Question

If 2 bulbs rated 2.5 W - 110 V and 100 W - 110 V are connected in series to a 220 V supply

• A. 2.5 W bulb will fuse
• B. 100 W bulb will fuse
• C. both will fuse
• D. both will not fuse

Answer 1

Answer: (A)

2. 5 W bulb will fuse

Answer 2

Explanation:

1. Calculate Resistance of Each Bulb:

   The resistance ($R$) of a bulb can be calculated using its power ($P$) and voltage ($V$) ratings with the formula $R = \frac{V^2}{P}$.

   • For the 2.5 W - 110 V bulb (Bulb 1): $R_1 = \frac{(110 \, \text{V})^2}{2.5 \, \text{W}} = \frac{12100}{2.5} = 4840 \, \Omega$
   • For the 100 W - 110 V bulb (Bulb 2): $R_2 = \frac{(110 \, \text{V})^2}{100 \, \text{W}} = \frac{12100}{100} = 121 \, \Omega$

2. Calculate Total Resistance in Series:

   When bulbs are connected in series, their resistances add up. $R_{total} = R_1 + R_2 = 4840 \, \Omega + 121 \, \Omega = 4961 \, \Omega$

3. Calculate Current in the Series Circuit:

   The total current ($I$) flowing through the series circuit is given by Ohm's Law: $I = \frac{V_{supply}}{R_{total}}$. $I = \frac{220 \, \text{V}}{4961 \, \Omega} \approx 0.04435 \, \text{A}$

   Since it's a series circuit, this current flows through both bulbs.

4. Calculate Voltage Across Each Bulb:

   Using Ohm's Law ($V = I \times R$), we can find the voltage across each bulb in the series circuit.

   • Voltage across Bulb 1 ($V_1$): $V_1 = I \times R_1 = 0.04435 \, \text{A} \times 4840 \, \Omega \approx 214.6 \, \text{V}$
   • Voltage across Bulb 2 ($V_2$): $V_2 = I \times R_2 = 0.04435 \, \text{A} \times 121 \, \Omega \approx 5.37 \, \text{V}$

5. Compare Calculated Voltage with Rated Voltage:

   A bulb fuses if the voltage across it significantly exceeds its rated voltage. Both bulbs are rated for 110 V.

   • For the 2.5 W bulb: The voltage across it ($V_1 \approx 214.6 \, \text{V}$) is much greater than its rated voltage (110 V). This excessive voltage will cause the bulb to draw too much current and fuse.
   • For the 100 W bulb: The voltage across it ($V_2 \approx 5.37 \, \text{V}$) is much less than its rated voltage (110 V). This bulb will not fuse; in fact, it will glow very dimly.

Therefore, the 2.5 W bulb will fuse.