13 Jun 2026

Verification of Thevenin's and Norton's Theorems

practical pg-iv electronics network-analysis thevenin norton

Aim

To verify Thevenin’s and Norton’s theorems for a linear DC resistive network.

Apparatus

DC supply, resistors, load resistor, voltmeter, ammeter, potentiometer or decade resistance box, and connecting leads.

Experimental arrangement

The original network is replaced by its equivalent source and resistance, and the load current is compared in all arrangements.

Theory

Any two-terminal network made from linear elements and sources has a terminal voltage-current relation of the form

\[V=V_{th}-IR_{th}.\]

The constant term is found by opening the load, for which $I=0$ and $V=V_{oc}=V_{th}$. The slope is the resistance seen from the terminals. To determine it directly, independent voltage sources are replaced by their internal resistance, ideally a short circuit, and independent current sources by an open circuit. Dependent sources, if present, must remain active and a test source must be used. Thevenin’s theorem therefore replaces the complete linear network by the open-circuit voltage $V_{th}$ in series with $R_{th}$.

At the opposite limiting condition, shorting the terminals makes $V=0$ and gives

\[I_{sc}=I_N=\frac{V_{th}}{R_{th}}.\]

Norton’s theorem represents the same terminal line by current source $I_N$ in parallel with $R_N$. Source transformation requires

\[R_N=R_{th},\qquad V_{th}=I_NR_N.\]

When a finite load $R_L$ is connected, the two equivalent forms predict

\[I_L=\frac{V_{th}}{R_{th}+R_L}=I_N\frac{R_N}{R_N+R_L}.\]

The theorem is verified by showing that the original network and both equivalents produce the same load current and voltage, not merely by comparing $V_{th}$ and $I_N$. The short-circuit current should be measured only when the source and network can safely supply it; otherwise it is calculated from the equivalent resistance.

Observations

Arrangement	Load resistance (ohm)	Load current (mA)
original network	1000	3.96
Thevenin equivalent	1000	3.94
Norton equivalent	1000	3.95

Calculation

For the trial network take the measured Thevenin parameters as $V_{th}=5.00$ V and $R_{th}=260\,\Omega$. The predicted load current for $R_L=1000\,\Omega$ is

\[I_L=\frac{V_{th}}{R_{th}+R_L}=\frac{5.00}{260+1000}=3.968\times10^{-3}\,\text{A}=3.97\,\text{mA}.\]

The equivalent Norton current is

\[I_N=\frac{V_{th}}{R_{th}}=\frac{5.00}{260}=19.23\,\text{mA},\qquad R_N=R_{th}=260\,\Omega.\]

Using the Norton form gives

\[I_L=I_N\frac{R_N}{R_N+R_L}=19.23\frac{260}{1260}=3.97\,\text{mA}.\]

The observed currents, 3.96, 3.94, and 3.95 mA, differ from this ideal value by less than one percent.

Result

The load currents in the original, Thevenin-equivalent, and Norton-equivalent circuits agree within experimental error.

Viva Questions

1. What is $V_{th}$? The open-circuit voltage at the output terminals.
2. How is $R_{th}$ found? Deactivate independent sources and find the resistance seen from the terminals.
3. What is the Norton current? The short-circuit current at the output terminals.

Maxima Code

Download the Thevenin-Norton calculation.