JEE Main · 2020 · 04 Sep 2020 · Shift 2 · Current Electricity

Four resistance

40Ω, 60Ω, 90Ω

, and

110Ω

make the arms of a quadrilateral ABCD. Across AC is a battery of emf 40 V and internal resistance negligible. The potential difference across BD in V is ____

Answer: 2

Solution

With the

40 V

battery across the diagonal

A C

, the current splits between the two arms. Along

A

B

C

the resistance is

40 + 60 = 100 Ω

, giving

i_{1} = \frac{40}{100} = 0.4 A

, while along

A

D

C

it is

90 + 110 = 200 Ω

, giving

i_{2} = \frac{40}{200} = 0.2 A

. Then

V_{B} = V_{A} - 40 i_{1} = V_{A} - 16 V

and

V_{D} = V_{A} - 90 i_{2} = V_{A} - 18 V

, so the potential difference across

B D

V_{B} - V_{D} = 2 V

JEE Main · 2020 · 04 Sep 2020 · Shift 2 · Integer answer

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