Electric current and current density
A fan, a phone charger and a bulb all work because charge flows through them. Link to earlier classes : electric current and circuits from Class 10.
Electric current is the rate of flow of charge through a cross-section of a conductor. By convention it points in the direction in which positive charge would move, which is opposite to the flow of electrons. Current is a scalar and its SI unit is the ampere (A), 1 A = 1 C s⁻¹.
Rule 1: Current — I = Q ÷ t (Q = charge passing in time t); in general I = dQ ÷ dt. The number of electrons crossing per second is I ÷ e.
- Easy : 12 C of charge passes in 4 s ⇒ I = 3 A.
- Medium : A current of 2 A for 10 minutes carries Q = 2 × 600 = 1200 C, which is 1200 ÷ 1.6 × 10⁻¹⁹ = 7.5 × 10²¹ electrons.
- Exam : If I = 3t (A), the charge in 4 s is the area under the I–t graph, Q = ½ × 4 × 12 = 24 C.
Rule 2: Current density — J = I ÷ A (A = cross-sectional area perpendicular to the flow). It is a vector along the current, measured in A m⁻².
- Easy : I = 5 A through A = 2.5 mm² = 2.5 × 10⁻⁶ m² ⇒ J = 2 × 10⁶ A m⁻².
- Medium : A thicker wire carrying the same current has a smaller J.
- Exam : The current through a surface is I = J A cos θ, where θ is the angle between J and the normal.
I = Q/t; conventional current is opposite to the electron flow; J = I/A.
Treating current as a vector. It has a direction along the wire but does not add by the vector rule.
Memory clue : "Current is charge per second; current density is current per area."
Spot the difference
| Quantity | Symbol | Unit | Type |
|---|---|---|---|
| Current | I | A | scalar |
| Current density | J | A m⁻² | vector |
Practice questions
- 1Write the SI unit of current in terms of C and s.
- 2Is the direction of conventional current the same as that of electron flow?
- 3Is current a scalar or a vector?
- 420 C of charge passes a point in 5 s. Find the current.
- 5Find the charge that flows in 10 minutes for a current of 2 A.
- 6Find the current density in a wire of cross-section 2.5 mm² carrying 5 A.
- 7A current of 0.8 A flows. How many electrons cross a section per second?
- 8A current varies as I = 3t amperes. Find the charge that flows in the first 4 s.
Show answers (try them yourself first)
- 11 A = 1 C s⁻¹.
- 2No, they are opposite.
- 3Scalar.
- 4I = 20 ÷ 5 = 4 A.
- 5Q = 2 × 600 = 1200 C.
- 6J = 5 ÷ 2.5 × 10⁻⁶ = 2 × 10⁶ A m⁻².
- 7n = 0.8 ÷ 1.6 × 10⁻¹⁹ = 5 × 10¹⁸ electrons per second.
- 8Q = area under the graph = ½ × 4 × 12 = 24 C.
Ohm's law, resistance and resistivity
A thin long wire of a heater gets hot, while the thick copper wire feeding it stays cool. Link to earlier classes : Ohm's law and the factors on which resistance depends, from Class 10.
Ohm's law says that, at a fixed temperature, the current through a conductor is proportional to the potential difference across it. The constant ratio V ÷ I is the resistance R, in ohm (Ω). The resistance of a wire depends on its material through its resistivity ρ (unit Ω m).
Rule 1: Ohm's law — V = I R (V = potential difference, I = current, R = resistance). The V–I graph of an ohmic conductor is a straight line through the origin.
- Easy : R = 20 Ω and I = 0.50 A ⇒ V = 10 V.
- Medium : A bulb of 484 Ω at 230 V draws I = 230 ÷ 484 = 0.475 A.
- Exam : The slope of the I–V graph is 1 ÷ R, so a steeper line means a smaller resistance.
Rule 2: Resistance of a wire — R = ρ l ÷ A (l = length, A = cross-sectional area). The resistivity ρ depends on the material and the temperature, not on the shape. Conductance is G = 1 ÷ R and conductivity is σ = 1 ÷ ρ.
