22  Electric Circuits

22.1 Syllabus inquiry question

  • How does charge flow in circuits to transfer energy?
Feynman Insight

From The Feynman Lectures on Physics, Vol II, Chapter 22:

Voltage is energy per charge. A circuit is a pathway that lets charges deliver that energy to devices and return to the source.

22.2 Learning Objectives

  • Define electric current and distinguish conventional current from electron flow.
  • Explain potential difference and electromotive force (emf).
  • Apply Ohm’s law to resistive circuits.
  • Compare series and parallel circuits.
  • Calculate electrical power and energy.

22.3 Content

22.3.1 Current and Charge Flow

Electric current is the rate of flow of charge:

\[I = \frac{Q}{t}\]

where: - \(I\) = current (A) - \(Q\) = charge (C) - \(t\) = time (s)

Current Direction
  • Conventional current flows from positive to negative (direction of positive charge movement)
  • Electron flow is opposite (electrons move from negative to positive)
  • Both describe the same physical process—the circuit works identically either way

22.3.2 Interactive: Simple Circuit

Visualising current flow in a basic circuit:

22.3.3 Potential Difference and EMF

Potential difference (voltage) is energy transferred per unit charge:

\[V = \frac{W}{q}\]

An electromotive force (emf) source (battery, generator, solar cell) provides energy to charges to drive current around the circuit.

Energy Transfer
  • The source (battery) converts chemical/mechanical energy to electrical energy
  • Components (resistors, lamps) convert electrical energy to heat, light, or motion
  • Voltage measures how much energy is transferred per coulomb of charge

22.3.4 Resistance and Ohm’s Law

Resistance measures how strongly a component opposes current flow:

\[R = \frac{V}{I}\]

For ohmic conductors (constant resistance):

\[V = IR\]

Resistors convert electrical energy into heat (thermal energy).

22.3.5 Interactive: Ohm’s Law Demonstration

22.3.6 Series Circuits

In a series circuit, components are connected end-to-end:

Series Circuit Rules
  • Current is the same through each component: \(I_{total} = I_1 = I_2 = ...\)
  • Voltage divides across components: \(V_{total} = V_1 + V_2 + ...\)
  • Resistance adds: \(R_{series} = R_1 + R_2 + ...\)

22.3.7 Interactive: Series Circuit

Two resistors in series:

22.3.8 Parallel Circuits

In a parallel circuit, components are connected across the same two points:

Parallel Circuit Rules
  • Voltage is the same across each branch: \(V_{total} = V_1 = V_2 = ...\)
  • Current divides between branches: \(I_{total} = I_1 + I_2 + ...\)
  • Resistance follows: \(\frac{1}{R_{parallel}} = \frac{1}{R_1} + \frac{1}{R_2} + ...\)

22.3.9 Interactive: Parallel Circuit

Two resistors in parallel:

22.3.10 Electrical Power and Energy

Electrical power is the rate of energy transfer:

\[P = IV\]

Alternative forms (using Ohm’s law):

\[P = I^2 R = \frac{V^2}{R}\]

Electrical energy over time:

\[E = Pt = IVt\]

Power Units
  • Power is measured in watts (W): 1 W = 1 J/s
  • Energy is measured in joules (J) or kilowatt-hours (kWh)
  • 1 kWh = 3.6 × 10⁶ J

22.4 Worked Examples

22.4.1 Example 1: Current from charge

A circuit transfers \(3.0\ \text{C}\) in \(2.0\ \text{s}\). Find the current.

Solution:

  1. Use \(I = Q/t\)

  2. \(I = 3.0/2.0\)

  3. \(I = 1.5\ \text{A}\)

22.4.2 Example 2: Ohm’s law

A resistor has \(12\ \text{V}\) across it and carries \(0.50\ \text{A}\). Find \(R\).

Solution:

  1. Use \(R = V/I\)

  2. \(R = 12/0.50\)

  3. \(R = 24\ \Omega\)

22.4.3 Example 3: Series resistance

Two resistors \(4.0\ \Omega\) and \(8.0\ \Omega\) are in series across \(12\ \text{V}\). Find the current.

Solution:

  1. Total resistance: \(R = R_1 + R_2 = 4.0 + 8.0 = 12\ \Omega\)

  2. Use Ohm’s law: \(I = V/R = 12/12 = 1.0\ \text{A}\)

  3. Current through both resistors is \(1.0\ \text{A}\)

22.4.4 Example 4: Parallel resistance and current

Two resistors \(6.0\ \Omega\) and \(3.0\ \Omega\) are in parallel across \(12\ \text{V}\).

