Three resistors having values R₁, R₂ and R₃ are connected in series to a battery. Suppose R₁ carries a current 2.0 A, R₂ has a resistance 3-0 ohm and R₃ dissipates 6.0 watt of power. Then voltage across R₃ is :
Given P = I² R₃, substituting values: 6 = (2.0)² R₃ ⟹ R₃ = 1.5Ω
Voltage across R₃, V₃ = IR₃ = 2.0 ×1.5 = 3V.
Answer: (c) 3V.
Class 12th Science Physics NCERT MCQ Questions
NCERT Books MCQ Questions Session 2024-2025.
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In a series circuit, the current remains the same across all resistors. Given that
R₁ carries 2.0 A,
R₃ also has 2.0 A. Power dissipated by
R₃ is given by
P = I² R. Solving 6 = (2)² R₃, we get
R₃ = 1.5 ohm. Voltage across
R₃ is V = IR = 2 × 1.5 = 3V. Answer: (c) 3 V.
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