For some purpose I need to derive the equivalent Norton circuit of an ideal independent voltage source. I found that it is impossible to do it. I describe the way I found it below and need comments, verifications, corrections, etc.
The ideal independent voltage source is represented as a Thevenin circuit.

  • Thevenin open-circuit voltage: For A and B being unconnected, it can easily be seen that the Thevenin open-circuit voltage, i.e. the voltage of A with respect to B, \${^{TH}}V_{OC}\$, is equal to the voltage of the voltage source.\begin{align*}{^{TH}V}_{OC}&=V\end{align*}
  • Thevenin short-circuit current: For A and B being short-circuited, there is no impedance in the circuit, so the Thevenin short-circuit current, i.e. the current from A to B, \${^{TH}}I_{SC}\$, is equal to infinity,\begin{align*}{^{TH}}I_{SC}&=\frac{V}{0}\cr&=\infty\end{align*}
  • Constructing its equivalent Norton circuit. The value of the of the ideal independent current source of a Norton circuit is the short-circuit current of its Thevenin counterpart,\begin{align*}I&={^{TH}}I_{SC}\cr&=\infty\end{align*} Its impedance is the impedance of its Thevenin counterpart, \$R=0\$.
  • Norton open-circuit voltage: Even no connection between A and B, the parallel impedance is zero then the voltage of A with respect to B, the Norton open-circuit voltage, \${^{NO}}V_{OC}\$, is a short-circuit voltage then it is zero,\begin{align*}{^{NO}}V_{OC}&=0\end{align*}
  • Norton short-circuit current: For A and B being short-circuited, the current from A to B, the Norton short-circuit current, \${^{NO}}I_{SC}\$, can easily be seen from the circuit that is equal to the current of the current source,\begin{align*}{^{NO}}I_{SC}&=I\cr&=\infty\end{align*}
  • The Thevenin open-circuit voltage (which represents the ideal independent voltage source), \${^{TH}}V_{OC}=V\$, is different from the Norton open-circuit voltage, \${^{NO}}V_{OC}=0\$. So an ideal independent voltage source has no equivalent Norton circuit. I can not convert an ideal independent voltage source to a Norton circuit.
    • 2
      \$\begingroup\$ If there is no R then I=0 and it cannot be converted \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Aug 11 at 12:15
    • 1
      \$\begingroup\$ Things that go to infinity or to zero often give rise to indeterminate cases. This is one of those. \$\endgroup\$ – Chu Aug 11 at 16:21

    An ideal voltage source can provide unlimited current, so it cannot be replaced with an ideal current source of any specified value. No matter what load current is drawn, the terminal voltage is always equal to the open-circuit voltage in this ideal case. So there is no meaningful value of short-circuit current, and Norton can't be applied.

    A practical voltage source has some non-zero source resistance, so in that case it is possible to calculate a Norton Equivalent current source. The resistor in the Thevenin model is actually required, it can't just be zeroed out.

    Division by zero is not a meaningful operation, because it leads to ambiguous or inconsistent results. Whenever you have a formula that has division by zero, more analysis is required.

    See https://math.stackexchange.com/questions/26445/division-by-0

    Instead of dividing by R=0, use an infinitesimally small R value and consider the Limit as R approaches 0. This models the practical voltage source approaching ideal. The product IR must equal V, so whichever non-zero R value is selected, it is always possible to calculate a specific I = V/R value that satisfies the Thevenin-Norton equivalence. However, there is a discontinuity at R=0. There is no meaningful way to produce an equivalent Norton current source for R=0.

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