# [Solved] Thevenin Equivalent Circuit of Ideal Independent Current Source

aharisjo Asks: Thevenin Equivalent Circuit of Ideal Independent Current Source
This is dual to <Norton Equivalent Circuit of Ideal Independent Voltage Source>. I have wrote it to answer a problem <click here> but I need to make it independent which requires corrections, verifications, comments, etc. In order to follow Norton to Thevenin convertion procedure the ideal independent current source to be converted is represented as a Norton circuit.

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} • Norton open-circuit voltage: For A and B being disconnected, no current flows in the circuit. The internal impedance of the current source is infinite (open circuit) so the open-circuit voltage, \$^{NO}V_{OC}\$, the voltage of A with respect to B is undetermined.begin{align*}^{NO}V_{OC}&=I_RRcr&=0timesinftycr&=textrm{NaN};textrm{(undetermined)}end{align*}
• Norton short-circuit current: For A and B being short-circuited, the short-circuit current, \$^{NO}I_{SC}\$, the current from A to B, is equal to the current of the independent current source,begin{equation*}^{NO}I_{SC}=Iend{equation*}

Then we can construct its equivalent Thevenin circuit. The voltage of the ideal independent voltage source of a Thevenin circuit is the open circuit voltage of its Norton counterpart,begin{align*}V&=^{NO}V_{OC}cr&=textrm{NaN};textrm{(undetermined)}end{align*} Its impedance is the impedance of its Norton counterpart, \$R=infty\$. • Thevenin open-circuit voltage: For A and B being open-circuited, there is no source which determines the open-circuit voltage, \$^{TH}V_{OC}\$, the voltage of A with respect to B. So it is undetermined.begin{align*}^{TH}V_{OC}&=textrm{NaN};textrm{(undetermined)}end{align*}
• Thevenin short-circuit current: For A and B being short-circuited, there is no source to determine the short-circuit current, \$^{TH}I_{SC}\$, the current from A to B, so it is zero or undetermined,begin{equation*}^{TH}I_{SC}=0;textrm{or};^{TH}I_{SC}=textrm{NaN}end{equation*}

The Norton short-circuit current (representing an ideal independent current source), \$^{TH}I_{SC}=I\$, is different from the short-circuit of its Thevenin equivalent, \$^{TH}I_{SC}=0\$ or \$^{TH}I_{SC}=textrm{NaN}\$. It implies that an ideal independent current source has no equivalent Thevenin circuit. So I can not convert an ideal independent current source to a Thevenin circuit. I need corrections, verifications, comments, etc.

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