Emf And Internal Resistance

Question

A cell of emf 6 volts and internal resistance 0.5 ohm is connected to an external resistor of 2.5 ohm, so what current flows?

RankGuru · Accepted Answer

A real cell has an internal resistance $r$ in series with the external load $R$, and the terminal circuit obeys $I = \frac{\varepsilon}{R + r}$, where $\varepsilon$ 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 $\varepsilon - Ir = 6 - 2 	imes 0.5 = 5$ V, which sensibly falls short of the 6 V emf because some voltage is dropped internally, and dividing $5 	ext{ V}$ by the $2.5 	ext{ ohm}$ load recovers the same 2 A current, verifying full consistency of the solution.

Emf And Internal Resistance

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