A current of A enters one corner one corner P of an equilateral triangle $P\mathcal{Q}R$having 3 wires of resistance 2$\ \Omega$each and leaves by the corner R. then the current $I_{1}$and $I_{2}$are

Question

Dr. Abubacker · Accepted Answer

(a) From Kirchhoff’s first law at junction P,
$I_{1} + I_{2} = 6\ \ldots(i)$
From Kirchhoff’s second law to the closed circuit PQRP,
$- 2I_{1} - 2I_{1} + 2I_{2} = 0$
$\Longrightarrow - 4I_{1} + 2I_{2} = 0$
$\Longrightarrow 2I_{1} - I_{2} = 0\ \ \ \ \ \ldots\ldots(ii)$
Adding Eqs. (i) and (ii), we get
$3I_{1} = 6 \Longrightarrow I_{1} = 2A$
From Eq. (i),
$I_{2} = 6 - 2 = 4A$

Q. A current of A enters one corner one corner P of an equilateral triangle $P\mathcal{Q}R$having 3 wires of resistance 2$\ \Omega$each and leaves by the corner R. then the current $I_{1}$and $I_{2}$are

Explanation

Reviewed by Dr. Abubacker

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