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Question

A circuit has a \( 30 \, \text{V} \) battery with \( 3 \, \Omega \) internal resistance and three
resistors \( 3 \, \Omega \), \( 6 \, \Omega \), \( 12 \, \Omega \) in parallel. What is the current
through the \( 6 \, \Omega \) resistor?

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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) + (1/12) = (4 + 2 + 1/12) = (7/12) ⇒ R_p = (12/7) ≈ 1.71 Ω . Total resistance: Rtₒtₐl = 3 + 1.71 = 4.71 Ω . Total current: I = (ε/Rtₒtₐl) = (30/4.71) ≈ 6.37 A . Voltage across parallel: V = I R_p = 6.37 × 1.71 ≈ 10.89 V

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