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

Answer 1

Answer: (A)

From Kirchhoff's first law at jucntion $P$
$I_1+I_2=6\dots (i)$
From Kirchhoff's second law to the closed circuit PQRP,
$-2I_1-2I_1+2I_2=0$
$\Rightarrow -4I_1+2I_2=0$
$\Rightarrow 2I_1-I_2=0$
Adding Eqs. (i) and (ii), we get
$3I_1=6$
$\Rightarrow I_1=2A$
From Eq. (i),
$I_2=6-2=4A$