Fig. shows three similar lamps L1, L2 and L3 connected across a power…

Let R be the resistance of each lamp. If E be the applied e.m.f., then the current in the circuit, I₁, is given by

I₁ = $\frac{E}{R+(R / 2)}=\frac{2 E}{3 R}$

Current flowing through L₂ or L₃ = $\frac{1}{2}\left[\frac{2 E}{3 R}\right]=\frac{E}{3 R}$

When L₃ is fused, the whole current flows through L₁ and L₂.

Thus I₂ = $\frac{E}{R+R}=\frac{E}{2 R}$

So, current through L₁ decreases and through L₂ increases. Hence option (b) is correct.