A $16\Omega$ wire is bend to form a square loop. A 9 V supply having internal resistance of $1\Omega$ is connected across one of its sides. The potential drop across the diagonals of the square loop is _____ $\times 10^{-1}$ V.

JEE Main · 2021 · 25 Jul 2021 · Shift 2 · Current Electricity

16Ω

wire is bend to form a square loop. A 9 V supply having internal resistance of

1Ω

is connected across one of its sides. The potential drop across the diagonals of the square loop is _____

\times 1 0^{- 1}

Answer: 45

Each side of the square carries

\frac{16}{4} = 4 Ω

. With the supply across one side, between the corners

A

and

B

, that side is in parallel with the other three sides in series:

R_{A B} = \frac{4 \times 12}{4 + 12} = 3 Ω

Including the internal resistance, the current drawn is

I = \frac{9}{1 + 3} = 2.25

A and the terminal voltage is

V_{A B} = 2.25 \times 3 = 6.75

V. Taking

V_{B} = 0

, the branch

A \to D \to C \to B

carries

\frac{6.75}{12} = 0.5625

A, so

V_{D} = 6.75 - 0.5625 \times 4 = 4.5 V, V_{C} = 4.5 - 0.5625 \times 4 = 2.25 V

The two diagonals therefore show the same potential difference,

V_{A} - V_{C} = V_{D} - V_{B} = 4.5

V. The drop across the diagonals is

45 \times 1 0^{- 1}

JEE Main · 2021 · 25 Jul 2021 · Shift 2 · Integer answer