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Question

A circuit has a \( 24 \, \text{V} \) battery with \( 3 \, \Omega \) internal resistance and three
resistors \( 6 \, \Omega \), \( 12 \, \Omega \), \( 24 \, \Omega \) in parallel. What is the current
through the \( 12 \, \Omega \) resistor?

Options

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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. Parallel resistance: (1/R_p) = (1/6) + (1/12) + (1/24) = (4 + 2 + 1/24) = (7/24) ⇒ R_p = (24/7) ≈ 3.43 Ω . Total resistance: Rtₒtₐl = 3 + 3.43 = 6.43 Ω . Total current: I = (ε/Rtₒtₐl) = (24/6.43) ≈ 3.73 A . Voltage across parallel: V = I R_p = 3.73 × 3.43 ≈ 12.79 V . Current through 12 Ω :

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