Current Electricity

Before you begin

Formulas you will need

Current, resistance and circuits V = IR · R = ρL ⁄ A · R ∝ 1 ⁄ r² for a wire of radius r
Series: R = R₁ + R₂ · Parallel: 1⁄R = 1⁄R₁ + 1⁄R₂
Terminal voltage: V = ε − Ir · Circuit current: I = ε ⁄ (R + r)
Power: P = VI = I²R = V² ⁄ R · Drift: I = nAev_d · σ = ne²τ ⁄ m

How to use this worksheet. For any wire question, remember that resistance depends on the square of the radius, not the radius itself — that single fact answers two of the previous year questions in Section D. For circuits with a battery, always add the internal resistance to the external resistance before dividing into the emf.

Section A · Level 1

Definitions and direct formulas

Eight questions. None should take more than twenty seconds.

The SI unit of electrical resistance is the:

Resistance is potential difference divided by current, so its unit is the volt per ampere, named the ohm.

The SI unit of resistivity is:

From R = ρL ÷ A, rearranging gives ρ = RA ÷ L, which has units of ohm times metre squared divided by metre, that is ohm metre.

The resistance of a uniform conductor is directly proportional to its:

R = ρL ÷ A, so resistance rises in direct proportion to length and falls in inverse proportion to area.

The SI unit of conductance is the:

Conductance is the reciprocal of resistance, measured in reciprocal ohms, which is given the name siemens.

Kirchhoff's junction rule is a statement of the conservation of:

The junction rule says the current entering a node equals the current leaving it, which is charge conservation. The loop rule is the one that expresses energy conservation.

The relation between current and drift velocity in a conductor is:

Current equals the number density of carriers times the cross-sectional area, the charge per carrier and the drift velocity, so I = nAev_d.

Which quantity remains the same for every component in a series circuit?

In series there is only one path, so the same current flows through every element. In parallel it is the voltage that is shared instead.

The power dissipated in a resistor R carrying current I is:

Power equals VI, and substituting V = IR gives I²R. The equivalent forms are VI and V² ÷ R.

Section B · Level 2

Networks and Ohm's law

Eight questions. Reduce the network before you substitute anything.

Resistors of 3 Ω and 6 Ω are connected in parallel. The equivalent resistance is:

Product over sum gives (3 × 6) ÷ 9 = 2 Ω. Note that a parallel combination is always smaller than the smallest resistor in it.

Q10

Three resistors of 6 Ω each are connected in parallel. The equivalent resistance is:

For n identical resistors in parallel the result is R ÷ n, so 6 ÷ 3 = 2 Ω. Option 1 is the series answer.

Q11

A potential difference of 12 V is applied across a 4 Ω resistor. The current is:

From V = IR, the current is 12 ÷ 4 = 3 A. Option 1 multiplies instead of dividing.

Q12

The power dissipated when 12 V is applied across a 4 Ω resistor is:

Using P = V² ÷ R gives 144 ÷ 4 = 36 W. You could equally use I²R with I = 3 A, giving 9 × 4 = 36 W.

Q13

If the length of a wire is doubled while its area stays the same, its resistance:

Resistance is directly proportional to length, so doubling the length doubles the resistance.

Q14

If the radius of a wire is doubled while its length stays the same, its resistance becomes:

Area is proportional to the square of the radius, and resistance is inversely proportional to area. Doubling the radius quadruples the area, so the resistance falls to one quarter. This is the key idea behind two of the previous year questions.

Q15

A 100 W bulb runs for 10 hours. The energy consumed is:

Energy in kilowatt hours is power in kilowatts times time in hours: 0.1 × 10 = 1 kWh.

Q16

A copper wire of length 2 m and cross-sectional area 1 × 10⁻⁶ m² has resistivity 1.7 × 10⁻⁸ Ω m. Its resistance is:

R = ρL ÷ A = (1.7 × 10⁻⁸ × 2) ÷ (1 × 10⁻⁶) = 3.4 × 10⁻⁸ ÷ 10⁻⁶ = 0.034 Ω. Careful tracking of the powers of ten is the whole task here.

Section C · Level 3

Cells, internal resistance and conduction

Seven questions. Add the internal resistance before dividing into the emf.

Q17

A battery of emf 12 V and internal resistance 1 Ω drives a 5 Ω external resistor. The current in the circuit is:

The total resistance is 5 + 1 = 6 Ω, so I = 12 ÷ 6 = 2 A. Option 1 comes from forgetting the internal resistance and dividing by 5 alone.

Q18

For the same circuit, the terminal voltage of the battery is:

Terminal voltage is V = ε − Ir = 12 − (2 × 1) = 10 V. It is always less than the emf when current is being drawn, and equals the emf only on open circuit.

