JEE MAIN - Physics (2022 - 26th July Morning Shift - No. 8)
The current I in the given circuit will be :
_26th_July_Morning_Shift_en_8_1.png)
10 A
20 A
4 A
40 A
Explicació
_26th_July_Morning_Shift_en_8_2.png)
The grouping of resistance is a wheatstone bridge
So, $${R_{net}} = 4\,\Omega $$
So, $$i = {V \over {{R_{net}}}} = 10\,A$$