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Let's treat the op-amp and its periphery as a "black box", leaving only the load resistor \$R_L\$ outside:

^{simulate this circuit – Schematic created using CircuitLab}

The "black boxes" above each have two terminals, X and Y, and you have no idea what's inside the box. Each box has some internal resistance, which we aim to measure.

We could measure it by changing the external resistance \$R_L\$, or by applying some external potential difference between X and Y, or even by forcing some current through the box. Whatever method we use, we shall measure the resulting change in voltage \$\Delta V_{XY}\$, and change in current \$\Delta I_O\$. Effective resistance \$R_{XY}\$ between X and Y inside the box will be, according to Ohm's law:

$$ R_{XY} = \frac{\Delta V_{XY}}{\Delta I_O} $$

For box B, you'd find that there's no change in voltage, since its internal voltage source will strongly oppose any change in \$V_{XY}\$, but current \$I_O\$ is able to rise and fall freely:

$$ R_{XY(B)} = \frac{\Delta V_{XY}}{\Delta I_O} = \frac{0}{\Delta I_O} = 0\Omega $$

In other words, box B (or any ideal voltage source) is said to have zero impedance.

Box C is a current source, and we would expect it to strongly oppose changes in current through it, \$\Delta I_O=0\$, while freely permitting \$V_{XY}\$ to change:

$$ R_{XY(C)} = \frac{\Delta V_{XY}}{0} = \infty \Omega $$

Ideal current sources, therefore, are said to have infinite source impedance.

Box D will have an impedance somewhere in between zero and infinity, because in this case it would be possible to impose any arbitrary voltage between X and Y, and it's also possible to inject an arbitrary current through the box. With either "perturbation" method, if you plotted \$I_O\$ on the horizontal axis, vs. its corresponding \$V_{XY}\$, the slope of that graph is \$R_{XY}\$. Just by inspection, you could predict that this slope would be:

$$ R_{XY(D)} = \frac{\Delta V_{XY}}{\Delta I_O} = 100\Omega $$

So, to answer your question, it remains to be determined which of the boxes B, C and D has behaviour which most closely resembles the behaviour of box A.

We are permitted to attach any load \$R_L\$ we desire and you could even apply some arbitrary potential difference between X and Y, and it would still keep \$I_O\$ constant. The op-amp will adjust \$V_{XY}\$ to whatever value is necessary to keep \$I_O\$ fixed. Thus \$\Delta I_O = 0\$, so:

$$ R_{XY(A)} = \frac{\Delta V_{XY}}{0} = \infty \Omega $$

This is consistent with the behaviour of the ideal current source in box C.

It's true that the op-amp is indeed a voltage source, using voltages to obtain this behaviour, but in the context of a two-terminal "black box", for which we only have access to the exposed terminals, nodes X and Y, any external "observer" circuitry would perceive it to behave like a current source, for which the impedance inside the box is effectively infinite. No amount of voltage applied, and no impedance of load attached (op-amp constraints notwithstanding, such as limited output voltage range and current limits) will result in a change in \$I_O\$.

Let's treat the op-amp and its periphery as a "black box", leaving only the load resistor \$R_L\$ outside:

^{simulate this circuit – Schematic created using CircuitLab}

The "black boxes" above each have two terminals, X and Y, and you have no idea what's inside the box. Each box has some internal resistance, which we aim to measure.

We could measure it by changing the external resistance \$R_L\$, or by applying some external potential difference between X and Y, or even by forcing some current through the box. Whatever method we use, we shall measure the resulting change in voltage \$\Delta V_{XY}\$, and change in current \$\Delta I_O\$. Effective resistance \$R_{XY}\$ between X and Y inside the box will be, according to Ohm's law:

$$ R_{XY} = \frac{\Delta V_{XY}}{\Delta I_O} $$

For box B, you'd find that there's no change in voltage, since its internal voltage source will strongly oppose any change in \$V_{XY}\$, but current \$I_O\$ is able to rise and fall freely:

$$ R_{XY(B)} = \frac{\Delta V_{XY}}{\Delta I_O} = \frac{0}{\Delta I_O} = 0\Omega $$

In other words, box B (or any ideal voltage source) is said to have zero impedance.

Box C is a current source, and we would expect it to strongly oppose changes in current through it, \$\Delta I_O=0\$, while freely permitting \$V_{XY}\$ to change:

$$ R_{XY(C)} = \frac{\Delta V_{XY}}{0} = \infty \Omega $$

Ideal current sources, therefore, are said to have infinite source impedance.