Solution:

  1. Parallel resistance: \(\frac{1}{R} = \frac{1}{6.0} + \frac{1}{3.0} = \frac{1}{6.0} + \frac{2}{6.0} = \frac{3}{6.0}\)

  2. \(R = 2.0\ \Omega\)

  3. Total current: \(I = V/R = 12/2.0 = 6.0\ \text{A}\)

  4. Individual branch currents: \(I_1 = 12/6.0 = 2.0\ \text{A}\), \(I_2 = 12/3.0 = 4.0\ \text{A}\)

22.4.5 Example 5: Electrical power

A \(60\ \text{W}\) lamp operates at \(12\ \text{V}\). Find the current and resistance.

Solution:

  1. From \(P = IV\): \(I = P/V = 60/12 = 5.0\ \text{A}\)

  2. From \(R = V/I\): \(R = 12/5.0 = 2.4\ \Omega\)

22.5 Common Misconceptions

Common Misconceptions

  • Misconception: Current is “used up” by components. Correction: Current is the same around a series loop. Energy is transferred, not charge consumed.

  • Misconception: Voltage flows through a circuit. Correction: Voltage is a potential difference (energy per charge), not a flowing substance. Current flows; voltage is measured between points.

  • Misconception: Parallel circuits reduce current. Correction: Parallel branches reduce total resistance, which increases total current from the source.

  • Misconception: Electrons move at near light speed. Correction: Drift speed is very slow (mm/s). Energy transfer happens through the electric field, which propagates at near light speed.

  • Misconception: A bigger resistor gets more current. Correction: A bigger resistor opposes current more, reducing the current through it (in series, it reduces total current).

22.6 Practice Questions

22.6.1 Easy (2 marks)

A current of \(0.80\ \text{A}\) flows for \(25\ \text{s}\). How much charge passes a point?

  • Use \(Q = It\) correctly (1)
  • Correct value with units: \(Q = 0.80 \times 25 = 20\) C (1)

Answer: 20 C

22.6.2 Medium (4 marks)

Two resistors \(4.0\ \Omega\) and \(8.0\ \Omega\) are in series across \(12\ \text{V}\). Find the total current and the voltage across each resistor.

  • Find total resistance: \(R = 4.0 + 8.0 = 12\ \Omega\) (1)
  • Use \(I = V/R = 12/12 = 1.0\) A (1)
  • Voltage across \(4.0\ \Omega\): \(V_1 = IR = 1.0 \times 4.0 = 4.0\) V (1)
  • Voltage across \(8.0\ \Omega\): \(V_2 = IR = 1.0 \times 8.0 = 8.0\) V (1)

Answer: Current = 1.0 A; \(V_1 = 4.0\) V; \(V_2 = 8.0\) V

22.6.3 Hard (5 marks)

A device rated \(18\ \text{W}\) is connected to \(12\ \text{V}\). Determine the current, resistance, and energy used in 5 minutes.

  • Use \(P = IV\) to find \(I = P/V = 18/12 = 1.5\) A (1)
  • Use \(R = V/I = 12/1.5 = 8.0\ \Omega\) (1)
  • Alternative: \(R = V^2/P = 144/18 = 8.0\ \Omega\) (1)
  • Energy: \(E = Pt = 18 \times 300 = 5400\) J (1)
  • Correct units throughout (1)

Answer: \(I = 1.5\) A; \(R = 8.0\ \Omega\); \(E = 5400\) J (or 5.4 kJ)

22.7 Multiple Choice Questions

Test your understanding with these interactive questions:

22.8 Quick Quiz: Electric Circuits

Test your circuits knowledge with this timed quiz:

22.9 Extended Response Practice

22.10 Summary

Key Takeaways
  • Current is charge per unit time: \(I = Q/t\)
  • Voltage is energy per charge: \(V = W/q\)
  • Ohm’s law: \(V = IR\) for ohmic conductors
  • Series circuits: same current, resistances add
  • Parallel circuits: same voltage, \(1/R_{total} = 1/R_1 + 1/R_2 + ...\)
  • Power: \(P = IV = I^2R = V^2/R\)
  • Energy: \(E = Pt\)

22.11 Self-Assessment

Check your understanding:

After studying this section, you should be able to: