Question

JEE MAINS PHYSICS QUESTION SEPTEMBER 2020

Answer 1

we have to find the current through 4Ω resistance.

solution : According to balanced Wheatstone's bridge,

there is no current through 5Ω resistors. so we can remove it from circuit.

now diagram of given resistors is shown in figure.

first find equivalent resistance,

1/Req = 1/(2Ω + 2Ω) + 1/(4Ω + 4Ω) + 1/(2Ω + 2Ω)

= 1/4 + 1/8 + 1/4

= 1/2 + 1/8

= 5/8

so, Req = 8/5

now current through circuit, i = V/Req

= 8/(8/5) = 5A

as potential of 2Ω resistor = potential of 4Ω resistor

⇒i₁ × (2Ω + 2Ω) = i₂ × (4Ω + 4Ω)

⇒i₁ = 2i₂

but i₁ = i₃ (current through resistance of ADB) = 2i₂......(1)

but i₁ + i₂ + i₃ = 5 .....(2)

from equations (1) and (2) we get, i₂ = 1A

therefore the current through 4 Ω is 1 Amp. so the correct option is (1) 1A