A \( 6 \, \text{V} \) battery with \( 0.5 \, \Omega \) internal resistance is connected to a \( 5.5 \, \Omega \) resisto
**EMF ε** is work done by non-electrostatic forces per unit charge, terminal voltage V = ε - I r, r internal resistance (Ω), I current (A). When external R = r, total resistance 2r, current I = ε/2r, power in external R is I²R = ε²/4r, total power ε²/2r, so half power dissipated externally, half internally. Total resistance: Rtₒtₐl = 5.5 + 0.5 = 6 Ω . Current: I = (ε/Rtₒtₐl) = (6/6) = 1 A . Power: P = I² R = 1² × 5.5 = 5.5 W . Applying I = n e A v_d, R = ρ l/A, R_t = R₀[1+αΔT], Kirchhoff's
Ref: NCERT > Physics Book > Current Electricity > EMF, Internal Resistance and Cells Combination
