UPSEE 2018 Solved Paper

Let I1 is current flowing through first loop.
∴ I1=

2

3+2

=

2

5

Aand I2 is current flowing through the second loop.
∴I2=

4

3+5

=

1

2

A
Let VA is potential of point A and VB is potential of point B.
Now, we apply Kirchhoff's law between points A and B,
∴VA−3I1−4+3I2=VB
⇒VA−VB=3×

2

5

+4−3×

1

2

=

12+40−15

10

=

37

10

⇒VA−VB=3.7V
Hence, (a) is the correct option.