A \( 8 \, \text{V} \) battery with \( 1 \, \Omega \) internal resistance is connected to a \( 7 \, \Omega \) resistor. W
**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 = 7 + 1 = 8 Ω . Current: I = (ε/Rtₒtₐl) = (8/8) = 1 A . Power: P = I² r = 1² × 1 = 1 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
