Emf And Internal Resistance

A real cell has an internal resistance $r$ in series with the external load $R$, and the terminal circuit obeys $I=\frac{\epsilon }{R+r}$, where $\epsilon$ is the emf, a treatment given in NCERT Class 12, Chapter 3 (Current Electricity). The total resistance the emf must drive current through is the sum $R+r=2.5+0.5=3$ ohm. Therefore $I=\frac{6}{3}=2$ A. Neglecting the internal resistance and using only the 2.5 ohm load gives $6/2.5=2.4$ A, which is why that option is a tempting but wrong distractor for students who overlook the cell's own resistance. The option 12 A wrongly multiplies emf by the total resistance instead of dividing, which reverses the operation entirely. The option 3 A ignores the external load and divides emf by twice the internal resistance in error. A plausibility check confirms the answer: the terminal voltage is $\epsilon -Ir=6-2\times 0.5=5$ V, which sensibly falls short of the 6 V emf because some voltage is dropped internally, and dividing $5\text{ V}$ by the $2.5\text{ ohm}$ load recovers the same 2 A current, verifying full consistency of the solution.