Practice question

**Internal resistance** causes voltage drop I r inside battery, so V = ε - I r decreases with I. For 16 V battery, r=2 Ω, I=2 A, V=16-4=12 V, external R = V/I =6 Ω. Measurement of V and I yields r = (ε - V)/I. Parallel resistance: (1/R_p) = (1/2) + (1/6) = (3 + 1/6) = (4/6) = (2/3) ⇒ R_p = (3/2) = 1.5 Ω . Total resistance: Rtₒtₐl = 1 + 1.5 = 2.5 Ω . Current: I = (ε/Rtₒtₐl) = (12/2.5) = 4.8 A . Applying I = n e A v_d, R = ρ l/A, R_t = R₀[1+αΔT], Kirchhoff's