Question

Two bulbs of 100w and 40w respectively connected in series across a 230v supply which bulb will glow bright and why?

Answer 1

When two bulbs (100W and 40W) are connected in series across a 230V supply, the 40W bulb will glow brighter. Here's why:

1. **Series Connection Basics**: In a series circuit, the same current flows through both bulbs. However, the voltage is distributed across them based on their resistance.

2. **Power and Resistance Relationship**:
   - The power rating of a bulb is given by \( P = \frac{V^2}{R} \), where \( P \) is power, \( V \) is voltage, and \( R \) is resistance.
   - Rearranging the formula, we get \( R = \frac{V^2}{P} \), meaning the resistance of a bulb is inversely proportional to its power rating.
   
   - Therefore, the 100W bulb has a lower resistance, and the 40W bulb has a higher resistance.

3. **Voltage Distribution in Series**: Since the bulbs are in series, the total voltage (230V) is divided between the two bulbs according to their resistance. The bulb with higher resistance (40W) will have a greater share of the voltage across it.

4. **Power Dissipation**: The power dissipated by each bulb in a series connection is \( P = I^2 R \). Since the current is the same for both, the bulb with the higher resistance (the 40W bulb) will dissipate more power and, thus, glow brighter.

In summary, the **40W bulb** will glow brighter because it has a higher resistance and will receive a greater share of the voltage across it in a series circuit.

Answer 2

When two bulbs of different wattages (100W and 40W) are connected in series across a 230V supply, the brightness of each bulb depends on the power they dissipate in the circuit.

Here’s a step-by-step explanation of which bulb will glow brighter:

1. **Determine the Resistance of Each Bulb:**
   - For a bulb, the power \( P \) and resistance \( R \) are related by the formula: \( P = \frac{V^2}{R} \), where \( V \) is the voltage across the bulb.
   - Rearranging this formula to find the resistance: \( R = \frac{V^2}{P} \).

   For the 100W bulb:
   \[
   R_{100W} = \frac{V^2}{P} = \frac{230^2}{100} = 529 \text{ ohms}
   \]

   For the 40W bulb:
   \[
   R_{40W} = \frac{V^2}{P} = \frac{230^2}{40} = 1322.5 \text{ ohms}
   \]

2. **Calculate the Total Resistance in Series:**
   \[
   R_{total} = R_{100W} + R_{40W} = 529 + 1322.5 = 1851.5 \text{ ohms}
   \]

3. **Find the Current Flowing Through the Circuit:**
   Using Ohm’s Law: \( I = \frac{V_{supply}}{R_{total}} \)
   \[
   I = \frac{230}{1851.5} \approx 0.124 \text{ A}
   \]

4. **Calculate the Voltage Drop Across Each Bulb:**
   - For the 100W bulb:
     \[
     V_{100W} = I \times R_{100W} = 0.124 \times 529 \approx 65.6 \text{ V}
     \]

   - For the 40W bulb:
     \[
     V_{40W} = I \times R_{40W} = 0.124 \times 1322.5 \approx 164.4 \text{ V}
     \]

5. **Determine the Power Dissipated by Each Bulb:**
   Power dissipation in each bulb is given by \( P = I^2 \times R \).

   - For the 100W bulb:
     \[
     P_{100W} = I^2 \times R_{100W} = (0.124)^2 \times 529 \approx 8.0 \text{ W}
     \]

   - For the 40W bulb:
     \[
     P_{40W} = I^2 \times R_{40W} = (0.124)^2 \times 1322.5 \approx 20.3 \text{ W}
     \]

**Conclusion:**
The 40W bulb will glow brighter because it dissipates more power (approximately 20.3W) compared to the 100W bulb (approximately 8.0W). Despite having a lower nominal power rating, the 40W bulb’s higher resistance causes it to drop more voltage and thus dissipate more power when connected in series with the 100W bulb.