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

• A

  the point $\mathrm{D}$ will be at a potential of $60 \mathrm{~V}$
• B

  the point $\mathrm{D}$ will be at a potential of $20 \mathrm{~V}$
• Correct answer
  C

  currents in the paths AD, DB and DC are in the ratio of $1: 2: 3$
• D

  currents in the paths $\mathrm{AB}, \mathrm{DB}$ and $\mathrm{DC}$ are in the ratio of $3: 2: 1$

Solution

Applying Kirchhoff's law at point D, we get $$ \begin{aligned} \mathrm{I}_{1} &=\mathrm{I}_{2}+\mathrm{I}_{3} \\ & \frac{\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{D}}}{10} =\frac{\mathrm{V}_{\mathrm{D}}-0}{20}+\frac{\mathrm{V}_{\mathrm{D}}-\mathrm{V}_{\mathrm{C}}}{30} \\ \text { or } & 70-\mathrm{V}_{\mathrm{D}} =\frac{\mathrm{V}_{\mathrm{D}}+\mathrm{V}_{\mathrm{D}}-10}{2}+\frac{70}{3} \\ \Rightarrow & \mathrm{V}_{\mathrm{D}} =40 \mathrm{~V} \\ \Rightarrow & \mathrm{i}_{1} =\frac{70-40}{10}=3 \mathrm{~A} \end{aligned} $$

$$ \mathrm{i}_{2}=\frac{40-0}{20}=2 \mathrm{~A} $$ and $$ \mathrm{i}_{3}=\frac{40-10}{30}=1 \mathrm{~A} $$

Hence, the correct answer is option: 'C'.