Question

In the given figure AB=BC=CA=L and DE=EF=FD=$\frac{L}{2}$.

Resistance per unit length of wire to make structure is $\lambda$ and resistance of wire connecting battery $\epsilon$ to end A and C is zero. Reading of Ammeter is

• A. $\frac{5\epsilon }{8\lambda L}$
• B. $\frac{3\epsilon }{2\lambda L}$
• C. $\frac{2\epsilon }{\lambda L}$
• D. $\frac{9\epsilon }{5\lambda L}$

Answer 1

The correct option is D $\frac{9\epsilon }{5\lambda L}$


$R_{eq}=\frac{5}{9}\lambda Lisi=\frac{\epsilon }{R_{eq}}=\frac{9\epsilon }{5\lambda L}$