Emf And Internal Resistance

A real cell carries internal resistance, so part of its EMF is lost driving current through the cell itself, and the terminal voltage available externally is (V = \varepsilon - Ir). The circuit current first follows from the total resistance: (I = \varepsilon/(R + r) = 12/(5.5 + 0.5) = 12/6 = 2) A. The terminal voltage equals the drop across the external resistor, (V = IR = 2 \times 5.5 = 11) V, or equivalently (V = \varepsilon - Ir = 12 - 2 \times 0.5 = 11) V. The value 12 V is incorrect because it ignores the internal-resistance drop and equates terminal voltage with EMF. The value 10 V overestimates the internal loss using twice the actual current. The value 6 V mistakenly applies the voltage-divider fraction inverted. This reflects the NCERT distinction between EMF and terminal potential difference. A sanity check confirms the result: since the internal resistance is small compared with the external load, the terminal voltage should fall only slightly below the 12 V EMF, exactly as obtained. This also illustrates why a freshly loaded battery measured on open circuit reads close to its EMF, but sags more noticeably as the load resistance is reduced and the drawn current grows, since the internal drop (Ir) then becomes a larger fraction of the total.