Question

In the circuit given here, the points $A,B$ and $C$ are $70V$, zero, $10V$ respectively. Then

• A. currents in the paths $AD,DB$ and $DC$ are in the ratio of $1:2:3$.
• B. currents in the paths $AB,DB$ and $DC$ are in the ratio of $3:2:1$.
• C. the point $D$ will be at a potential of $60V$.
• D. the point $D$ will be at a potential of $20V$.

Answer 1

Answer: (B)

Applying Kirchhoff's law at point $D$, we get
$I_1=I_2+I_3$
$\frac{V_A-V_D}{10}=\frac{V_D-0}{20}+\frac{V_D-V_C}{30}$
or $70-V_D=\frac{V_D}{2}+\frac{V_D-10}{3}$
$\Rightarrow V_D=40V$
$\Rightarrow i_1=\frac{70-40}{10}=3A$

$i_2=\frac{40-0}{20}=2A$
and $i_3=\frac{40-10}{30}=1A$