Question

Find the potential at node A.

• A. $-\frac{5}{3}\text{V}$
• B. $+\frac{5}{3}\text{V}$
• C. $+\frac{4}{3}\text{V}$
• D. $-\frac{2}{3}\text{V}$

Answer 1

The correct option is A $-\frac{5}{3}\text{V}$

Let the potential at point $A$ be $x$


Writing the path equation relating potentials all four branches,

$x-20-i_1\left(1\right)=10$

$i_1=x-30$

$x-15-i_2\left(2\right)=20$

$i_2=\frac{x-35}{2}$

$x-i_3\left(2\right)-5=-50$

$i_3=\frac{x+45}{2}$

$x-i_4\left(1\right)=-30$

$i_4=x+30$

Applying KCL at the junction i.e at point $A$,

$i_1+i_2+i_3+i_4=0$

Applying kirchhoff current law at junciton A

$(x-30)+\left(\frac{x-35}{2}\right)+\left(\frac{x+45}{2}\right)+(x+30)=0$

$\Rightarrow \frac{2x-60+x-35+x+45+2x+60}{2}=0$

$\Rightarrow 6x+10=0$

$\Rightarrow x=-\frac{5}{3}$

Thus potential at $A$ is $x=-\frac{5}{3}\text{V}$

Why this question ? $\underset{–}{\text{Tip}}$: In such problems always identify the junction first, and apply KCL. Also assign the potential to unknown point and write the equation for traversing along a given path in circuit.