Current and circuits practice questions.

A uniform wire of length 2.0 m and cross-sectional area 1.0 × 10⁻⁶ m² has a resistance of 0.034 Ω. The resistivity of the metal is:

• 3.4 × 10⁻⁸ Ω m
• 1.7 × 10⁻⁸ Ω m
• 6.8 × 10⁴ Ω m
• 8.5 × 10⁻⁹ Ω m

Answer: 1.7 × 10⁻⁸ Ω m

ρ = RA/L = (0.034 × 1.0 × 10⁻⁶)/2.0 = 1.7 × 10⁻⁸ Ω m (typical for copper). Forgetting to divide by L, or swapping A and L, gives the distractors.

As the current in a filament lamp increases, its resistance increases. Which statement best explains this behaviour?

• The number of free conduction electrons in the metal filament decreases significantly at high current
• The filament gets shorter as it heats up, reducing the current
• Electrons move faster, so they experience less resistance
• The filament's temperature rises, lattice ions vibrate more and scatter electrons more frequently

Answer: The filament's temperature rises, lattice ions vibrate more and scatter electrons more frequently

Higher current dissipates more power, raising the filament temperature; greater amplitude of ion vibration increases electron scattering, so resistance rises — the lamp is non-ohmic.

A 4.0 kΩ resistor and an 8.0 kΩ resistor are connected in series across a 12 V supply of negligible internal resistance. The potential difference across the 8.0 kΩ resistor is:

• 8.0 V
• 4.0 V
• 12 V
• 6.0 V

Answer: 8.0 V

In a potential divider the p.d. splits in proportion to resistance: V = 12 × 8.0/(4.0 + 8.0) = 8.0 V. Choosing 4.0 V comes from reading off the wrong resistor.

A 6.0 Ω resistor is connected in series with a parallel combination of a 6.0 Ω resistor and a 3.0 Ω resistor. The total resistance of the network is:

• 15 Ω
• 1.5 Ω
• 8.0 Ω
• 9.0 Ω

Answer: 8.0 Ω

The parallel pair gives (6.0 × 3.0)/(6.0 + 3.0) = 2.0 Ω, then adding the series 6.0 Ω gives 8.0 Ω. Adding all three (15 Ω) or treating all as parallel (1.5 Ω) are typical errors.

A resistor has 6.0 V across it and carries 2.0 A. What is its resistance (in ohms)?

Answer: 3

Ohm's law: R = V/I = 6.0/2.0 = 3.0 ohms.

A device operates at 12 V and draws 2.0 A. What is its power (in watts)?

Answer: 24

Electrical power P = VI = 12 × 2.0 = 24 W. Also P = I^2R = V^2/R.