A \( 15 \, \text{V} \) battery with negligible internal resistance is connected to a \( 5 \, \Omega \) and \( 10 \, \Ome
**Electrical power** P = V I = I² R = V²/R (W), energy E = P t = I² R t (J), heating effect Joule's law H = I² R t. When internal r equals external R, total resistance 2R, I = ε/2R, power in external = I²R = ε²/4R, total = ε²/2R, fraction external = 1/2, illustrating maximum power transfer when R = r. Total resistance: R = 5 + 10 = 15 Ω . Current: I = (V/R) = (15/15) = 1 A . Power: P = I² R = 1² × 5 = 5 W . Applying I = n e A
Ref: NCERT > Physics Book > Current Electricity > Electrical Power, Energy and Heating Effect
