Question

In the given circuit, potential of junction $A$ is :

• A. $1.73\text{V}$
• B. $4\text{V}$
• C. $3.82\text{V}$
• D. $2.5\text{V}$

Answer 1

The correct option is C $3.82\text{V}$

Let the voltage of point $A$ be $V_A$ and currents are flowing in the circuit as shown in the figure below.


Here we assume that potential of point $A$ is higher than $2\text{V}$.

Now applying KVL between points of potential $8\text{V}$ and $2\text{V}$ :
$8-2=2(i_1+i_2)+4i_2$

$6=2i_1+6i_2\Rightarrow 3=i_1+3i_2$ ....$\left(1\right)$

Applying KVL between points of potential $8\text{V}$ and $-6\text{V}$ :
$8+6=2(i_1+i_2)+6i_1$

$14=8i_1+2i_2\Rightarrow 7=4i_1+i_2$ ....$\left(2\right)$

By $\left(1\right)$ & $\left(2\right)$

$i_2=\frac{5}{11}$

Applying KVL across $4\Omega$ resistance :
$V_A-2=4\times \frac{5}{11}\Rightarrow V_A=2+\frac{20}{11}=\frac{42}{11}=3.82V$


$\text{Alternate solution}:$

Applying KCL at the junction $A$ :

$\frac{8-V_A}{2}=\frac{V_A-(-6)}{6}+\frac{V_A-2}{4}$

$\Rightarrow 6(8-V_A)=2(V_A+6)+3(V_A-2)$

$\Rightarrow 11V_A=42$

Or, $V_A=\frac{42}{11}=3.82\text{V}$