Practice question
Question
Two cells of emf \( 2.5 \, \text{V} \) and \( 3.5 \, \text{V} \) with internal resistances \( 0.5 \,
\Omega \) and \( 1 \, \Omega \) are connected in series. What is the equivalent emf?
Explanation
**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. For series: εₑq = ε₁ + ε₂ = 2.5 + 3.5 = 6.0 V . Applying I = n e A v_d, R = ρ l/A, R_t = R₀[1+αΔT], Kirchhoff's ΣI=0, ΣV=0, R_eq series/parallel, V = ε - I r and P = I²R, evaluation yields 6.0 V,
Discussion
Comments
Please log in to join the discussion.
Login to commentNo comments yet. Be the first to start the discussion.