Class 12 Exam > Class 12 Questions > When two bulbs A (40 W, 220 V) and B (60 W, 2...
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When two bulbs A (40 W, 220 V) and B (60 W, 220 V) are connected in series, which bulb is brighter.
- a)bulb A
- b)bulb B
- c)both
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
When two bulbs A (40 W, 220 V) and B (60 W, 220 V) are connected in se...
As
P = V x i = i2R = V2/R
R for 40 W bulb
R40 = V2/P = (200)2/40 = 1210Ω
R for 60 W bulb
R60 = V2/P = (200)2/60 = 806.67Ω
⇒ R40>R60
When two bulbs are connected in series, current flow through them is same and power is calculated using the following formula:
P = i2R or P ∞ R
Since, R40 > R60 therefore, P40 > P60 Therefore, power dissipation is large in 40 W bulb as compare to 60 W bulb. 40 W bulb is brighter than 60 W.
P = V x i = i2R = V2/R
R for 40 W bulb
R40 = V2/P = (200)2/40 = 1210Ω
R for 60 W bulb
R60 = V2/P = (200)2/60 = 806.67Ω
⇒ R40>R60
When two bulbs are connected in series, current flow through them is same and power is calculated using the following formula:
P = i2R or P ∞ R
Since, R40 > R60 therefore, P40 > P60 Therefore, power dissipation is large in 40 W bulb as compare to 60 W bulb. 40 W bulb is brighter than 60 W.
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When two bulbs A (40 W, 220 V) and B (60 W, 220 V) are connected in se...
Understanding Bulbs in Series
When two bulbs are connected in series, the same current flows through both bulbs. This affects their brightness, which is determined by the power consumed by each bulb.
Power and Resistance Calculation
- The power rating of a bulb indicates its resistance. Power (P) is given by the formula P = V^2 / R, where V is the voltage and R is the resistance.
- For Bulb A (40 W, 220 V):
- Resistance (R_A) = V^2 / P = (220^2) / 40 = 1210 ohms.
- For Bulb B (60 W, 220 V):
- Resistance (R_B) = V^2 / P = (220^2) / 60 = 806.67 ohms.
Current in Series Circuit
- In a series circuit, the total resistance (R_total) is the sum of the individual resistances:
- R_total = R_A + R_B = 1210 + 806.67 = 2016.67 ohms.
- The total current (I) flowing through the circuit can be calculated using Ohm's law (I = V / R_total):
- I = 220 / 2016.67 = 0.109 A.
Power Dissipation in Each Bulb
- The power consumed by each bulb can be calculated using P = I^2 * R:
- For Bulb A:
- P_A = (0.109)^2 * 1210 = 13.03 W.
- For Bulb B:
- P_B = (0.109)^2 * 806.67 = 9.70 W.
Conclusion: Brightness Comparison
- Since power dissipation determines brightness, Bulb A (13.03 W) consumes more power than Bulb B (9.70 W).
- Therefore, Bulb A will be brighter when the two bulbs are connected in series.
Thus, the correct answer is option 'A' - Bulb A.
When two bulbs are connected in series, the same current flows through both bulbs. This affects their brightness, which is determined by the power consumed by each bulb.
Power and Resistance Calculation
- The power rating of a bulb indicates its resistance. Power (P) is given by the formula P = V^2 / R, where V is the voltage and R is the resistance.
- For Bulb A (40 W, 220 V):
- Resistance (R_A) = V^2 / P = (220^2) / 40 = 1210 ohms.
- For Bulb B (60 W, 220 V):
- Resistance (R_B) = V^2 / P = (220^2) / 60 = 806.67 ohms.
Current in Series Circuit
- In a series circuit, the total resistance (R_total) is the sum of the individual resistances:
- R_total = R_A + R_B = 1210 + 806.67 = 2016.67 ohms.
- The total current (I) flowing through the circuit can be calculated using Ohm's law (I = V / R_total):
- I = 220 / 2016.67 = 0.109 A.
Power Dissipation in Each Bulb
- The power consumed by each bulb can be calculated using P = I^2 * R:
- For Bulb A:
- P_A = (0.109)^2 * 1210 = 13.03 W.
- For Bulb B:
- P_B = (0.109)^2 * 806.67 = 9.70 W.
Conclusion: Brightness Comparison
- Since power dissipation determines brightness, Bulb A (13.03 W) consumes more power than Bulb B (9.70 W).
- Therefore, Bulb A will be brighter when the two bulbs are connected in series.
Thus, the correct answer is option 'A' - Bulb A.
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When two bulbs A (40 W, 220 V) and B (60 W, 220 V) are connected in series, which bulb is brighter.a)bulb Ab)bulb Bc)bothd)none of theseCorrect answer is option 'A'. Can you explain this answer? for Class 12 2025 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about When two bulbs A (40 W, 220 V) and B (60 W, 220 V) are connected in series, which bulb is brighter.a)bulb Ab)bulb Bc)bothd)none of theseCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Class 12 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for When two bulbs A (40 W, 220 V) and B (60 W, 220 V) are connected in series, which bulb is brighter.a)bulb Ab)bulb Bc)bothd)none of theseCorrect answer is option 'A'. Can you explain this answer?.
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