Linear Circuit Analysis


Norton's' Theorem

Norton's theorem states that any two-terminal linear circuit containing only voltage sources, current sources and resistors (or impedances in the case of AC circuits) can be replaced by an equivalent combination of a current source $I_{N}$ in parallel with a resistor $R_{N}$ (or impedance $Z_{N}$ in the case of AC circuits).

IN RN A B
Fig. 1. Any linear circuit with two terminals can be replaced with the Norton equivalent circuit.

Norton's theorem and its dual, Thévenin's theorem, are widely used to simplify circuit analysis and study a circuit's initial-condition and steady-state response. Norton's theorem may in some cases be more convenient to use than Kirchhoff's circuit laws when analyzing linear circuits.

Calculating the Norton equivalent circuit

Calculating the Norton equivalent circuit, or simply the Norton equivalent of a circuit with two terminals means computing the Norton current and Norton resistance of the circuit.

The Norton current can be computed using any of the following methods:

  • $I_{N} = I_{s.c}$ - the Norton current is equal to the short-circuit current at the output terminals of the original circuit. To compute $I_{s.c.}$ one can use any of the methods of linear circuit analysis such as nodal analysis, mesh analysis, or current and voltage division.
  • If the Thévenin voltage $V_{Th}$ and Thévenin resistance $R_{Th}$ are known and $R_{Th} \neq 0$, one can calculate the Norton current using $I_{N} = V_{Th}/R_{Th}$ .

The Norton resistance can be computed using any of the following methods:

  • $R_{N}$ can be computed by deactivating all independent sources and using the test voltage or test current methods using Ohm's law, $R_{N}=\frac{V_{test}}{I_{test}}$. To use the test voltage method, we connect a test voltage source at the output terminals (usually, but not necessarily, taken as $V_{test}=1~V$) and compute current $I_{test}$. To use the test current method, we connect a test current source at the output terminals (usually, but not necessarily, taken as $I_{test}=1~A$) and compute voltage $V_{test}$.
  • If the circuit does not contain any dependent sources, $R_{N}$ can be computed by deactivating all independent sources and looking at the resistance seen from the output terminals (in this case, $R_{N}$ can often be computed using series and parallel simplifications of resistors).
  • If $I_N$ and $V_{Th}$ are known and $I_{N} \neq 0$, one can calculate the Norton resistance using $R_{N} = \frac{V_{Th}}{I_{N}}$. However, please note that if $I_{N}=0$, we cannot divide $\frac{V_{Th}}{I_{N}}$; this does not mean that $R_{N}$ is equal to $0$ (or $\infty$), but it only means we need to use other methods to compute $R_{N}$.

Table 1 shows possible ways to compute $R_{Th}$ and $I_{N}$. However, please note that these are not the only ways to compute the Norton components, and you can often come up with alternative ways.

Table 1. How to compute $I_{N}$ and $I_{N}$ based on the elements that exist in the circuit.
If the circuits contains only... You should...
Resistors
  1. Compute $R_{N}$ using resistor simplification techniques. Alternatively, (for instance, if you end up with delta-wye transformation and you forgot how to do it...) you can can use the $V_{test}/I_{test}$ method.1)
  2. $I_{N} = 0$
Resistors and independent sources
  1. Deactivate all the independent sources in the circuit2) and comptue $R_{N}$ using resistor simplification techniques or the $V_{test}/I_{test}$ method.1)
  2. Compute the short-circuit current $I_{sc}$, which gives you $I_{N} = I_{sc}$. Alternatively, if the Norton resistance is not zero, you can compute the open-circuit voltage $V_{oc}$, and set $I_{N} = V_{oc}/R_{N}$
Resistors and dependent sources
  1. Compute $R_{N}$ using the $V_{test}/I_{test}$ method.1)
  2. $I_{N} = 0$
Resistors and independent and dependent sources
  1. Deactivate all the independent sources in the circuit2) and compute $R_{N}$ using the $V_{test}/I_{test}$ method.1)
  2. Compute the short-circuit current $I_{sc}$, which gives you $I_{N} = I_{sc}$. Alternatively, if the Norton resistance is not zero, you can compute the open-circuit voltage $V_{oc}$, and set $I_{N} = V_{oc}/R_{N}$

1) You can use either the test voltage or test current method. They should both give you the same result.
2) To deactivate the independent sources in the circuit, you need to replace all the independent voltage sources with short circuits (wires) and all the independent current sources with open circuits (remove them). Make sure you do not modify the dependent sources.

Notes
  • Sometimes, the Norton voltage and Norton resistance can be computed simultaneously by performing successive source transformations and using series and parallel combinations of resistors, voltage sources and current sources.
  • The Noton current of a circuit that does not contain any independent voltage and current sources is always 0.
  • The Thévenin and Norton resistances, the Thévenin voltage and the Norton current satisfy the following relationships: $$\begin{equation}R_{Th}=R_N\end{equation}$$ $$\begin{equation}V_{Th}=I_N R_{Th}\end{equation}$$Because of these relationships, it is usually necessary to find only two quantities because the other two can be calculated afterwards.
  • To deactivate a current source we remove it from the circuit (replace it with open-circuit or break); to deactivate a voltage source we replace it with a short-circuit (or wire).
  • The above techniques can also be used to compute the Norton current and Norton impedance $Z_{N}$ in linear AC circuits.
Sample Solved Problems
  • DC Norton & Thévenin circuts (analytical)
    6050
    6051
    6052
    6053
  • DC Norton & Thévenin circuits (numerical)
    6000
See also
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