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If two bulbs of wattage 25 and 30, each rated at 220 volts, are connected in series with a 440 volt supply, which bulb will fuse
- a)25 W bulb
- b)30 W bulb
- c)Neither of them
- d)Both of them
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
If two bulbs of wattage 25 and 30, each rated at 220 volts, are connec...
Safe current through 25 W bulb
Safe current through 30W blub
This is greater than the safe current for 25W bulb, so it will fuse
Most Upvoted Answer
If two bulbs of wattage 25 and 30, each rated at 220 volts, are connec...
Understanding the Problem
When two bulbs are connected in series, the same current flows through both bulbs. The total voltage from the supply is divided between the bulbs based on their resistances.
Calculating Resistance of Each Bulb
- The formula for resistance (R) is given by:
R = V^2 / P
where V is the rated voltage and P is the power (wattage).
- For the 25 W bulb:
R1 = (220^2) / 25 = 1936 ohms
- For the 30 W bulb:
R2 = (220^2) / 30 = 1613.33 ohms
Total Resistance in Series
- Total Resistance (R_total) = R1 + R2 = 1936 + 1613.33 = 3549.33 ohms
Current in the Circuit
- Using Ohm's Law, the current (I) can be calculated as:
I = V_total / R_total
I = 440 / 3549.33 = 0.124 A (approximately)
Voltage Drop Across Each Bulb
- Voltage drop across the 25 W bulb (V1):
V1 = I * R1 = 0.124 * 1936 = 240.64 V
- Voltage drop across the 30 W bulb (V2):
V2 = I * R2 = 0.124 * 1613.33 = 200.12 V
Conclusion: Which Bulb Will Fuse?
- The 25 W bulb is rated for 220 V but experiences approximately 240.64 V, exceeding its rating.
- The 30 W bulb receives approximately 200.12 V, which is within its limits.
Thus, the 25 W bulb will fuse, confirming that the correct answer is option 'A'.
When two bulbs are connected in series, the same current flows through both bulbs. The total voltage from the supply is divided between the bulbs based on their resistances.
Calculating Resistance of Each Bulb
- The formula for resistance (R) is given by:
R = V^2 / P
where V is the rated voltage and P is the power (wattage).
- For the 25 W bulb:
R1 = (220^2) / 25 = 1936 ohms
- For the 30 W bulb:
R2 = (220^2) / 30 = 1613.33 ohms
Total Resistance in Series
- Total Resistance (R_total) = R1 + R2 = 1936 + 1613.33 = 3549.33 ohms
Current in the Circuit
- Using Ohm's Law, the current (I) can be calculated as:
I = V_total / R_total
I = 440 / 3549.33 = 0.124 A (approximately)
Voltage Drop Across Each Bulb
- Voltage drop across the 25 W bulb (V1):
V1 = I * R1 = 0.124 * 1936 = 240.64 V
- Voltage drop across the 30 W bulb (V2):
V2 = I * R2 = 0.124 * 1613.33 = 200.12 V
Conclusion: Which Bulb Will Fuse?
- The 25 W bulb is rated for 220 V but experiences approximately 240.64 V, exceeding its rating.
- The 30 W bulb receives approximately 200.12 V, which is within its limits.
Thus, the 25 W bulb will fuse, confirming that the correct answer is option 'A'.
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If two bulbs of wattage 25 and 30, each rated at 220 volts, are connected in series with a 440 volt supply, which bulb will fusea)25 W bulbb)30 W bulbc)Neither of themd)Both of themCorrect answer is option 'A'. Can you explain this answer?
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