23  Circuit Analysis

23.1 Syllabus inquiry question

  • How can conservation laws be used to analyse complex circuits?
Feynman Insight

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

A circuit only works because energy and charge are conserved. Kirchhoff’s laws are conservation principles written in electrical language.

23.2 Learning Objectives

  • Apply Kirchhoff’s current and voltage laws.
  • Use sign conventions to write loop equations.
  • Calculate equivalent resistance in mixed circuits.
  • Determine terminal voltage with internal resistance.
  • Solve currents in simple multi-loop circuits.

23.3 Content

23.3.1 Kirchhoff’s Current Law (KCL)

At any junction, the total current entering equals the total current leaving:

\[\sum I_{in} = \sum I_{out}\]

Conservation of Charge

KCL is a direct expression of charge conservation. Charge cannot accumulate at a junction—what flows in must flow out.

23.3.2 Interactive: Current at a Junction

Visualising current conservation at a junction:

Key observation: If 3.0 A enters and 1.5 A leaves through one branch, the other branch must carry 1.5 A.

23.3.3 Kirchhoff’s Voltage Law (KVL)

Around any closed loop, the sum of voltage gains and drops is zero:

\[\sum \Delta V = 0\]

Conservation of Energy

KVL is a direct expression of energy conservation. A charge gains energy from the source and loses it across components—the total around any closed path is zero.

23.3.4 Sign Conventions for KVL

When traversing a loop: - EMF source: + if traversed from − to +, − if from + to − - Resistor: − if traversing in the direction of current, + if opposite

23.3.5 Interactive: Voltage Around a Loop

23.3.6 Equivalent Resistance in Mixed Circuits

Many circuits combine series and parallel elements. Simplify step by step:

  1. Identify purely series or parallel groups
  2. Reduce each group to a single equivalent resistor
  3. Repeat until one equivalent resistance remains

23.3.7 Interactive: Mixed Circuit

A resistor in series with a parallel pair:

23.3.8 Internal Resistance and Terminal Voltage

Real sources have internal resistance (\(r\)) that causes voltage loss inside the source:

\[I = \frac{\mathcal{E}}{R + r}\]

Terminal voltage (voltage available to the external circuit):

\[V_{terminal} = \mathcal{E} - Ir\]

Understanding Internal Resistance
  • When \(I = 0\) (open circuit), \(V_{terminal} = \mathcal{E}\)
  • When current flows, some voltage is “lost” across internal resistance
  • Higher current → greater voltage drop → lower terminal voltage

23.3.9 Interactive: Battery with Internal Resistance

23.3.10 Strategy for Loop Analysis

For complex circuits with multiple loops:

  1. Assign loop currents and directions (clockwise or anticlockwise)
  2. Write KVL equations for each independent loop
  3. Use KCL to relate currents in shared branches
  4. Solve the simultaneous equations

23.4 Worked Examples

23.4.1 Example 1: Equivalent resistance

A \(2.0\ \Omega\) resistor is in series with a parallel pair of \(6.0\ \Omega\) and \(3.0\ \Omega\).

Solution:

  1. Find parallel equivalent: \(\frac{1}{R_p} = \frac{1}{6.0} + \frac{1}{3.0} = \frac{1}{6.0} + \frac{2}{6.0} = \frac{3}{6.0}\)

    \(R_p = 2.0\ \Omega\)

  2. Add series resistance: \(R_{total} = 2.0 + 2.0 = 4.0\ \Omega\)

23.4.2 Example 2: Internal resistance

A \(9.0\ \text{V}\) battery has internal resistance \(1.0\ \Omega\) and supplies a \(4.0\ \Omega\) load.

Solution:

  1. Total resistance: \(R_{total} = R + r = 4.0 + 1.0 = 5.0\ \Omega\)

  2. Current: \(I = \mathcal{E}/R_{total} = 9.0/5.0 = 1.8\ \text{A}\)

  3. Terminal voltage: \(V_{terminal} = \mathcal{E} - Ir = 9.0 - (1.8)(1.0) = 7.2\ \text{V}\)

  4. Alternatively: \(V_{terminal} = IR = 1.8 \times 4.0 = 7.2\ \text{V}\)

23.4.3 Example 3: KCL at a junction

At a junction, \(2.5\ \text{A}\) and \(1.5\ \text{A}\) flow in. One branch carries \(3.0\ \text{A}\) out. Find the current in the other branch.

Solution:

  1. Apply KCL: \(\sum I_{in} = \sum I_{out}\)

  2. \(2.5 + 1.5 = 3.0 + I_2\)

  3. \(I_2 = 4.0 - 3.0 = 1.0\ \text{A}\) (flowing out)

23.4.4 Example 4: Two-loop circuit

Left loop: \(12\ \text{V}\) source and \(4.0\ \Omega\) resistor. Right loop: \(6.0\ \text{V}\) source and \(3.0\ \Omega\) resistor. The loops share a \(2.0\ \Omega\) resistor. Find the currents.

