Question

A current of $6A$ enters one corner $P$ of an equilateral triangle $PQR$ having $3$ wires of resistances $2\Omega$ each and leaves by the corner $R$. Then, the current $I_1$ and $I_2$ are

• A. $1A,2A$
• B. $2A,3A$
• C. $2A,4A$
• D. $4A,2A$

Answer 1

The correct option is B $2A,3A$

From Kirchoff's junction law

(at junction $P$)

$6-I_1-I_2=0$

$\Rightarrow$ $I_1+I_2=\mathrm{6.....}\left(i\right)$

Applying kirchoff's loop law in teh loop 1 shown

$2I_1+2I_1-2I_2=0$

$\Rightarrow$ $2I_1-I_2=\mathrm{0.....}\left(ii\right)$

adding (i) and (ii)

$3I_1=6\Rightarrow$ $I_1=2A$

and $I_2=2I_1$

$I_2=2\times 2=4$

$I_2=4A$