- Easy : Copper (ρ = 1.7 × 10⁻⁸ Ω m), l = 10 m, diameter 1 mm ⇒ A = 7.85 × 10⁻⁷ m² and R = 1.7 × 10⁻⁸ × 10 ÷ 7.85 × 10⁻⁷ = 0.22 Ω.
- Medium : A wire is stretched to twice its length. The volume stays fixed, so the area halves, and R becomes 4 times.
- Exam : A wire of 12 Ω bent into a circle has two equal 6 Ω halves in parallel between opposite points, so the resistance there is 3 Ω.
In the figure : V–I graphs of an ohmic resistor (straight line) and a non-ohmic device such as a diode (curve).
V = IR; R = ρl/A; stretching a wire to n times its length makes R become n² times.
Saying that resistance depends on the voltage applied. For an ohmic conductor R stays the same.
Memory clue : "Longer, thinner, hotter: more resistance (rho l over A)."
Spot the difference
| Quantity | Depends on | Unit |
|---|---|---|
| Resistance R | material, l, A, temperature | Ω |
| Resistivity ρ | material, temperature only | Ω m |
Practice questions
- 1State Ohm's law.
- 2Write the SI unit of resistivity.
- 3Name the factors on which the resistance of a wire depends.
- 4Find the voltage across a 20 Ω resistor carrying 0.50 A.
- 5Find the resistance of 10 m of copper wire of diameter 1 mm (ρ = 1.7 × 10⁻⁸ Ω m).
- 6A wire is stretched to double its length. By what factor does its resistance change?
- 7Two wires of the same material have lengths in the ratio 1 : 2 and radii in the ratio 1 : 2. Find the ratio of their resistances.
- 8A wire of 12 Ω is bent into a circle. Find the resistance between two diametrically opposite points.
Show answers (try them yourself first)
- 1At constant temperature the current through a conductor is proportional to the potential difference across it (V ∝ I).
- 2Ω m.
- 3Length, area of cross-section, material and temperature.
- 4V = IR = 10 V.
- 5R = ρl ÷ A ≈ 0.22 Ω.
- 6It becomes 4 times.
- 7R ∝ l ÷ r², so R₁ ÷ R₂ = (1 ÷ 2) × (4 ÷ 1) = 2 : 1.
- 8Two 6 Ω halves in parallel give 3 Ω.
Drift velocity and mobility
A bulb lights the moment the switch is pressed, yet the electrons creep along very slowly. Link to earlier classes : free electrons in metals from Class 10.
Free electrons in a metal move randomly at high speed. When an electric field is applied, each electron also acquires a small average velocity opposite to the field, the drift velocity v_d. The relaxation time τ is the average time between two collisions, and the mobility μ is the drift velocity per unit field.
Rule 1: Drift velocity and current — v_d = (e E ÷ m) τ and I = n e A v_d (n = free electrons per unit volume, A = area, E = field, m = electron mass). Mobility μ = v_d ÷ E, in m² V⁻¹ s⁻¹.
- Easy : Copper has n = 8.5 × 10²⁸ m⁻³. For I = 2 A and A = 1 mm², v_d = 2 ÷ (8.5 × 10²⁸ × 1.6 × 10⁻¹⁹ × 10⁻⁶) = 1.5 × 10⁻⁴ m s⁻¹.
- Medium : For E = 0.10 V m⁻¹ and τ = 2 × 10⁻¹⁴ s, v_d = 1.6 × 10⁻¹⁹ × 0.10 × 2 × 10⁻¹⁴ ÷ 9.1 × 10⁻³¹ = 3.5 × 10⁻⁴ m s⁻¹, so μ = 3.5 × 10⁻³ m² V⁻¹ s⁻¹.
- Exam : At 1.5 × 10⁻⁴ m s⁻¹ an electron needs about 1.9 hours to drift 1 m, yet the bulb lights at once because the electric field is set up along the whole wire at nearly the speed of light.
Rule 2: Resistivity from the microscopic view — ρ = m ÷ (n e² τ). A smaller τ (more collisions, for example at higher temperature) gives a larger resistivity.
