The magnitude of $i$ in ampere unit is
Diffcult
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(a) Applying Kirchoff's law in following figure.
At junction $A$ :
$i + {i_1} + {i_2} = 1$ .... $(i)$
For Loop $(i)$
$ - \,60\,i + (15 + 5){i_1} = 0$
$==>$ ${i_1} = 3i$ ...$(ii)$
For loop $(2)$
$-(15 + 5) i_1 + 10 i_2 = 0$
$==>$ $i_2 = i_1 = (3 i) = 6i$
On solving equation $(i), (ii)$ and $(iii)$ we get $i$ =$ 0.1\, A$
Short Trick : Branch current $=$
$\,{\rm{main current }}\left( {\frac{{{\rm{Resistance \,of \,opposite\, branch}}}}{{{\rm{Total\, resistance}}}}} \right)$
$==>$ $i = 1 \times \left[ {\frac{{\frac{{20}}{3}}}{{\frac{{20}}{3} + 60}}} \right]$
$=$ $0.1\, A$
At junction $A$ :
$i + {i_1} + {i_2} = 1$ .... $(i)$
For Loop $(i)$
$ - \,60\,i + (15 + 5){i_1} = 0$
$==>$ ${i_1} = 3i$ ...$(ii)$
For loop $(2)$
$-(15 + 5) i_1 + 10 i_2 = 0$
$==>$ $i_2 = i_1 = (3 i) = 6i$
On solving equation $(i), (ii)$ and $(iii)$ we get $i$ =$ 0.1\, A$
Short Trick : Branch current $=$
$\,{\rm{main current }}\left( {\frac{{{\rm{Resistance \,of \,opposite\, branch}}}}{{{\rm{Total\, resistance}}}}} \right)$
$==>$ $i = 1 \times \left[ {\frac{{\frac{{20}}{3}}}{{\frac{{20}}{3} + 60}}} \right]$
$=$ $0.1\, A$
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