Question

JEE MAINS PHYSICS QUESTION SEPTEMBER 2020

Answer 1

Answer:

c) 36 v

Explanation:

Answer 2

Four resistance 90Ω, 10Ω, 90Ω, and 110Ω make the arms of a quadrilateral ABCD. Across AC is a battery of emf 40V and internal resistance negligible.

The potential difference between points A and B is ....

solution : resistors AB and BC are joined in series so, R₁ = AB + BC = 90 + 10 = 100Ω

also resistors AD and DC are joined in series

so, R₂ = AD + DC = 110 + 90 = 200Ω

here it is clear that R₁ and R₂ are in parallel combination.

Req = 1/R₁ + 1/R₂ = 1/100 + 1/200 = 3/200

so, Req = 200/3 Ω

now current through circuit, i = V/Req

= 40/(200/3) = 120/200 = 0.6 A

so, potential across R₁ = potential across R₂

⇒i₁ × 100 = i₂ × 200

⇒i₁ = 2i₂

but i = i₁ + i₂ = 0.6 A

so, i₁ = 0.4 A and i₂ = 0.2A

now potential across 90Ω (AB) = 90 × 0.4 = 36V

therefore the potential difference between points A and B is 36V i.e., option (3) is correct