Question

A current of $6 \, A$ 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,
$ -2 I_{1}-2 I_{1}+2 I_{2}=0 $
$\Rightarrow -4 I_{1}+2 I_{2}=0 $
$\Rightarrow 2 I_{1}-I_{2}=0$
Adding Eqs. (i) and (ii), we get
$3 I_{1}=6$
$\Rightarrow I_{1}=2 A$
From Eq. (i),
$I_{2}=6-2=4 A$