- Easy : Copper with τ = 2.5 × 10⁻¹⁴ s gives ρ = 9.1 × 10⁻³¹ ÷ (8.5 × 10²⁸ × (1.6 × 10⁻¹⁹)² × 2.5 × 10⁻¹⁴) = 1.7 × 10⁻⁸ Ω m.
- Medium : If the current in the same wire is doubled, v_d is doubled.
- Exam : ρ does not depend on E, which is the reason why Ohm's law holds for metals.
I = neAv_d; v_d is tiny (about 10⁻⁴ m s⁻¹); ρ = m/(ne²τ).
Thinking that electrons travel from the switch to the bulb in a fraction of a second. The signal, not the electron, travels fast.
Memory clue : "I = n e A v-drift : a crowd moving slowly still carries a big current."
Spot the difference
| Speed | Typical value |
|---|---|
| Random thermal speed of electrons | about 10⁵ m s⁻¹ |
| Drift velocity | about 10⁻⁴ m s⁻¹ |
| Speed of the electric signal | about 3 × 10⁸ m s⁻¹ |
Practice questions
- 1Write the formula connecting current and drift velocity.
- 2Write the SI unit of mobility.
- 3What is meant by relaxation time?
- 4A copper wire (n = 8.5 × 10²⁸ m⁻³, A = 1 mm²) carries 2 A. Find the drift velocity.
- 5Find the drift velocity for E = 0.10 V m⁻¹ and τ = 2 × 10⁻¹⁴ s.
- 6Find the mobility for the situation of question 5.
- 7Using ρ = m ÷ (ne²τ) with τ = 2.5 × 10⁻¹⁴ s, find the resistivity of copper.
- 8How does the drift velocity change if the current in a wire is doubled?
Show answers (try them yourself first)
- 1I = n e A v_d.
- 2m² V⁻¹ s⁻¹.
- 3The average time between two successive collisions of an electron.
- 4v_d = I ÷ (neA) ≈ 1.5 × 10⁻⁴ m s⁻¹.
- 5v_d = eEτ ÷ m ≈ 3.5 × 10⁻⁴ m s⁻¹.
- 6μ = v_d ÷ E ≈ 3.5 × 10⁻³ m² V⁻¹ s⁻¹.
- 7ρ ≈ 1.7 × 10⁻⁸ Ω m.
- 8It doubles.
Temperature dependence of resistance and limits of Ohm's law
A bulb draws a large current at the instant of switching on, when the filament is cold, and a smaller current once it glows. Link to earlier classes : heating effect of current from Class 10.
The resistivity of a metal rises as it gets hotter, because the ions vibrate more and the electrons collide more often. The fractional change per degree is the temperature coefficient of resistivity α. Semiconductors behave oppositely: their resistivity falls as temperature rises.
Rule 1: Resistance and temperature — R = R₀ [1 + α (T − T₀)] (R₀ = resistance at reference temperature T₀). So α = (R − R₀) ÷ (R₀ ΔT), in °C⁻¹.
- Easy : R₀ = 100 Ω at 20 °C and α = 0.004 °C⁻¹ ⇒ R at 70 °C = 100 × (1 + 0.004 × 50) = 120 Ω.
- Medium : A wire has 50 Ω at 0 °C and 70 Ω at 100 °C, so α = 20 ÷ (50 × 100) = 0.004 °C⁻¹.
- Exam : A wire of 40 Ω at 20 °C with α = 0.005 °C⁻¹ has 40 × (1 + 0.005 × 50) = 50 Ω at 70 °C.
Rule 2: Where Ohm's law fails — Ohm's law fails when the V–I graph is not a straight line, or when the same V gives different I for the two directions, or when more than one value of V gives the same I. Examples are a diode, an LED, a thermistor and a gas-discharge tube.
- Easy : Nichrome and manganin have a very small α, so they are used for heating coils and standard resistors.
- Medium : Silicon and germanium have a negative α.
- Exam : A bulb filament shows a curved V–I graph because it heats up and its resistance rises with the current.
R = R₀[1 + αΔT]; metals: α positive; semiconductors: α negative; alloys: α very small.