Solution:

Let mesh currents be \(I_1\) (left, clockwise) and \(I_2\) (right, clockwise).

  1. Left loop KVL: \(12 - 4I_1 - 2(I_1 - I_2) = 0\)

    \(12 - 6I_1 + 2I_2 = 0\) … (1)

  2. Right loop KVL: \(6 - 3I_2 - 2(I_2 - I_1) = 0\)

    \(6 + 2I_1 - 5I_2 = 0\) … (2)

  3. From (1): \(I_2 = (6I_1 - 12)/2 = 3I_1 - 6\)

  4. Substitute into (2): \(6 + 2I_1 - 5(3I_1 - 6) = 0\)

    \(6 + 2I_1 - 15I_1 + 30 = 0\)

    \(-13I_1 = -36\)

    \(I_1 = 2.77\ \text{A}\)

  5. \(I_2 = 3(2.77) - 6 = 2.31\ \text{A}\)

  6. Current through shared resistor: \(I_1 - I_2 = 0.46\ \text{A}\) (downward)

23.5 Common Misconceptions

Common Misconceptions

  • Misconception: KCL and KVL are additional rules separate from conservation laws. Correction: They directly express conservation of charge (KCL) and energy (KVL).

  • Misconception: The sign of voltage drops doesn’t matter in KVL. Correction: Direction and sign conventions are essential for correct equations. Be consistent.

  • Misconception: Internal resistance can be ignored. Correction: Internal resistance affects current and reduces terminal voltage, especially at high currents.

  • Misconception: Equivalent resistance is always larger than individual resistances. Correction: Parallel connections reduce total resistance below the smallest individual resistance.

  • Misconception: Complex circuits can’t be solved without mesh analysis. Correction: Many circuits can be simplified using series/parallel rules before applying Kirchhoff’s laws.

23.6 Practice Questions

23.6.1 Easy (2 marks)

At a junction, \(0.40\ \text{A}\) enters. One branch carries \(0.15\ \text{A}\). Find the current in the other branch.

  • Apply KCL: \(I_{in} = I_1 + I_2\) (1)
  • Correct value: \(I_2 = 0.40 - 0.15 = 0.25\) A (1)

Answer: 0.25 A

23.6.2 Medium (4 marks)

A \(12\ \text{V}\) source with \(0.50\ \Omega\) internal resistance supplies a \(5.0\ \Omega\) load. Find the current and terminal voltage.

  • Total resistance: \(R = 5.0 + 0.50 = 5.5\ \Omega\) (1)
  • Current: \(I = 12/5.5 = 2.18\) A (or 2.2 A) (1)
  • Terminal voltage: \(V = \mathcal{E} - Ir = 12 - (2.18)(0.50) = 10.9\) V (1)
  • OR: \(V = IR = 2.18 \times 5.0 = 10.9\) V (1)
  • Correct units (1)

Answer: Current = 2.2 A; Terminal voltage = 10.9 V (or 11 V)

23.6.3 Hard (5 marks)

A circuit has two loops that share a \(1.0\ \Omega\) resistor. The left loop has a \(10\ \text{V}\) source and \(2.0\ \Omega\) resistor. The right loop has a \(6.0\ \text{V}\) source and \(3.0\ \Omega\) resistor. Use KVL to determine the loop currents.

  • Set up left loop KVL: \(10 - 2I_1 - 1(I_1 - I_2) = 0\) (1)
  • Set up right loop KVL: \(6 - 3I_2 - 1(I_2 - I_1) = 0\) (1)
  • Simplify to: \(10 - 3I_1 + I_2 = 0\) and \(6 + I_1 - 4I_2 = 0\) (1)
  • Solve simultaneous equations correctly (1)
  • Answers: \(I_1 \approx 4.2\) A, \(I_2 \approx 2.5\) A (accept reasonable rounding) (1)

Answer: \(I_1 = 4.2\) A (left loop); \(I_2 = 2.5\) A (right loop)

23.7 Multiple Choice Questions

Test your understanding with these interactive questions:

23.8 Summary

Key Takeaways
  • KCL (junction rule): \(\sum I_{in} = \sum I_{out}\) — conservation of charge
  • KVL (loop rule): \(\sum \Delta V = 0\) — conservation of energy
  • Mixed circuits: simplify using series/parallel rules first
  • Internal resistance: \(V_{terminal} = \mathcal{E} - Ir\)
  • Loop analysis: assign currents, write KVL for each loop, solve simultaneous equations

23.9 Self-Assessment

Check your understanding:

After studying this section, you should be able to: