Network theorems

Network theorems let us analyse complicated linear circuits without writing every mesh or nodal equation. Superposition handles several sources one at a time; Thevenin and Norton reduce an entire network, as seen from two terminals, to one source and one impedance; and the maximum power transfer theorem tells what load gets the most power. They apply to linear networks and are used constantly in both DC and AC (phasor) analysis.

Key concepts

• Superposition theorem: in a linear network with several independent sources, the response (current or voltage) in any element is the algebraic sum of the responses caused by each source acting alone, with the other independent sources "killed":
  • Ideal voltage source → replaced by a short circuit (keeping its internal resistance).
  • Ideal current source → replaced by an open circuit.
  • Dependent sources are not removed.
  • Applies to voltage and current, not to power (power is a square-law quantity). Valid only for linear circuits.
• Thevenin's theorem: any linear two-terminal network can be replaced by a voltage source $V_{th}$ in series with a resistance (impedance) $R_{th}$.
  • $V_{th}$ = open-circuit voltage across the terminals.
  • $R_{th}$ = resistance seen from the terminals with independent sources killed; or $R_{th} = V_{oc}/I_{sc}$; or, with dependent sources, apply a test source and find V/I.
  • Load current: $I_L = \frac{V_{th}}{R_{th} + R_L}$. Very useful when the load changes.
• Norton's theorem: the same network can be replaced by a current source $I_N$ (short-circuit current) in parallel with $R_N$.
  • $R_N = R_{th}$; $I_N = V_{th}/R_{th}$; conversion between Thevenin and Norton forms is a source transformation.
  • Load current: $I_L = I_N\frac{R_N}{R_N + R_L}$.
• Maximum power transfer theorem:
  • DC/resistive: maximum power is delivered to a load when $R_L = R_{th}$.
  • AC: when the load impedance is the complex conjugate of the source impedance, $Z_L = Z_{th}^*$ (i.e. $R_L = R_{th}$, $X_L = -X_{th}$). If only a resistive load is allowed, $R_L = |Z_{th}|$.
  • At maximum power transfer the efficiency is only 50 % — half the power is lost in the source resistance. So power systems do not operate at this condition (they aim for high efficiency); it matters in communication and electronic circuits (antenna matching, amplifiers, speakers).
• Other related theorems (useful for MCQs): reciprocity (in a linear bilateral network, swapping source and ammeter positions gives the same reading), Millman's theorem (parallel voltage sources combined into one), Tellegen's theorem (sum of powers in all branches = 0), compensation and substitution theorems.
• Conditions for applicability: Thevenin, Norton and superposition need linearity; reciprocity also needs bilateral elements and a single source.

High-yield points

• Superposition: short ideal voltage sources, open ideal current sources, keep dependent sources; not valid for power.
• $V_{th}$ = open-circuit voltage; $I_N$ = short-circuit current; $R_{th} = R_N = V_{oc}/I_{sc}$.
• Maximum power (DC): $R_L = R_{th}$, $P_{max} = V_{th}^2/(4R_{th})$.
• AC maximum power: $Z_L = Z_{th}^*$; for a purely resistive load $R_L = |Z_{th}|$.
• Efficiency at maximum power transfer = 50 %.

Formulas

$I_L = \frac{V_{th}}{R_{th}+R_L},\qquad I_N = \frac{V_{th}}{R_{th}},\qquad R_N = R_{th}$

$P_{max} = \frac{V_{th}^2}{4R_{th}}\quad (\text{at } R_L = R_{th}),\qquad \eta = \frac{R_L}{R_L + R_{th}} = 50\,\%$

Worked example

A 12 V source with 4 Ω in series feeds a network where a 12 Ω resistor is connected across the terminals A-B (parallel with the load). Find Thevenin equivalent at A-B, maximum power and the load for it.

1. $V_{th}$ = open-circuit voltage = $12\times\frac{12}{4 + 12} = 9$ V.
2. $R_{th}$ = 4 ∥ 12 = 48/16 = 3 Ω.
3. Maximum power at $R_L = 3$ Ω: $P_{max} = \frac{9^2}{4\times3} = \frac{81}{12} = 6.75$ W.
4. Norton: $I_N = 9/3 = 3$ A (= 12 V / 4 Ω short-circuit current ✓), $R_N = 3$ Ω.

Exam tips

• Superposition cannot be used to compute power directly.
• Kill voltage sources by short circuit, current sources by open circuit; keep dependent sources.
• $R_{th} = V_{oc}/I_{sc}$ — handy shortcut, also works with dependent sources.
• Maximum power transfer → 50 % efficiency; AC case needs conjugate matching.
• Norton and Thevenin resistances are equal.

Exam-style questions

From Calmon's question bank, which has 25 questions on this topic.

1. An AC source has Thevenin impedance $Z_{th} = (3 + j4)\,\Omega$. If the load is a pure resistance, the value giving maximum power transfer is:

• A. $7\,\Omega$
• B. $4\,\Omega$
• C. $5\,\Omega$
• D. $3\,\Omega$

Answer: C. For a purely resistive load, maximum power occurs at $R_L = |Z_{th}| = \sqrt{3^2+4^2} = 5\,\Omega$.

2. A 24 V source in series with $6\,\Omega$ is connected across a $3\,\Omega$ resistor; a $4\,\Omega$ resistor then connects the top of the $3\,\Omega$ resistor to terminal a (terminal b is the bottom). The Thevenin equivalent at a–b is:

• A. 24 V, $13\,\Omega$
• B. 8 V, $2\,\Omega$
• C. 8 V, $6\,\Omega$
• D. 16 V, $6\,\Omega$

Answer: C. $V_{th} = 24\times\dfrac{3}{6+3} = 8$ V (no current in the $4\,\Omega$); $R_{th} = 4 + (6\parallel3) = 4 + 2 = 6\,\Omega$.

Practise AElE01 questions →