Practice question
Question
Two cells of emf \( 4 \, \text{V} \) and \( 5 \, \text{V} \) with internal resistances \( 0.5 \, \Omega
\) and \( 1 \, \Omega \) are connected in series with a \( 5.5 \, \Omega \) resistor. What is the
current through the circuit?
Explanation
**Cells combination** series ε_eq = Σ ε_i, r_eq = Σ r_i, parallel for identical cells ε_eq = ε, r_eq = r/n, n number of cells. Maximum current when external R = r_eq, power transfer theorem, explaining why matching resistances maximizes power. Equivalent emf: εₑq = 4 + 5 = 9 V . Total resistance: Rtₒtₐl = 0.5 + 1 + 5.5 = 7 Ω . Current: I = (εₑq/Rtₒtₐl) = (9/7) ≈ 1.29 A . 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
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