Question

A current of 6 A enters one corner P of an equilateral triangle PQR having 3 wires of resistance 2 $\Omega$ each and leaves by the corner R. The currents $i_1$ in ampere is ________ . 

Hint:

The current division rule is a shortcut for parallel circuits. The current in one branch is the total current multiplied by the ratio of the other branch's resistance to the total resistance of the parallel combination: $i_1 = I_{total} \times \frac{R_2}{R_1+R_2}$.

Answer 1

Correct Answer: 2

The total current of 6 A enters at point P and leaves at point R.
The current splits at junction P into two paths to reach R.
Path 1: The direct wire from P to R. The resistance of this path is $R_{PR} = 2 \Omega$. Let the current be $i_2$.
Path 2: The path through wires PQ and then QR. The resistances are in series. The resistance of this path is $R_{PQR} = R_{PQ} + R_{QR} = 2 \Omega + 2 \Omega = 4 \Omega$. The current in this path is $i_1$.
These two paths, PR and PQR, are in parallel between points P and R. Therefore, the voltage drop across both paths is the same.
$V_{PR} = i_1 \cdot R_{PQR} = i_2 \cdot R_{PR}$.
$i_1 \cdot (4) = i_2 \cdot (2) \implies i_2 = 2i_1$.
The total current entering at P is the sum of the currents in the two paths:
$I_{total} = i_1 + i_2 = 6$ A.
Substitute $i_2 = 2i_1$ into the equation for the total current:
$i_1 + 2i_1 = 6 \implies 3i_1 = 6$.
$i_1 = \frac{6}{3} = 2$ A.
The current $i_1$ in the arm PQ is 2 A.