Use Thevenin equivalent circuits

WHAT is a Thevenin equivalent?

WHY does this work? Because the network is linear, the terminal voltage $V$ and terminal current $I$ obey a straight-line relationship: $V = V_{th} - I\,R_{th}$ A straight line needs only two parameters: an intercept ($V_{th}$) and a slope ($-R_{th}$). That is all the outside world can ever measure — so that is all the equivalent needs.

HOW to derive $V = V_{th} - I R_{th}$ from scratch

Why this step? Superposition is only legal because the circuit is linear — that is the single assumption the whole theorem rests on.

HOW to find the two numbers (recipe)

1. $V_{th}$ — disconnect the load. Compute the open-circuit voltage across the terminals using any method (node/mesh/divider).
2. $R_{th}$ —
   • No dependent sources? Deactivate all independent sources (V-source → wire, I-source → gap) and compute the equivalent resistance looking in.
   • Dependent sources present? You cannot deactivate those. Instead apply a test source $V_{test}$, find $I_{test}$, and take $R_{th}=V_{test}/I_{test}$.
3. Redraw: $V_{th}$ in series with $R_{th}$, then reconnect the load.

Worked Example 1 — a divider with a load

Source $V_S=12\text{ V}$, $R_1=4\,\Omega$ (series), $R_2=2\,\Omega$ (to ground) form the terminals across $R_2$. Find the Thevenin equivalent, then the current into a load $R_L=6\,\Omega$.

Step 1 — $V_{th}$ (open circuit). $V_{th}=V_S\frac{R_2}{R_1+R_2}=12\cdot\frac{2}{6}=4\text{ V}$ Why? With load removed, $R_1$ and $R_2$ form a plain voltage divider.

Step 2 — $R_{th}$. Kill $V_S$ (short it). Then $R_1$ and $R_2$ are in parallel as seen from the terminals: $R_{th}=R_1\| R_2=\frac{4\cdot2}{4+2}=1.33\,\Omega$ Why parallel? Shorting $V_S$ ties the top of $R_1$ to ground, so both resistors now run from the terminal node to ground — that is the parallel configuration.

Step 3 — reconnect load. $I_L=\frac{V_{th}}{R_{th}+R_L}=\frac{4}{1.33+6}=0.545\text{ A}$

Worked Example 2 — verify with short-circuit current

Take Example 1's original circuit and short the output terminals. The short bypasses $R_2$, so current from $V_S$ flows only through $R_1$: $I_{sc}=\frac{V_S}{R_1}=\frac{12}{4}=3\text{ A}$ Check: $R_{th}=\dfrac{V_{th}}{I_{sc}}=\dfrac{4}{3}=1.33\,\Omega.$ ✅ Matches Step 2.

Why this matters: two independent methods agree → confidence the equivalent is right.

Worked Example 3 — maximum power transfer

Using Example 1's equivalent, what $R_L$ draws the most power? Power delivered: $P_L=I_L^2R_L=\left(\frac{V_{th}}{R_{th}+R_L}\right)^2 R_L$ Differentiate and set $dP_L/dR_L=0$: $\frac{d}{dR_L}\frac{R_L}{(R_{th}+R_L)^2}=\frac{(R_{th}+R_L)^2-R_L\cdot2(R_{th}+R_L)}{(R_{th}+R_L)^4}=0$ $\Rightarrow R_{th}+R_L-2R_L=0\Rightarrow \boxed{R_L=R_{th}}$ So max power at $R_L=1.33\,\Omega$, giving $P_{max}=\dfrac{V_{th}^2}{4R_{th}}=\dfrac{16}{4(1.33)}=3\text{ W}$.

Why $R_L=R_{th}$? Thevenin collapses the whole circuit to one $R_{th}$; matching the load to it is the classic impedance-matching result.

Flashcards

Connections

• Norton Equivalent Circuits — dual form: $I_N=I_{sc}$, $R_N=R_{th}$; convert via $V_{th}=I_N R_{th}$.
• Superposition Theorem — the tool used to derive the terminal law.
• Maximum Power Transfer Theorem — direct consequence ($R_L=R_{th}$).
• Voltage and Current Dividers — used to compute $V_{th}$.
• Source Transformation — swap between Thevenin and Norton forms.
• Nodal and Mesh Analysis — general methods for finding $V_{oc}$ and $I_{sc}$.

Concept Map

Hinglish (regional understanding)