Five bulbs are connected in series across 220-Volt line. If one bulb is fused the remaining bulbs are again connected across the same line. Which one of the arrangements will be more illuminated? Justify.

The average power absorbed by each light bulb is directly proportional to the amount of light it emits.

Case I: 

Let $R_1 = R_2 = R_3 = R_4 = R_5 = 100\; \Omega$

Then total resistance $R = 500\;\Omega$

Now, Power $(P_1) =$$\frac{V^2}{R} = \frac{(220)^2}{500}$$= 96.8\;\;Watt$

Case II:

Let $R_1 = R_2 = R_3 = R_4 = 100\; \Omega$

Then total resistance $R = 400\;\Omega$

Now, Power $(P_1) =$$\frac{V^2}{R} = \frac{(220)^2}{400}$$= 121\;\;Watt$

Since $P_2 > P_1$ 

Therefore the second case with four light bulbs draws more power and will produce more light.