Question

A wire of resistance $9\Omega$ is broken in two parts. The length ratio being 1 : 2. Two pieces are connected in parallel. The net resistance will be

• A. $2\Omega$
• B. $8\Omega$
• C. $4\Omega$
• D. $6\Omega$

Answer 1

The correct option is A $2\Omega$

Since $R=\frac{\rho l}{A}$, the resistance of indivisual pieces will be $3\Omega$ and $6\Omega$.

Now the net resistance $\left(R_{eq}\right)=\frac{R_1{R}_2}{R_1+R_2}=\frac{3\times 6}{3+6}=2\Omega$

Therefore, A is correct option.