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Two wires of equal length and equal diameter and having resistivities ρ₁ and ρ₂ are connected in series. The equivalent resistivity of the combination is
Options
(a) ρ₁+ρ₂/2
(b) ρ₁+ρ₂
(c) ρ₁ρ₂/ρ₁+ρ₂
(d) √ρ₁ρ₂
Correct Answer:
ρ₁+ρ₂/2
Explanation:
The resistance of two wires are
R₁ = ρ₁ (l / A) and R₂ = ρ₂ (l / A)
Now, equivalent resistance of series connection of wire
Req = R₁ + R₂ = ρ₁ (l / A) + ρ₂ (l / A) = (l / A) (ρ₁ + ρ₂)
The equivalent resistance Req can be given by Req = ρ(eq) (l / A)
⇒ ρ(eq) (l / A) = (l / A) (ρ₁ + ρ₂)
Hence, ρ(eq) = (ρ₁ + ρ₂) / 2
Related Questions: - A particle is kept at rest at the top of a sphere of diameter 42m.When disturbed slightly
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Topics: Current Electricity
(136)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A particle is kept at rest at the top of a sphere of diameter 42m.When disturbed slightly
- With what minimum acceleration can a fireman slide down a rope while breaking
- Two capacitors of capacitances 2 μF and 3 μF are joined in series.
- When the energy of the incident radiation is increased by 20%,
- If the image formed by a convex mirror of focal length 30 cm is a quarter of the size
Topics: Current Electricity (136)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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