Question

In the given network. The currents $i_1,i_2,i_3$ are respectively:

• A. $\frac{5}{2}A,\frac{5}{8}A,\frac{15}{8}A$
• B. $\frac{5}{3}A,\frac{5}{8}A,\frac{12}{8}A$
• C. $\frac{5}{2}A,\frac{15}{8}A,\frac{5}{3}A$
• D. $\frac{5}{3}A,\frac{5}{2}A,\frac{15}{3}A$

Answer 1

Answer: (A)

Using the junction rule the currents in the branches are Loop $ADCA,-4\left(i_1-i_2\right)+2\left(i_2+i_3-i_1\right)-i_1+10=0$
$\Rightarrow 7i_1-6i_2-2i_3=\mathrm{10...}(i)$

Loop $ABCA$
$-4i_2-2\left(i_2+i_3\right)-i_1+10=0$
$\Rightarrow i_1+6i_2+2i_3=\mathrm{10...}(ii)$
Loop $BCDEB$, $5-2\left(i_2+i_3\right)-2\left(i_2+i_3-i_1\right)=0$
$\Rightarrow -2i_1+4i_2+4i_3=\mathrm{5...}(iii)$
Solving (i), (ii), (iii), we get
$i_1=\frac{5}{2}A,i_2=\frac{5}{8}A,i_3=\frac{15}{8}A$