Terminal Voltage And Internal Resistance

The open-circuit terminal voltage equals the emf because no current flows, so $\epsilon =12$ V, and once current is drawn the terminal voltage falls according to $V=\epsilon -Ir$, a relationship developed in NCERT Class 12, Chapter 3 (Current Electricity). The voltage lost inside the cell is $\epsilon -V=12-10.5=1.5$ V, and this drop occurs across the internal resistance carrying 3 A. Hence $r=\frac{\epsilon -V}{I}=\frac{1.5}{3}=0.5$ ohm. The option 4 ohm mistakenly divides the terminal voltage difference by a smaller misread current figure rather than the actual 3 A. The option 3.5 ohm divides the full 10.5 V by 3 A, wrongly treating the entire loaded terminal voltage as if it appeared across the internal resistance. The option 1.5 ohm reports the internal voltage drop as if it were directly the resistance, forgetting the crucial step of dividing by the current. A magnitude check confirms plausibility: a small internal resistance of 0.5 ohm producing a modest 1.5 V drop at 3 A is typical for a healthy battery, and the units of volts divided by amperes correctly yield ohms, so both the sign and magnitude of the answer are physically reasonable.