Q4. Assertion: the potential difference across the terminals of a cell falls as more current is drawn from it. Reason: part of the emf is used to drive the current through the cell’s internal resistance, V = ε − Ir. Are both true, and does the reason explain the assertion?
Both are true, and the reason explains the assertion (NCERT Eq. (3.38), page 94).
Q5. A source of emf 6.0 V and internal resistance 1.0 Ω is connected to a 5.0 Ω resistor. Find the potential difference across the resistor.
Vext = εR/(R + r) = 6.0 × 5.0/6.0 = 5.0 V (NCERT Summary point 11, page 103).
Q6. For the constant-emf model with ε > 0 and internal resistance r > 0, which external resistance R takes the most power, and how much?
P = IR = εR/(R + r). Divide top and bottom by R: the bottom becomes (R + r)/R = (R − r)/R + 4r, which is least, 4r, when R = r. So R = r takes the most power, P = ε/(4r) (reasoned from NCERT Eq. (3.40), page 95, and Eq. (3.33), page 93). The cell then wastes as much power inside it, Ir, so the efficiency V/ε = R/(R + r) is only 50 %.