JEE MAIN - Physics (2021 - 25th July Evening Shift - No. 20)
A 16 $$\Omega$$ wire is bend to form a square loop. A 9V 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
Answer
45
Explanation
Here assume current as
_25th_July_Evening_Shift_en_20_2.png)
By KVL in outer loop
9 $$-$$ 12i $$-$$ 4i = 0
16i = 9
8i = $${9 \over 2}$$ = 4.5
= 45 $$\times$$ 10-1
By KVL in outer loop
9 $$-$$ 12i $$-$$ 4i = 0
16i = 9
8i = $${9 \over 2}$$ = 4.5
= 45 $$\times$$ 10-1
Comments (0)
