A conductor with rectangular cross-section has dimensions a × 2 a × 4 a as shown in the figure. Resistance across A B is x , across C D is y and across E F is z . Then - .

We know that R = ρ l A x = ρ 4 a a × 2 a = 2 ρ a y = ρ a 2 a × 4 a = ρ 8 a z = ρ 2 a a × 4 a = ρ 2 a ∴ x > z > y

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PhysicsCurrent ElectricityEquivalent ResistanceEasy2 minPYQ_2020

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A conductor with rectangular cross-section has dimensions $(a \times 2 a \times 4 a)$ as shown in the figure. Resistance across $A B$ is $x$ , across $C D$ is $y$ and across $E F$ is $z$ . Then -
.

Options:

Answer:

Solution:

We know that $R = \frac{ρ l}{A}$

$x = \frac{ρ (4 a)}{a \times 2 a} = \frac{2 ρ}{a}$

$y = \frac{ρ (a)}{2 a \times 4 a} = \frac{ρ}{8 a}$

$z = \frac{ρ (2 a)}{a \times 4 a} = \frac{ρ}{2 a}$

$∴ x > z > y$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:Current ElectricitySubtopic:Equivalent Resistance

⏱ 2mℹ️ Source: PYQ_2020

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