Question

In the network given points one of potentials of $70\text{V},0\text{V}$ and $10\text{V}$ respectively. The ratio of current through AD, DB and DC is

• A. $1:2:3$
• B. $2:3:1$
• C. $3:2:1$
• D. $1:2:1$

Answer 1

The correct option is C $3:2:1$

At junction D,

$i_1=i_2+i_3$

$\frac{V_{\text{A}}-V_{\text{D}}}{10}=\frac{V_{\text{D}}-V_{\text{B}}}{20}+\frac{V_{\text{D}}-V_{\text{C}}}{30}$

$\frac{70-V_{\text{D}}}{10}=\frac{V_{\text{D}}-0}{20}+\frac{V_{\text{D}}-10}{30}$



$420-6V_{\text{D}}=3V_{\text{D}}+2V_{\text{D}}-20$

$440=11V_{\text{D}}$

$V_D=40\text{V}$

$i_1:i_2:i_3=\frac{V_{\text{A}}-V_{\text{D}}}{10}:\frac{V_{\text{D}}-V_{\text{B}}}{20}:\frac{V_{\text{D}}-V_{\text{C}}}{30}$

$=\frac{70-40}{10}=\frac{40}{20}=\frac{40-10}{30}$

$=3:2:1$