Believing that Ohm's law is true for every conductor. It is a property of some materials only.
Memory clue : "Metal heats, resistance rises. Semiconductor heats, resistance falls."
Spot the difference
| Material | α | Use |
|---|---|---|
| Copper, silver | positive, moderate | wires |
| Nichrome, manganin | positive, very small | heaters, standard resistors |
| Silicon, germanium | negative | thermistors, sensors |
Practice questions
- 1Does the resistivity of a metal rise or fall on heating?
- 2Why are alloys used for standard resistors?
- 3Name two devices that do not obey Ohm's law.
- 4A 100 Ω wire at 20 °C has α = 0.004 °C⁻¹. Find its resistance at 70 °C.
- 5A wire has 50 Ω at 0 °C and 70 Ω at 100 °C. Find α.
- 6What is the sign of α for silicon?
- 7A 40 Ω wire at 20 °C has α = 0.005 °C⁻¹. Find its resistance at 70 °C.
- 8Why is the current in a bulb highest at the moment of switching on?
Show answers (try them yourself first)
- 1It rises.
- 2Their α is very small, so the resistance stays nearly constant when the temperature changes.
- 3Any two of: diode, LED, thermistor, gas-discharge tube.
- 4120 Ω.
- 5α = 0.004 °C⁻¹.
- 6Negative.
- 7R = 40 × (1 + 0.005 × 50) = 50 Ω.
- 8The filament is cold and its resistance is low, so I = V ÷ R is large until it heats up.
Electrical energy and power
An electric heater is hot and a mobile charger is warm, but a tube light stays cool: they use power at different rates. Link to earlier classes : Joule's heating law from Class 10.
When a current I flows under a potential difference V, the source does work and the conductor gets heated (Joule heating). Power is the rate of using energy; the electrical energy is commonly billed in kilowatt-hour (kWh), where 1 kWh = 3.6 × 10⁶ J.
Rule 1: Power and energy — P = V I = I² R = V² ÷ R and E = P t.
- Easy : A 60 W, 230 V bulb draws I = 60 ÷ 230 = 0.26 A and has R = 230² ÷ 60 = 882 Ω.
- Medium : A 1.5 kW heater used for 4 hours consumes 6 kWh. At ₹8 per unit the cost is ₹48.
- Exam : A 1000 W immersion heater warms 2 kg of water from 20 °C to 70 °C (c = 4200 J kg⁻¹ °C⁻¹) in t = 2 × 4200 × 50 ÷ 1000 = 420 s = 7 minutes, if there is no loss.
Rule 2: Bulbs in series and power loss in lines — At the same voltage, the higher-wattage bulb has the smaller resistance. In series the same current flows, so the bulb of larger resistance dissipates more power. Transmitting at high voltage cuts the I²R loss in the line.
- Easy : 100 W and 60 W bulbs (230 V) have R = 529 Ω and 882 Ω.
- Medium : In series they carry 0.163 A, giving 14.1 W in the 100 W bulb and 23.4 W in the 60 W bulb, so the 60 W bulb glows brighter.
- Exam : To send 10 kW through a line of 2 Ω, the loss is (50 A)² × 2 = 5000 W at 200 V but only (5 A)² × 2 = 50 W at 2000 V.
P = VI = I²R = V²/R; E = Pt; 1 kWh = 3.6 × 10⁶ J.
Using P = V²/R for bulbs in series. In series the current is common, so use P = I²R.
Memory clue : "Series: same I, use I²R. Parallel: same V, use V²/R."
Spot the difference
| Circuit | Same quantity | Brighter bulb |
|---|---|---|
| Series | current | higher resistance (lower wattage) |
| Parallel | voltage | lower resistance (higher wattage) |
Practice questions
- 1How many joules are there in 1 kWh?
- 2Write three expressions for electrical power.
- 3Which bulb has the larger resistance at 230 V: 100 W or 60 W?
- 4Find the current and resistance of a 60 W, 230 V bulb.
- 5A 1.5 kW heater runs for 4 hours. Find the energy used and the cost at ₹8 per unit.
