Practice question

Given: A 12 V battery with negligible internal resistance is connected to a 3 Ω and 6 Ω resistor in series. What is the power dissipated in the 6 Ω resistor? These values define the system as per NCERT data. Formula: Total resistance: R = 3 + 6 = 9 Ω. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Current: I = V/R = 12/9 = 4/3 A . Power: P = I² R = (4/3)² × 6 = 16/9 × 6 = 32/3 approx 10.67 W . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.