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If 25 W, 220 V and 100 W, 220 V bulbs are connected in series across a 440 V line, then
- a)only 25 W bulb will fuse
- b)only 100 W bulb will fuse
- c)both bulbs will fuse
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
If 25 W, 220 V and 100 W, 220 V bulbs are connected in series across ...
Solution:
When two bulbs are connected in series across a voltage source, the same current flows through both bulbs. From Ohm's law, we know that the power dissipated by a bulb is given by P = VI, where V is the voltage across the bulb and I is the current flowing through it.
Let us assume that the 25 W bulb has a resistance of R1 and the 100 W bulb has a resistance of R2. Then, the voltage drop across the 25 W bulb will be (R1 / (R1 + R2)) x 440 V, and the voltage drop across the 100 W bulb will be (R2 / (R1 + R2)) x 440 V.
Since the power dissipated by a bulb is proportional to the square of the voltage drop across it, we can write:
Power dissipated by 25 W bulb = (R1 / (R1 + R2))^2 x 440^2 / R1 x 25
Power dissipated by 100 W bulb = (R2 / (R1 + R2))^2 x 440^2 / R2 x 100
Now, we know that the total power dissipated by the bulbs is given by:
Total power = (220^2 / (R1 + R2)) x (1/25 + 1/100)
We can simplify this expression to get:
Total power = 484 / (R1 + R2)
Now, we can see that if the 25 W bulb fuses, its resistance becomes very large, and the total resistance of the circuit becomes very large. This means that the current flowing through the 100 W bulb becomes very small, and the power dissipated by it becomes much less than 100 W. Therefore, only the 25 W bulb will fuse in this case.
Hence, option A is the correct answer.
When two bulbs are connected in series across a voltage source, the same current flows through both bulbs. From Ohm's law, we know that the power dissipated by a bulb is given by P = VI, where V is the voltage across the bulb and I is the current flowing through it.
Let us assume that the 25 W bulb has a resistance of R1 and the 100 W bulb has a resistance of R2. Then, the voltage drop across the 25 W bulb will be (R1 / (R1 + R2)) x 440 V, and the voltage drop across the 100 W bulb will be (R2 / (R1 + R2)) x 440 V.
Since the power dissipated by a bulb is proportional to the square of the voltage drop across it, we can write:
Power dissipated by 25 W bulb = (R1 / (R1 + R2))^2 x 440^2 / R1 x 25
Power dissipated by 100 W bulb = (R2 / (R1 + R2))^2 x 440^2 / R2 x 100
Now, we know that the total power dissipated by the bulbs is given by:
Total power = (220^2 / (R1 + R2)) x (1/25 + 1/100)
We can simplify this expression to get:
Total power = 484 / (R1 + R2)
Now, we can see that if the 25 W bulb fuses, its resistance becomes very large, and the total resistance of the circuit becomes very large. This means that the current flowing through the 100 W bulb becomes very small, and the power dissipated by it becomes much less than 100 W. Therefore, only the 25 W bulb will fuse in this case.
Hence, option A is the correct answer.
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Community Answer
If 25 W, 220 V and 100 W, 220 V bulbs are connected in series across ...
As for an electric appliance R = (Vs2 / W), so for the same specified voltage Vs,
i.e. R25 = 4R with R100 = R
Now in series, potential divides in proportion to resistance.
So, V1 = 
i.e. V25 = 4/5 x 440 = 352 V and V2 = 
i.e. V100 = 1/5 x 440 = 88 V
From this, it is clear that voltage across 100 W bulb (= 88 V) is less than specified (220 V), while that across 25 W bulb (= 352 V) is greater than specified (220 V). So, 25 W bulb will fuse.
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If 25 W, 220 V and 100 W, 220 V bulbs are connected in series across a 440 V line, thena)only 25 W bulb will fuseb)only 100 W bulb will fusec)both bulbs will fused)None of theseCorrect answer is option 'A'. Can you explain this answer?
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