Current through bulb 4 is maximum

Step 1: Resistance of the bulbs

$$R_1 = \frac{110^2}{60} \approx 202 \, \Omega, \quad R_2 = \frac{110^2}{100} = 121 \, \Omega.$$

Step 2: Total current in the circuit

In series, the same current flows through both bulbs:

$$I = \frac{V_{\text{total}}}{R_1 + R_2} = \frac{220}{202 + 121} \approx 0.682 \, \text{A}.$$

Step 3: Voltage across each bulb

• Voltage across the 60 W bulb:

$$V_1 = I \cdot R_1 = 0.682 \cdot 202 \approx 137.6 \, \text{V}.$$

• Voltage across the 100 W bulb:

$$V_2 = I \cdot R_2 = 0.682 \cdot 121 \approx 82.5 \, \text{V}.$$

Step 4: Compare with rated voltage

• The 60 W bulb experiences
  $137.6 \, \text{V}$ 

  , exceeding its $110 \, \text{V}$ 

  rating, so it fuses.

• The 100 W bulb experiences
  $82.5 \, \text{V}$ 

  , within its $110 \, \text{V}$ 

  rating.

Conclusion

The 60 W bulb will fuse.

$$\boxed{\text{Answer: 60 W bulb.}}$$

To find the resistance

$R$

that must be placed in series with the bulb, let's analyze the problem step by step.

Given:

1. Power of the bulb (
   $P$ 

   ) = 500 W,

2. Voltage rating of the bulb (
   $V_b$ 

   ) = 100 V,

3. Supply voltage (
   $V_s$ 

   ) = 200 V.

Step 1: Resistance of the bulb

The resistance of the bulb (

$R_b$

) can be calculated using the formula:

$$R_b = \frac{V_b^2}{P}.$$

Substitute the values:

$$R_b = \frac{100^2}{500} = \frac{10000}{500} = 20 \, \Omega.$$

Step 2: Total current through the circuit

The bulb is rated to draw 500 W at 100 V. Thus, the current through the bulb is:

$$I = \frac{P}{V_b}.$$

Substitute the values:

$$I = \frac{500}{100} = 5 \, \text{A}.$$

Step 3: Voltage drop across the series resistor

The total supply voltage is 200 V, and the bulb operates at 100 V. Therefore, the voltage drop across the series resistor

$R$

is:

$$V_R = V_s - V_b.$$

Substitute the values:

$$V_R = 200 - 100 = 100 \, \text{V}.$$

Step 4: Resistance of the series resistor

Using Ohm's law, the resistance of the series resistor is:

$$R = \frac{V_R}{I}.$$

Substitute the values:

$$R = \frac{100}{5} = 20 \, \Omega.$$

Final Answer:

The resistance that must be placed in series with the bulb is:

$$\boxed{20 \, \Omega}.$$