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Question

A circuit has a \( 8 \, \text{V} \) battery with \( 1 \, \Omega \) internal resistance and two
resistors \( 3 \, \Omega \) and \( 6 \, \Omega \) in parallel. What is the total current?

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Explanation

**Kirchhoff's loop rule** (energy conservation) states algebraic sum of potential differences around closed loop zero, Σ ε - Σ I R =0, ensures total voltage drop across series resistors equals source emf. Junction rule (charge conservation) Σ I_in = Σ I_out ensures current continuity. Parallel resistance: (1/R_p) = (1/3) + (1/6) = (2 + 1/6) = (3/6) = 0.5 ⇒ R_p = 2 Ω . Total resistance: Rtₒtₐl = 1 + 2 = 3 Ω . Current: I = (ε/Rtₒtₐl) = (8/3) ≈ 2.67 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,

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