Box D will have an impedance somewhere in between zero and infinity, because in this case it would be possible to impose any arbitrary voltage between X and Y, and it's also possible to inject an arbitrary current through the box. With either "perturbation" method, if you plotted \$I_O\$ on the horizontal axis, vs. its corresponding \$V_{XY}\$, the slope of that graph is \$R_{XY}\$. Just by inspection, you could predict that this slope would be:

$$ R_{XY(D)} = \frac{\Delta V_{XY}}{\Delta I_O} = 100\Omega $$

For more information, and a graphical treatment of using voltage and current changes, and the slope of the characteristic V-I graph to represent impedance, see this answer.

To answer your question, it remains to be determined which of the boxes B, C and D has behaviour which most closely resembles the behaviour of box A.

We are permitted to attach any load \$R_L\$ we desire and you could even apply some arbitrary potential difference between X and Y, and it would still keep \$I_O\$ constant. The op-amp will adjust \$V_{XY}\$ to whatever value is necessary to keep \$I_O\$ fixed. Thus \$\Delta I_O = 0\$, so:

$$ R_{XY(A)} = \frac{\Delta V_{XY}}{0} = \infty \Omega $$

This is consistent with the behaviour of the ideal current source in box C.

It's true that the op-amp is indeed a voltage source, using voltages to obtain this behaviour, but in the context of a two-terminal "black box", for which we only have access to the exposed terminals, nodes X and Y, any external "observer" circuitry would perceive it to behave like a current source, for which the impedance inside the box is effectively infinite. No amount of voltage applied, and no impedance of load attached (op-amp constraints notwithstanding, such as limited output voltage range and current limits) will result in a change in \$I_O\$.

Let's treat the op-amp and its periphery as a "black box", leaving only the load resistor \$R_L\$ outside:

^{simulate this circuit – Schematic created using CircuitLab}

The "black boxes" above each have two terminals, X and Y, and you have no idea what's inside the box. Each box has some internal resistance, which we aim to measure.

We could measure it by changing the external resistance \$R_L\$, or by applying some external potential difference between X and Y, or even by forcing some current through the box. Whatever method we use, we shall measure the resulting change in voltage \$\Delta V_{XY}\$, and change in current \$\Delta I_O\$. Effective resistance \$R_{XY}\$ between X and Y inside the box will be, according to Ohm's law:

$$ R_{XY} = \frac{\Delta V_{XY}}{\Delta I_O} $$

For box B, you'd find that there's no change in voltage, since its internal voltage source will strongly oppose any change in \$V_{XY}\$, but current \$I_O\$ is able to rise and fall freely:

$$ R_{XY(B)} = \frac{\Delta V_{XY}}{\Delta I_O} = \frac{0}{\Delta I_O} = 0\Omega $$

In other words, box B (or any ideal voltage source) is said to have zero impedance.

Box C is a current source, and we would expect it to strongly oppose changes in current through it, \$\Delta I_O=0\$, while freely permitting \$V_{XY}\$ to change:

$$ R_{XY(C)} = \frac{\Delta V_{XY}}{0} = \infty \Omega $$

Ideal current sources, therefore, are said to have infinite source impedance.

Box D will have an impedance somewhere in between zero and infinity, because in this case it would be possible to impose any arbitrary voltage between X and Y, and it's also possible to inject an arbitrary current through the box. With either "perturbation" method, if you plotted \$I_O\$ on the horizontal axis, vs. its corresponding \$V_{XY}\$, the slope of that graph is \$R_{XY}\$. Just by inspection, you could predict that this slope would be:

$$ R_{XY(D)} = \frac{\Delta V_{XY}}{\Delta I_O} = 100\Omega $$

So, to answer your question, it remains to be determined which of the boxes B, C and D has behaviour which most closely resembles the behaviour of box A.

We are permitted to attach any load \$R_L\$ we desire and you could even apply some arbitrary potential difference between X and Y, and it would still keep \$I_O\$ constant. The op-amp will adjust \$V_{XY}\$ to whatever value is necessary to keep \$I_O\$ fixed. Thus \$\Delta I_O = 0\$, so:

$$ R_{XY(A)} = \frac{\Delta V_{XY}}{0} = \infty \Omega $$

This is consistent with the behaviour of the ideal current source in box C.

It's true that the op-amp is indeed a voltage source, using voltages to obtain this behaviour, but in the context of a two-terminal "black box", for which we only have access to the exposed terminals, nodes X and Y, any external "observer" circuitry would perceive it to behave like a current source, for which the impedance inside the box is effectively infinite.

