd = I/(neA). Doubling the radius makes A four times larger, so vd falls to v/4.
Answer: (C)
In a metre bridge, the balance point is at 25 cm from the left end. If the resistances in the two gaps are interchanged, the new balance point is at:
- 25 cm
- 50 cm
- 75 cm
- 100 cm
Show solution
R/S = 25/75. After swapping, S/R = 75/25, so the balance point is at 75 cm. This is a common check of the bridge.
Answer: (C)
15. Quick Revision Summary
- I = neAvd; vd = eEτ/m; ρ = m/(ne²τ).
- R = ρl/A; stretched n times ⇒ n²R; RT = R₀(1 + αΔT).
- Series adds R; parallel adds 1/R.
- I = E/(R + r); V = E − Ir; Pmax = E²/4r at R = r.
- Kirchhoff: junction (charge), loop (energy); bridge balance P/Q = R/S.