Question

A current of $ 6A$ 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______.

Answer 1

Step 1. Given data

Current entering P is $i_t=6A$

Resistance of each side of triangle is $2\Omega$

Step 2. Solving for $i_1$

The diagram can be redrawn as:

Since total current will be constant $i_t=i_1+i_2$

Through the arm $PQ$ and $QR$same current flows, as they are in series, equivalent resistance is $R_{PQ-QR}=2+2=4\Omega$

The voltage across parallel arms is same, implies that the voltage across resistance of $4\Omega$ and $2\Omega$ is $V_1$

$$\begin{gathered} i_t=i_1+i_2 \\ \Rightarrow 6=\frac{V_1}{4}+\frac{V_1}{2} \\ \Rightarrow 6=\frac{3V_1}{4} \\ \Rightarrow V_1=8V \\ \Rightarrow i_1=\frac{V_1}{4}=\frac{8}{4}=2A \end{gathered}$$

Hence the correct answer is $2A$.