Question:
In the circuit, given in the figure currents in different branches and value of one resistor are shown. Then potential at point $\mathrm{B}$ with respect to the point $\mathrm{A}$ is :
Correct Option: 1
Solution:
Let us asssume the potential at $\mathrm{A}=\mathrm{V}_{\mathrm{A}}=0$
Now at junction $\mathrm{C}$, According to $\mathrm{KCL}$
$\mathrm{i}_{1}+\mathrm{i}_{3}=\mathrm{i}_{2}$
$1 \mathrm{~A}+\mathrm{i}_{3}=2 \mathrm{~A}$
$\mathrm{i}_{3}=2 \mathrm{~A}$
Now Analyse potential along $\mathrm{ACDB}$
$\mathrm{v}_{\mathrm{A}}+1+\mathrm{i}_{3}(2)-2=\mathrm{v}_{\mathrm{B}}$
$0+1+2(1)-2=\mathrm{v}_{\mathrm{B}}$
$\mathrm{v}_{\mathrm{B}}=3-2$
$\mathrm{v}_{\mathrm{B}}=1 \mathrm{Amp}$