In the circuit, given in the figure currents in different branches and value of one resistor are shown. Then potential at point $B$ with respect to the point $A$ is$.......V$
A$+1$
B$-1$
C$-2$
D$+2$
JEE MAIN 2020, Diffcult
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A$+1$
a Let us asssume the potential at $A=V_{A}=0$
Now at junction $C,$ According to $KCL$
$i_{1}+i_{3}=i_{2}$
$1 A + i _{3}=2 A$
$i _{3}=2 A$
Now Analyse potential along $ACDB$
$v_{A}+1+i_{3}(2)-2=v_{B}$
$0+1+2(1)-2=v_{B}$
$v_{B}=3-2$
$v_{B}=1 Amp$
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