Let's treat the op-amp and its periphery as a "black box", leaving only the load resistor \$R_L\$ outside:

^{simulate this circuit – Schematic created using CircuitLab}

The "black boxes" above each have two terminals, X and Y, and you have no idea what's inside the box. Each box has some internal resistance, which we aim to measure.

We could measure it by changing the external resistance \$R_L\$, or by applying some external potential difference between X and Y, or even by forcing some current through the box. Whatever method we use, we shall measure the resulting change in voltage \$\Delta V_{XY}\$, and change in current \$\Delta I_O\$. Effective resistance \$R_{XY}\$ between X and Y inside the box will be, according to Ohm's law:

$$ R_{XY} = \frac{\Delta V_{XY}}{\Delta I_O} $$

For box B, you'd find that there's no change in voltage, since its internal voltage source will strongly oppose any change in \$V_{XY}\$, but current \$I_O\$ is able to rise and fall freely:

$$ R_{XY(B)} = \frac{\Delta V_{XY}}{\Delta I_O} = \frac{0}{\Delta I_O} = 0\Omega $$

In other words, box B (or any ideal voltage source) is said to have zero impedance.

Box C is a current source, and we would expect it to strongly oppose changes in current through it, \$\Delta I_O=0\$, while freely permitting \$V_{XY}\$ to change:

$$ R_{XY(C)} = \frac{\Delta V_{XY}}{0} = \infty \Omega $$

Ideal current sources, therefore, are said to have infinite source impedance.

Box D will have an impedance somewhere in between zero and infinity, because in this case it would be possible to impose any arbitrary voltage between X and Y, and it's also possible to inject an arbitrary current through the box. With either "perturbation" method, if you plotted \$I_O\$ on the horizontal axis, vs. its corresponding \$V_{XY}\$, the slope of that graph is \$R_{XY}\$. Just by inspection, you could predict that this slope would be:

$$ R_{XY(D)} = \frac{\Delta V_{XY}}{\Delta I_O} = 100\Omega $$

So, to answer your question, it remains to be determined which of the boxes B, C and D has behaviour which most closely resembles the behaviour of box A.

We are permitted to attach any load \$R_L\$ we desire and you could even apply some arbitrary potential difference between X and Y, and it would still keep \$I_O\$ constant. The op-amp will adjust \$V_{XY}\$ to whatever value is necessary to keep \$I_O\$ fixed. Thus \$\Delta I_O = 0\$, so:

$$ R_{XY(A)} = \frac{\Delta V_{XY}}{0} = \infty \Omega $$

This is consistent with the behaviour of the ideal current source in box C.

It's true that the op-amp is indeed a voltage source, using voltages to obtain this behaviour, but in the context of a two-terminal "black box", for which we only have access to the exposed terminals, nodes X and Y, any external "observer" circuitry would perceive it to behave like a current source, for which the impedance inside the box is effectively infinite. No amount of voltage applied, and no impedance of load attached (op-amp constraints notwithstanding, such as limited output voltage range and current limits) will result in a change in \$I_O\$.

To treat the op-amp and its periphery as a source of some kind, you'd place it all inside a black box, leaving only the load resistor \$R_L\$ outside:

^{simulate this circuit – Schematic created using CircuitLab}

The "black boxes" above each have two terminals, X and Y, and you have no idea what's inside the box. Each box has some internal resistance, which we aim to measure.

We could measure it by changing the external resistance \$R_L\$, or by applying some external potential difference between X and Y, or even by forcing some current through the box. Whatever method we use, we shall measure the resulting change in voltage \$\Delta V_{XY}\$, and change in current \$\Delta I_O\$. Effective resistance \$R_{XY}\$ between X and Y inside the box will be, according to Ohm's law:

$$ R_{XY} = \frac{\Delta V_{XY}}{\Delta I_O} $$

For box B, you'd find that there's no change in voltage, since its internal voltage source will strongly oppose any change in \$V_{XY}\$, but current \$I_O\$ is able to rise and fall freely:

$$ R_{XY(B)} = \frac{\Delta V_{XY}}{\Delta I_O} = \frac{0}{\Delta I_O} = 0\Omega $$

In other words, box B (or any ideal voltage source) is said to have zero impedance.

Box C is a current source, and we would expect it to strongly oppose changes in current through it, \$\Delta I_O=0\$, while freely permitting \$V_{XY}\$ to change:

$$ R_{XY(C)} = \frac{\Delta V_{XY}}{0} = \infty \Omega $$

Ideal current sources, therefore, are said to have infinite source impedance.

