STANDARD Physics Combination of Resistors MCQ Question
Three resistances 2 Ω, 4 Ω, 5 Ω are combined in series and this combination is connected to a battery of 12V emf and negligible internal resistance. The potential drop across these resistances are
(5.45, 4.36, 2.18) V
(2.18, 5.45, 4.36) V
(4.36, 2.18, 5.45) V
(2.18,4.36,5.45) V
Correct Answer
Detailed Explanation
The total resistance in the circuit is R = 2 + 4 + 5 = 11 Ω. The current I = V/R = 12/11 A. The potential drop across each resistor is calculated as V = IR. For 2 Ω, V₁ = 2.18 V; for 4 Ω, V₂ = 4.36 V; for 5 Ω, V₃ = 5.45 V.
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