UPSEE 2019 Solved Paper

According to the question, we draw the circuit as belowLet the current in the circuit is I,
Now applying KVL rule,
6−1I−2I−3=0
⇒ 6−3=3I
⇒I=

3

3

=1A
As VAB is the potential across battery 6V and resistance of 1Ω.
So, VAB=6−1I
⇒VAB=6−1=5V
Hence, the potential difference across A and B is 5 V.