Question

In the network shown, points $A$, $B$ and $C$ are potentials of $10$ V, $5$ V and $-10$ V respectively then

• A. Point $D$ is at a potential of $5$ V
• B. Point $D$ is at a potential of $55$ V
• C. Point $D$ is at a potential of $-5.5$ V
• D. Point $D$ is at a potential of $5.5$ V

Answer 1

The correct option is A Point $D$ is at a potential of $5$ V

Let, current through branch AD, BD and CD be $I_1,I_2$ and $I_3$respectively.
Using KCL for given network,
$I_1+I_2+I_3=0$

$\frac{V_D-10}{10}+\frac{V_D-5}{20}+\frac{V_D+10}{30}=0$

$6(V_D-10)+3(V_D-5)+2(V_D+10)=0$

$11V_D-55=0$

$\therefore V_D=5$ Volt