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9702 · 9.1

Electric current — practice questions

Practice and worked examples for 9702 Electric current. Short previews only — attempt the full question in MarkScheme against the official scheme.

Worked example 1

A wire has a cross-sectional area of 2.0 × 10⁻⁶ m² and carries a current of 4.0 A. If the number density of free electrons is 8.5 × 10²⁸ m⁻³, calculate the mean drift velocity of the electrons. (Elementary charge e = 1.60 × 10⁻¹⁹ C)

Show solution outline
  1. Identify the given values: A = 2.0 × 10⁻⁶ m² I = 4.0 A n = 8.5 × 10²⁸ m⁻³ q = e = 1.60 × 10⁻¹⁹ C
  2. Recall the formula for current in terms of drift velocity: I = Anvq.
  3. Rearrange the formula to solve for drift velocity (v): v = I / (Anq).
  4. Substitute the values into the rearranged formula: v = 4.0 / ((2.0 × 10⁻⁶) × (8.5 × 10²⁸) × (1.60 × 10⁻¹⁹))
  5. Calculate the result: v = 4.0 / (2.72 × 10⁴) ≈ 1.47 × 10⁻⁴ m/s (to 3 s.f.)
  6. State the answer: The mean drift velocity of the electrons is approximately 1.47 × 10⁻⁴ m/s.

Worked example 2

A current of 250 mA flows through a resistor for 4.0 minutes. Calculate (a) the total charge that passes through the resistor, and (b) the number of electrons that pass through the resistor in this time. (Elementary charge e = 1.60 × 10⁻¹⁹ C).

Show solution outline
  1. First, convert all units to SI units. Current I = 250 mA = 250 × 10⁻³ A = 0.250 A Time Δt = 4.0 minutes = 4.0 × 60 s = 240 s
  2. Part (a): Calculate the total charge (ΔQ). Use the formula I = ΔQ / Δt, rearranged to ΔQ = I × Δt. ΔQ = 0.250 A × 240 s ΔQ = 60 C
  3. Part (b): Calculate the number of electrons (N). The total charge is the number of electrons multiplied by the charge of one electron: ΔQ = N × e. Rearrange to find N: N = ΔQ / e. N = 60 C / (1.60 × 10⁻¹⁹ C) N = 3.75 × 10²⁰
  4. State the final answers: (a) The total charge is 60 C. (b) The number of electrons is 3.75 × 10²⁰.
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