Current electricity

NUST NET (Engineering) · Physics · Electricity. A short explanation of the idea, the rules to remember, the mistake to avoid, a worked example and practice questions with answers.

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The idea

Current is the rate of flow of charge (I = Q/t). The resistance of a wire depends on its material, length and thickness: a longer or thinner wire resists more. For metals resistance rises with temperature; for semiconductors it falls, because heating frees more charge carriers.

A real cell has internal resistance. Its EMF is the energy given to each coulomb, but some is lost inside the cell, so the terminal potential difference is less than the EMF whenever current flows.

Harder networks are solved with Kirchhoff's rules, which are conservation of charge and of energy. The Wheatstone bridge and the potentiometer are null methods: at balance no current flows through the galvanometer.

Rules to remember

Common mistake

Leaving out the internal resistance. The current is E/(R + r), not E/R, and the p.d. across the terminals is E − Ir, which is less than the EMF.

Worked example

A battery of EMF 9.0 V and internal resistance 1.0 Ω is connected to a 6.0 Ω and a 3.0 Ω resistor in parallel. What is the potential difference across its terminals?

  1. External resistance: R = (6.0 × 3.0)/(6.0 + 3.0) = 18/9 = 2.0 Ω.
  2. Current: I = E/(R + r) = 9.0/(2.0 + 1.0) = 3.0 A.
  3. Terminal p.d.: V = E − Ir = 9.0 − (3.0 × 1.0) = 6.0 V (the same as IR = 3.0 × 2.0).

Answer: 6.0 V

Practice questions

Try each one, then open the answer.

1. A coil has a resistance of 10.0 Ω at 0 °C and 12.0 Ω at 100 °C. The temperature coefficient of resistance of its material is

  1. A
    2.0 × 10−3 K−1
  2. B
    1.7 × 10−3 K−1
  3. C
    2.0 × 10−2 K−1
  4. D
    0.20 K−1
Show answer

Answer: A. α = (Rt − R0)/(R0t) = (12.0 − 10.0)/(10.0 × 100) = 2.0 × 10−3 K−1. Dividing by 12.0 Ω instead of R0 gives 1.7 × 10−3 K−1.

2. A 4.0 Ω resistor and a 12 Ω resistor are connected in parallel to a battery. The 4.0 Ω resistor dissipates 36 W. The power dissipated in the 12 Ω resistor is

  1. A
    12 W
  2. B
    108 W
  3. C
    36 W
  4. D
    4.0 W
Show answer

Answer: A. For the 4.0 Ω resistor V2/R = 36 W gives V2 = 144, so V = 12 V across both. The 12 Ω resistor dissipates 144/12 = 12 W. 108 W wrongly takes P ∝ R, which holds only for resistors in series (same current).

3. The SI unit of resistivity is

  1. A
    Ω m−1
  2. B
    Ω
  3. C
    Ω m
  4. D
    Ω m2
Show answer

Answer: C. ρ = RA/L has the unit Ω × m2 ÷ m = Ω m.

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