Question

In the circuit, given in the figure currents in different branches and the value of one resistor are shown. Then potential at point B with respect to point A is:

• A. $+2\mathrm{V}$
• B. $-2\mathrm{V}$
• C. $+1\mathrm{V}$
• D. $-1\mathrm{V}$

Answer 1

The correct option is C

$+1\mathrm{V}$

Step 1: Given data

${\mathrm{i}}_1=1\mathrm{Ampere}$

${\mathrm{i}}_2=2\mathrm{Ampere}$

$\mathrm{R}=2\mathrm{\Omega }$

Assume that potential at $A$, ${\mathrm{V}}_{\mathrm{A}}=0$.

Step 2: Find the current $\left({\mathrm{i}}_3\right)$

Apply Kirchhoff's Law

Now at junction $C$, According to Kirchhoff's Law

$$\begin{gathered} {\mathrm{i}}_1+{\mathrm{i}}_3={\mathrm{i}}_2 \\ 1+{\mathrm{i}}_3=2 \\ {\mathrm{i}}_3=1 \end{gathered}$$

Step 3: Calculate the potential at point $B$ with respect to point $A$

Now analyze potential along $ACDB$

$$\begin{gathered} {\mathrm{V}}_{\mathrm{A}}+{\mathrm{i}}_1+{\mathrm{i}}_3\times \mathrm{R}-2={\mathrm{V}}_{\mathrm{B}} \\ 0+1+2+1\times 2-2={\mathrm{V}}_{\mathrm{B}} \\ {\mathrm{V}}_{\mathrm{B}}=3-2 \\ {\mathrm{V}}_{\mathrm{B}}=1\mathrm{Volt} \end{gathered}$$

Hence, the option $\left(\mathrm{C}\right)$ is correct.