Module B.5 IBDP SL/HL Track

Current and Circuits

Examine electric current paths, electromotive force, internal resistance, and network circuit layouts.

Current and Circuits

Introduction

Electric circuits are energy-transfer systems built from moving charge. The guide emphasizes that cells provide emf, components shape current paths, and resistance explains why electrical energy is often dissipated as thermal energy.

Guide Focus

  • Use current, potential difference, resistance, resistivity, and power relationships.
  • Analyse series and parallel resistor combinations.
  • Include emf and internal resistance in practical cell models.

Key Concepts

1. Current and potential difference

Direct current is the rate of flow of charge: I = delta q / delta t. Potential difference is work done per unit charge: V = W / q.

2. Resistance and resistivity

Resistance is R = V / I. For a uniform conductor, resistivity is rho = RA / L, so material, length, and cross-sectional area all affect resistance.

3. Ohmic and non-ohmic behaviour

A metal conductor at constant temperature is treated as ohmic: V is proportional to I. Lamps, diodes, thermistors, and LDRs can show non-ohmic behaviour because their resistance changes with operating conditions.

4. Power in resistors

Electrical power dissipated by a resistor can be written as P = IV = I^2R = V^2 / R.

5. Series, parallel, and internal resistance

In series, current is the same and voltages add: Rs = R1 + R2 + … In parallel, voltage is the same and currents add: 1/Rp = 1/R1 + 1/R2 + … A real cell is modelled by emf epsilon and internal resistance r: epsilon = I(R + r).

Common Mistakes

  • Adding parallel resistors as if they were in series.
  • Treating terminal potential difference as always equal to emf.
  • Forgetting that ideal ammeters have zero resistance and ideal voltmeters have infinite resistance.

Exam Tips

  • Redraw complicated circuits with junctions clearly labelled.
  • Use current rules for series paths and voltage rules for parallel branches.
  • For cells with internal resistance, calculate lost volts Ir before comparing with terminal voltage.

Practice Questions

Question 1 (Multiple Choice)

Two identical resistors are connected in parallel to an ideal cell. Compared with one resistor alone, the total resistance is:

A. Half as large. B. Twice as large. C. The same. D. Four times as large.

Correct Answer: A

Solution Architecture

For two equal resistors R in parallel, 1/Rp = 1/R + 1/R = 2/R, so Rp = R/2.


Question 2 (Structured Paper 2 Style)

A cell of emf 9.0 V and internal resistance 0.50 ohm is connected to an external resistor of 4.0 ohm.

(a) Calculate the current. [2 marks]

(b) Calculate the terminal potential difference. [2 marks]

Paper 2 Structured Problem

Markscheme Breakdown

Part (a) Solution:

epsilon = I(R + r), so I = 9.0 / (4.0 + 0.50) = 2.0 A.

Part (b) Solution:

Terminal pd = IR = 2.0 x 4.0 = 8.0 V, equivalently epsilon - Ir = 9.0 - 1.0 = 8.0 V.