Thevenin and Norton equivalents

Thevenin's theorem. Looking into any pair of terminals of a linear resistive network that contains independent sources, the network behaves exactly like one voltage source $V_{th}$ in series with one resistance $R_{th}$, as far as anything connected externally at those terminals is concerned. Two measurements pin the equivalent down completely:

where: $V_{oc}$ is the open-circuit voltage across the terminals with the external load removed, $I_{sc}$ is the short-circuit current that would flow if the terminals were shorted, and the sources-deactivated resistance is found by shorting every independent voltage source and opening every independent current source, then computing the resulting resistance looking into the terminals — exactly the same deactivation rule used in the superposition theorem.

Norton's theorem. The identical network, seen from the same two terminals, is equally well replaced by a single current source $I_N$ in parallel with a resistance $R_N$:

where: $R_N$ is numerically identical to $R_{th}$ — Thevenin and Norton describe the same physical network, only re-drawn — and $I_N$ is exactly the short-circuit current $I_{sc}$ from the Thevenin picture. Because $V=IR$ links a series source-plus-resistance to a parallel current-source-plus-resistance in general, converting between the two forms is a single division or multiplication once $R_{th}(=R_N)$ is known; no new circuit analysis is needed. The figure below shows the reduction: a messy original network collapses to a two-element Thevenin form (or, equivalently, a two-element Norton form) at the same pair of terminals A–B.

Maximum power transfer. A load $R_L$ connected across a Thevenin equivalent draws current $I = V_{th}/(R_{th}+R_L)$, so it dissipates power $P_L = I^2R_L$. Treating $R_L$ as the variable and maximising $P_L$ gives the single best-known corollary of this pair of theorems:

where: $R_L$ is the load resistance the designer can choose, $R_{th}$ is fixed by the source network, and $P_{max}$ is the largest power any resistive load can ever extract from that network. Choosing $R_L$ any larger or smaller than $R_{th}$ always delivers strictly less power to the load — the full algebra is worked out below.

A network has a $24\ \text{V}$ source in series with a $6\ \Omega$ resistor, feeding terminals A–B across which a $12\ \Omega$ resistor is permanently connected in parallel. Find the Thevenin and Norton equivalents at A–B, the current through a load $R_L=4\ \Omega$ connected there, and the load resistance and power for maximum power transfer.

Step 1. Find $V_{th}$. Remove any external load at A–B; the circuit left is just the $24\ \text{V}$ source with series $6\ \Omega$ feeding the $12\ \Omega$ resistor, a simple series loop: $I = \dfrac{24}{6+12} = \dfrac{24}{18} = 1.333\ \text{A}$, so $V_{th} = I\times12 = 1.333\times12 = 16\ \text{V}$ — the voltage across the $12\ \Omega$ resistor, which equals the open-circuit terminal voltage.

Step 2. Find $R_{th}$. Deactivate the $24\ \text{V}$ source (replace it with a short). Looking into A–B, the $6\ \Omega$ and $12\ \Omega$ resistors are now in parallel: $R_{th} = \dfrac{6\times12}{6+12} = \dfrac{72}{18} = 4\ \Omega$. The Thevenin equivalent is $V_{th}=16\ \text{V}$ in series with $R_{th}=4\ \Omega$.

Step 3. Convert to Norton form. $R_N = R_{th} = 4\ \Omega$, and $I_N = V_{th}/R_{th} = 16/4 = 4\ \text{A}$ — a $4\ \text{A}$ current source in parallel with $4\ \Omega$. Check: shorting the Norton terminals gives $I_{sc}=4\ \text{A}$, matching $I_{sc}=V_{th}/R_{th}=16/4=4\ \text{A}$ from the Thevenin side.

Step 4. Connect $R_L=4\ \Omega$ to the Thevenin equivalent: total resistance $R_{th}+R_L=4+4=8\ \Omega$, so $I_L = \dfrac{V_{th}}{R_{th}+R_L} = \dfrac{16}{8} = 2\ \text{A}$.

Step 5. Apply the maximum-power-transfer result from the derivation: the load draws maximum power when $R_L=R_{th}=4\ \Omega$ — which is exactly the $R_L$ used in Step 4, so this network's answer doubles as its own maximum-power case. $P_{max} = \dfrac{V_{th}^2}{4R_{th}} = \dfrac{16^2}{4\times4} = \dfrac{256}{16} = \mathbf{16\ W}$.

Competitive stretch

Cross-check $P_{max}$ directly from Step 4's current: $P_L = I_L^2R_L = 2^2\times4 = 16\ \text{W}$, matching Step 5 exactly. Now compute the efficiency at this operating point: the source's total generated power is $P_{total}=I_L^2(R_{th}+R_L)=2^2\times8=32\ \text{W}$, so the load receives $16/32=50\%$ of it — at $R_L=R_{th}$, the source resistance itself always dissipates the other half, so maximum power transfer never means maximum efficiency.

• Computing $R_{th}$ without deactivating the independent sources first — $R_{th}$ is specifically the sources-off resistance seen from the terminals, not the resistance with sources still live.
• Deactivating sources the wrong way — voltage sources become short circuits, current sources become open circuits, the same rule used in superposition.
• Confusing $V_{th}$ with the terminal voltage under load — $V_{th}$ is the open-circuit voltage, measured only with the external load removed.
• Assuming $R_L=R_{th}$ maximises efficiency as well as power — exactly half the total generated power is lost in $R_{th}$ itself at that operating point, so efficiency there is only 50 percent.

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