Box D will have an impedance somewhere in between zero and infinity, because in this case it would be possible to impose any arbitrary voltage between X and Y, and it's also possible to inject an arbitrary current through the box. With either "perturbation" method, if you plotted \$I_O\$ on the horizontal axis, vs. its corresponding \$V_{XY}\$, the slope of that graph is \$R_{XY}\$. Just by inspection, you could predict that this slope would be:

$$ R_{XY(D)} = \frac{\Delta V_{XY}}{\Delta I_O} = 100\Omega $$

So, to answer your question, it remains to be determined which of the boxes B, C and D has behaviour which most closely resembles the behaviour of box A.

We aren't able to impose a a voltage, because that would upset the biasing of the internals, but we are permitted to attach any load \$R_L\$ we desire and you could actually inject current too, and it would still work, for reasons I won't go into here. As explained in the article, \$I_O\$ is independent of \$R_L\$, and the op-amp will adjust \$V_{XY}\$ to whatever value is necessary to keep \$I_O\$ fixed. Thus \$\Delta I_O = 0\$, so:

$$ R_{XY(A)} = \frac{\Delta V_{XY}}{0} = \infty \Omega $$

This is consistent with the behaviour of the ideal current source in box C.

The only caveat is that you aren't able to alter potentials \$V_X\$ or \$V_Y\$ as you could for box C, because the op-amp requires the ability to decide those values for itself. In all other respects, though, for box A, the ratio of change of voltage vs. change of current, is \$\infty\$.

It's true that the op-amp is indeed a voltage source, using voltages to obtain this behaviour, but in the context of a two-terminal "black box", for which we only have access to the exposed nodes terminals (X and Y), any external "observer" circuitry would perceive it to behave like a current source, for which the impedance inside the box is effectively infinite.

Let's treat the op-amp and its periphery as a "black box", leaving only the load resistor \$R_L\$ outside:

^{simulate this circuit – Schematic created using CircuitLab}

The "black boxes" above each have two terminals, X and Y, and you have no idea what's inside the box. Each box has some internal resistance, which we aim to measure.

We could measure it by changing the external resistance \$R_L\$, or by applying some external potential difference between X and Y, or even by forcing some current through the box. Whatever method we use, we shall measure the resulting change in voltage \$\Delta V_{XY}\$, and change in current \$\Delta I_O\$. Effective resistance \$R_{XY}\$ between X and Y inside the box will be, according to Ohm's law:

$$ R_{XY} = \frac{\Delta V_{XY}}{\Delta I_O} $$

For box B, you'd find that there's no change in voltage, since its internal voltage source will strongly oppose any change in \$V_{XY}\$, but current \$I_O\$ is able to rise and fall freely:

$$ R_{XY(B)} = \frac{\Delta V_{XY}}{\Delta I_O} = \frac{0}{\Delta I_O} = 0\Omega $$

In other words, box B (or any ideal voltage source) is said to have zero impedance.

Box C is a current source, and we would expect it to strongly oppose changes in current through it, \$\Delta I_O=0\$, while freely permitting \$V_{XY}\$ to change:

$$ R_{XY(C)} = \frac{\Delta V_{XY}}{0} = \infty \Omega $$

Ideal current sources, therefore, are said to have infinite source impedance.

Box D will have an impedance somewhere in between zero and infinity, because in this case it would be possible to impose any arbitrary voltage between X and Y, and it's also possible to inject an arbitrary current through the box. With either "perturbation" method, if you plotted \$I_O\$ on the horizontal axis, vs. its corresponding \$V_{XY}\$, the slope of that graph is \$R_{XY}\$. Just by inspection, you could predict that this slope would be:

$$ R_{XY(D)} = \frac{\Delta V_{XY}}{\Delta I_O} = 100\Omega $$

So, to answer your question, it remains to be determined which of the boxes B, C and D has behaviour which most closely resembles the behaviour of box A.

We are permitted to attach any load \$R_L\$ we desire and you could even apply some arbitrary potential difference between X and Y, and it would still keep \$I_O\$ constant. The op-amp will adjust \$V_{XY}\$ to whatever value is necessary to keep \$I_O\$ fixed. Thus \$\Delta I_O = 0\$, so:

$$ R_{XY(A)} = \frac{\Delta V_{XY}}{0} = \infty \Omega $$

This is consistent with the behaviour of the ideal current source in box C.

It's true that the op-amp is indeed a voltage source, using voltages to obtain this behaviour, but in the context of a two-terminal "black box", for which we only have access to the exposed terminals, nodes X and Y, any external "observer" circuitry would perceive it to behave like a current source, for which the impedance inside the box is effectively infinite.