Question

In the circuit given here, the points $A, B$ and $C$ are $70 V$, zero, $10 \, V$ 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 $60\, V$.
• D. the point $D$ will be at a potential of $20\, V$.

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 $70-V_{D}=\frac{V_{D}}{2}+\frac{V_{D}-10}{3}$
$\Rightarrow V_{D}=40\, V$
$\Rightarrow i_{1}=\frac{70-40}{10}=3\, A$

$i_{2} =\frac{40-0}{20}=2\, A$
and $i_{3} =\frac{40-10}{30}=1\, A$