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Q.
In the given network. The currents $i_{1}, i_{2}, i_{3}$ are respectively:
Current Electricity
Solution:
Using the junction rule the currents in the branches are Loop $A D C A,-4\left(i_{1}-i_{2}\right)+2\left(i_{2}+i_{3}-i_{1}\right)-i_{1}+10=0$
$\Rightarrow 7 i_{1}-6 i_{2}-2 i_{3}=10...(i)$
Loop $A B C A$
$-4 i_{2}-2\left(i_{2}+i_{3}\right)-i_{1}+10=0$
$\Rightarrow i_{1}+6 i_{2}+2 i_{3}=10...(ii)$
Loop $B C D E B$, $5-2\left(i_{2}+i_{3}\right)-2\left(i_{2}+i_{3}-i_{1}\right)=0$
$\Rightarrow -2 i_{1}+4 i_{2}+4 i_{3}=5...(iii)$
Solving (i), (ii), (iii), we get
$i_{1}=\frac{5}{2} A , i_{2}=\frac{5}{8} A , i_{3}=\frac{15}{8} A$
