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Question

Two cells of emf \( 3 \, \text{V} \) and \( 5 \, \text{V} \) with internal resistances \( 1 \, \Omega
\) and \( 3 \, \Omega \) are connected in series with a \( 4 \, \Omega \) resistor. What is the current
through the circuit?

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Explanation

**Series combination** R_eq = R₁+R₂+..., same current I through each, voltage divides proportionally V_i = I R_i. Parallel combination 1/R_p = 1/R₁+1/R₂+..., same voltage V across each, current divides inversely, equivalent R_p = (R₁ R₂)/(R₁+R₂) for two resistors. Equivalent emf: εₑq = 3 + 5 = 8 V . Total resistance: Rtₒtₐl = r₁ + r₂ + R = 1 + 3 + 4 = 8 Ω . Current: I = (εₑq/Rtₒtₐl) = (8/8) = 1 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

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