Q19

Three cells of 2 V each, with internal resistance 0.5 Ω each, are joined in series across a 3.5 Ω resistor. The current is:

In series the emfs add to 6 V and the internal resistances add to 1.5 Ω. Total resistance is 3.5 + 1.5 = 5 Ω, so I = 6 ÷ 5 = 1.2 A. Option 2 ignores the internal resistances.

Q20

Two wires of the same material and length have radii in the ratio 1 : 2. The ratio of their resistances is:

Resistance goes as one over the radius squared. With radii in the ratio 1 : 2, the resistances are in the ratio 1/1 : 1/4, which is 4 : 1. The thinner wire has the greater resistance.

Q21

The conductivity σ of a conductor with n free electrons per unit volume, electron charge e, mass m and mean collision time τ is:

From the free electron model, σ = ne²τ ÷ m. Conductivity rises with carrier density and with the mean time between collisions, and falls with electron mass.

Q22

Two identical resistors dissipate the most total power for a given supply voltage when connected:

For a fixed supply voltage, P = V² ÷ R, so the lowest total resistance dissipates the most power. Parallel halves the resistance while series doubles it, so parallel dissipates four times the power of series.

Q23

The resistance of a pure metallic conductor as its temperature rises:

In metals, higher temperature means more frequent collisions and a shorter mean free time, so resistance rises. Semiconductors behave the opposite way, which is why they are the answer to the Ohm's law question in Section D.

Section D · Previous year questions

Straight from the AAI ATC papers

Seven questions, every one exactly as it appeared. All seven come from the two 2025 shifts.

Q24PYQ 2025 · Shift 1

In a closed circuit, a battery of emf 10 V and internal resistance of 0.4 Ω is connected to an external resistor such that the current flowing through the circuit is 2 A. What will be the value of the external resistor?

From ε = I(R + r), we get 10 = 2(R + 0.4), so R + 0.4 = 5 and R = 4.6 Ω. Option 1 is what you get by ignoring the internal resistance entirely — which is precisely why it is there.

Q25PYQ 2025 · Shift 1

Which of the following represents the correct relation between the emf ε and the terminal voltage V of a cell supplying current I through internal resistance r?

Some of the emf is dropped across the internal resistance, so the terminal voltage is the emf minus that drop: V = ε − Ir. Option 1 would apply only while the cell is being charged rather than discharged.

Q26PYQ 2025 · Shift 2

A battery with an EMF of 25 V and an internal resistance of 2 Ω is connected to a 4 Ω and a 12 Ω resistor in parallel. What will be the total current drawn from the battery?

The parallel pair reduces to (4 × 12) ÷ 16 = 3 Ω. Adding the internal resistance gives 3 + 2 = 5 Ω, so the current is 25 ÷ 5 = 5 A. Two steps, and the internal resistance must be included in the second.

Q27PYQ 2025 · Shift 2

Two wires of equal length and having radius 1 cm and 4 cm respectively are made of the same material. The ratio of the resistance of the wires is given by:

Resistance is inversely proportional to the square of the radius. With radii 1 and 4, the squares are 1 and 16, so the resistances are in the ratio 1/1 : 1/16, which is 16 : 1. Options 1 and 2 come from using the radius rather than its square.

Q28PYQ 2025 · Shift 1

Two cylindrical wires, A and B, made of the same material with equal lengths but different radii r₁ and r₂, are first connected in series and then in parallel. What is the ratio of the effective resistance in the series combination to that in the parallel combination?

Write R₁ = k ÷ r₁² and R₂ = k ÷ r₂². In series the total is k(r₁² + r₂²) ÷ (r₁²r₂²). In parallel it is k ÷ (r₁² + r₂²). Dividing the first by the second gives (r₁² + r₂²)² ÷ (r₁²r₂²), which is the whole bracket squared. Checking with r₁ = 1 and r₂ = 2 gives 6.25 either way.

Q29PYQ 2025 · Shift 1

A reverse bias voltage of 10 V is applied across a p-n junction diode such that the current flowing through the diode in the reverse bias configuration is 2 µA. What will be the resistance of the diode in this configuration?

R = V ÷ I = 10 ÷ (2 × 10⁻⁶) = 5 × 10⁶ Ω, which is 5000 kΩ or 5 MΩ. The huge value is the point — a reverse biased diode barely conducts. The wrong options are all off by powers of ten.

Q30PYQ 2025 · Shift 2

Which of the following materials exhibits behaviour that violates Ohm's law under specific conditions?

Ohm's law requires a linear relation between current and voltage. Three of the options describe exactly that linear behaviour, so they obey the law. Only the semiconductor, whose current-voltage curve bends, violates it. Read the descriptions rather than the material names.

Formulas you will need

Definitions and direct formulas

Networks and Ohm's law

Cells, internal resistance and conduction

Straight from the AAI ATC papers

↓ Free download — Current Electricity PDF

Seven questions from one chapter in a single year.