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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
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Topics: Current Electricity
(136)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The period of a simple pendulum inside a stationary lift is T. The lift accelerates
- When a spring is extended by 2 cm energy stored is 100 J. When extended further by 2 cm
- The ratio of the numerical value of the average velocity and average speed
- λ₁ and λ₂ are used to illuminate the slits. β₁ and β₂ are the corresponding fringe
- Two equal and opposite charges of masses m₁ and m₂ are accelerated
Topics: Current Electricity (136)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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