BITSAT 2024 Shift 1 Paper

where, equivalent emf of the combination,
Eeq‌=‌

E1R1+E2R2

R1+R2

‌=‌

2×4+2×4

4+4

=2V
and Req=‌

R1R2

R1+R2

=‌

4×4

4+4

=‌

16

8

=2Ω
Net emf of the circuit,
E′‌=E2+E‌eq ‌
‌=2+2
‌=4V
∴ Current flowing through branch AB,
I=‌

E′

Req

=‌

4

2

=2A
Clearly from figure, it is clear that direction of current is from point A to B clockwise is loop.