Model the source with internal resistance

A practical energy source can be represented as an ideal emf $E$ in series with internal resistance $r$, connected to a load resistance $R$.

The current supplied to the load is $I = \frac{E}{R+r}.$

Condition for constant current

For the source to supply a constant current into the load, the current should remain nearly independent of the load resistance $R$.

From $I = \frac{E}{R+r},$ the current will be approximately constant if changes in $R$ do not significantly affect the denominator.

This happens when $r \gg R.$

Then, $I = \frac{E}{R+r} \approx \frac{E}{r},$ which is nearly constant and almost independent of the load resistance.

Check the options

• A: very large as compared to the load resistance
  If $r \gg R$, then $I \approx \dfrac{E}{r}$, nearly constant. Correct.

• B: equal to the resistance of the load
  Then $I = \dfrac{E}{2R}$, which depends on $R$. Incorrect.

• C: non-zero but less than the resistance of the load
  Then current strongly depends on $R$. Incorrect.

• D: zero
  Then $I = \dfrac{E}{R}$, which varies directly with load resistance. This corresponds to a constant voltage source, not a constant current source. Incorrect.

Conclusion

The source behaves like a constant current source when its internal resistance is very large compared to the load resistance.

Therefore, the correct option is $\boxed{A}.$

Comparison with stored answer

The stored correct answer is D, but from circuit theory, $r=0$ gives an ideal constant voltage source, not a constant current source. Hence I do not agree with the stored answer.