- 6A 100 W and a 60 W bulb (both 230 V) are joined in series to 230 V. Which one glows brighter?
- 7How long will a 1000 W heater take to warm 2 kg of water from 20 °C to 70 °C (c = 4200 J kg⁻¹ °C⁻¹)?
- 8Compare the line loss when 10 kW is sent through a 2 Ω line at 200 V and at 2000 V.
Show answers (try them yourself first)
- 13.6 × 10⁶ J.
- 2P = VI = I²R = V²/R.
- 3The 60 W bulb.
- 4I = P ÷ V ≈ 0.26 A; R = V² ÷ P ≈ 882 Ω.
- 5E = 1.5 × 4 = 6 kWh; cost = 6 × 8 = ₹48.
- 6The 60 W bulb (23.4 W against 14.1 W), since it has the larger resistance.
- 7t = mcΔT ÷ P = 2 × 4200 × 50 ÷ 1000 = 420 s = 7 min.
- 8At 200 V : I = 50 A and loss = 5000 W. At 2000 V : I = 5 A and loss = 50 W.
Cells, emf and internal resistance
A phone battery lasts less when it is old because its internal resistance has grown. Link to earlier classes : cells and batteries from Class 10.
The emf (ε) of a cell is the potential difference between its terminals when no current is drawn (open circuit). The terminal voltage V is the potential difference across it when current flows; it is lower than ε because of the cell's internal resistance r.
Rule 1: Terminal voltage and current — V = ε − I r while discharging, and I = ε ÷ (R + r) for an external resistance R. While charging, V = ε + I r.
- Easy : ε = 12 V, r = 0.5 Ω, I = 4 A ⇒ V = 12 − 2 = 10 V.
- Medium : ε = 6 V, r = 1 Ω, R = 5 Ω ⇒ I = 6 ÷ 6 = 1 A and V = 5 V.
- Exam : A cell of ε = 10 V gives a terminal voltage of 9.6 V at 2 A, so r = (10 − 9.6) ÷ 2 = 0.2 Ω.
Rule 2: Cells in series and in parallel — Series (n identical cells) : ε_eq = n ε and r_eq = n r. Parallel : ε_eq = ε and r_eq = r ÷ n.
- Easy : Four cells (1.5 V, 0.2 Ω each) in series: ε = 6 V and r = 0.8 Ω.
- Medium : With R = 3.2 Ω in series, I = 6 ÷ (3.2 + 0.8) = 1.5 A.
- Exam : The same four cells in parallel give ε = 1.5 V, r = 0.05 Ω and I = 1.5 ÷ 3.25 = 0.46 A, so series is better when R is large.
V = ε − Ir; I = ε/(R + r); series adds ε and r; parallel keeps ε and divides r.
Treating the emf as the voltage across the cell in a closed circuit. The terminal voltage is smaller.
Memory clue : "Emf is the cell's pay; internal resistance takes a cut: V = ε − Ir."
Spot the difference
| Quantity | Open circuit | Closed circuit |
|---|---|---|
| Terminal voltage | V = ε | V = ε − Ir |
| Current | zero | ε ÷ (R + r) |
Practice questions
- 1Define the emf of a cell.
- 2When is the terminal voltage equal to the emf?
- 3Write the terminal voltage of a cell that is being charged.
- 4A 12 V cell of internal resistance 0.5 Ω delivers 4 A. Find its terminal voltage.
- 5A 6 V cell with r = 1 Ω is joined to a 5 Ω resistor. Find the current and the terminal voltage.
- 6Four cells (1.5 V, 0.2 Ω each) in series are connected to a 3.2 Ω resistor. Find the current.
- 7The same four cells are connected in parallel to the same 3.2 Ω resistor. Find the current.
- 8A 10 V cell shows a terminal voltage of 9.6 V at 2 A. Find r.
Show answers (try them yourself first)
- 1The potential difference between the terminals of a cell when no current is drawn.
- 2When the circuit is open (no current).
- 3V = ε + Ir.
- 4V = 12 − 4 × 0.5 = 10 V.
