Power of Electrical Circuit: Practice Problem & Solution

Solution Explained:

To solve this problem, we apply the core principles of Power of Electrical Circuit. Understanding the underlying formula is key to arriving at the correct answer below:

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.

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