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Question

A circuit has a \( 28 \, \text{V} \) battery with \( 4 \, \Omega \) internal resistance and three
resistors \( 4 \, \Omega \), \( 8 \, \Omega \), \( 16 \, \Omega \) in parallel. What is the current
through the \( 8 \, \Omega \) resistor?

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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. Parallel resistance: (1/R_p) = (1/4) + (1/8) + (1/16) = (4 + 2 + 1/16) = (7/16) ⇒ R_p = (16/7) ≈ 2.29 Ω . Total resistance: Rtₒtₐl = 4 + 2.29 = 6.29 Ω . Total current: I = (ε/Rtₒtₐl) = (28/6.29) ≈ 4.45 A . Voltage across parallel: V = I R_p = 4.45 × 2.29 ≈ 10.19 V . Current through

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