- 5I = 6 ÷ (5 + 1) = 1 A; V = IR = 5 V.
- 6ε = 6 V, r = 0.8 Ω, I = 6 ÷ 4 = 1.5 A.
- 7ε = 1.5 V, r = 0.05 Ω, I = 1.5 ÷ 3.25 ≈ 0.46 A.
- 8r = (10 − 9.6) ÷ 2 = 0.2 Ω.
Kirchhoff's rules and the Wheatstone bridge
The wiring of a house has many branches, and the current splits and joins at junctions. Link to earlier classes : series and parallel circuits from Class 10.
Kirchhoff's junction rule (KCL) says the sum of currents entering a junction equals the sum leaving it (charge is conserved). Kirchhoff's loop rule (KVL) says the algebraic sum of potential changes around any closed loop is zero (energy is conserved). A Wheatstone bridge uses a balance condition to measure an unknown resistance.
Rule 1: Kirchhoff's rules — Σ I (in) = Σ I (out) at a junction, and Σ ε − Σ I R = 0 around a loop. Going through a resistor along the current is a fall of IR; going through a cell from − to + is a rise of ε.
- Easy : At a junction 3 A and 2 A enter and 4 A leaves, so the third wire carries 1 A out of the junction.
- Medium : A 10 V cell in series with 2 Ω and 3 Ω gives 10 − 2I − 3I = 0, so I = 2 A.
- Exam : The loop rule of a circuit with the internal resistance r gives ε − Ir − IR = 0, the same as I = ε ÷ (R + r).
Rule 2: Wheatstone bridge — The bridge is balanced (no current in the galvanometer) when P ÷ Q = R ÷ S. In a metre bridge, R ÷ S = l ÷ (100 − l), where l is the balancing length in cm.
- Easy : P = 10 Ω, Q = 15 Ω, R = 20 Ω ⇒ S = 30 Ω.
- Medium : In a metre bridge the balance point is at 40 cm with S = 15 Ω in the right gap, so R = 15 × 40 ÷ 60 = 10 Ω.
- Exam : For the balanced bridge above, the galvanometer arm can be removed. (10 + 15) = 25 Ω is in parallel with (20 + 30) = 50 Ω, giving 25 × 50 ÷ 75 = 16.7 Ω.
In the figure : A Wheatstone bridge with arms P, Q, R, S and a galvanometer G between the middle points.
KCL : charge conserved; KVL : energy conserved; balance : P/Q = R/S.
Getting the signs wrong in the loop rule. Fix a direction of travel and keep to it.
Memory clue : "Junction: in = out. Loop: ups = downs. Bridge: P over Q equals R over S."
Spot the difference
| Rule | Conserves | Applied at |
|---|---|---|
| Junction rule | charge | a junction |
| Loop rule | energy | a closed loop |
Practice questions
- 1State Kirchhoff's junction rule.
- 2State Kirchhoff's loop rule.
- 3What happens in the galvanometer when a Wheatstone bridge is balanced?
- 4At a junction 3 A and 2 A enter and 4 A leaves through one wire. Find the current in the remaining wire.
- 5A 10 V cell is in series with 2 Ω and 3 Ω resistors. Find the current.
- 6A balanced bridge has P = 10 Ω, Q = 15 Ω and R = 20 Ω. Find S.
- 7In a metre bridge the null point is at 40 cm and the known resistance in the right gap is 15 Ω. Find the unknown resistance.
- 8Find the equivalent resistance of the balanced bridge of question 6.
Show answers (try them yourself first)
- 1The sum of currents entering a junction equals the sum of currents leaving it.
- 2The algebraic sum of the potential changes around a closed loop is zero.
- 3There is no current in it.
- 45 A enters and 4 A leaves, so 1 A leaves through the remaining wire.
- 5I = 10 ÷ 5 = 2 A.
- 6S = Q R ÷ P = 15 × 20 ÷ 10 = 30 Ω.
- 7R = 15 × 40 ÷ 60 = 10 Ω.
- 825 Ω and 50 Ω in parallel: 25 × 50 ÷ 75 ≈ 16.7 Ω.