
Two electric bulbs marked 25 W -220 V and 100 W- 220V are connected in series to a 440 V supply. Which of the bulbs will fuse?
A. Both
B. 100 W
C. 25 W
D. neither
Answer
563.4k+ views
Hint: The power and voltage ratings for both the bulbs have been provided to us. By using these, we can find the maximum current that can flow through that particular bulb and determine if it fuses or not.
Formula used:
The power consumed can be written as:
$P = V I = \dfrac{V^2}{R}$.
Complete answer:
We first find the current ratings for the two bulbs.
For the bulb of 25 W power,
$I = \dfrac{P}{V} = \dfrac{25}{220}$ A.
For the bulb of 100 W power,
$I = \dfrac{P}{V} = \dfrac{100}{220} = \dfrac{5}{11} $ A.
Now, we need to determine the current flowing in the condition when both the bulbs are connected in series.
When the two bulbs are connected in series, their resistances simply add and when the equivalent is divided by the supply voltage, we obtain the current.
To obtain the resistance, we use the expression,
$R = \dfrac{V^2}{P}$.
In the series circuit, the equivalent resistance will be:
$R = \dfrac{(220)^2}{25} + \dfrac{(220)^2}{100} = (220)^2 \left( \dfrac{100 + 25 }{25 \times 100} \right) = (220)^2 \left( \dfrac{125 }{2500} \right) $
Dividing this with the supply voltage, we get:
$I = \dfrac{440}{(220)^2\left( \dfrac{125 }{2500} \right)} = \dfrac{2 \times 2500}{220 \times 125} = \dfrac{2}{11}$A.
Now, compare this with the current ratings for the two bulbs. This current clearly exceeds the current rating of 25 W bulb. Therefore, this bulb definitely fuses. The 100 W bulb will not fuse though as the current rating for it is more than the current flowing in the series circuit.
Therefore, the correct answer is option (B).
Note:
A bulb fuses when a current flows in it more than its rated capacity in the circuit. We went for comparing the currents everywhere because the definition of fusing is closely related to current flow, so that is the quantity to be compared. One should mind here that with the given ratings of power and voltage, we can only tell the limits for the quantities and not the exact quantities. Power rating though gives the exact power consumed by the bulb in unit time.
Formula used:
The power consumed can be written as:
$P = V I = \dfrac{V^2}{R}$.
Complete answer:
We first find the current ratings for the two bulbs.
For the bulb of 25 W power,
$I = \dfrac{P}{V} = \dfrac{25}{220}$ A.
For the bulb of 100 W power,
$I = \dfrac{P}{V} = \dfrac{100}{220} = \dfrac{5}{11} $ A.
Now, we need to determine the current flowing in the condition when both the bulbs are connected in series.
When the two bulbs are connected in series, their resistances simply add and when the equivalent is divided by the supply voltage, we obtain the current.
To obtain the resistance, we use the expression,
$R = \dfrac{V^2}{P}$.
In the series circuit, the equivalent resistance will be:
$R = \dfrac{(220)^2}{25} + \dfrac{(220)^2}{100} = (220)^2 \left( \dfrac{100 + 25 }{25 \times 100} \right) = (220)^2 \left( \dfrac{125 }{2500} \right) $
Dividing this with the supply voltage, we get:
$I = \dfrac{440}{(220)^2\left( \dfrac{125 }{2500} \right)} = \dfrac{2 \times 2500}{220 \times 125} = \dfrac{2}{11}$A.
Now, compare this with the current ratings for the two bulbs. This current clearly exceeds the current rating of 25 W bulb. Therefore, this bulb definitely fuses. The 100 W bulb will not fuse though as the current rating for it is more than the current flowing in the series circuit.
Therefore, the correct answer is option (B).
Note:
A bulb fuses when a current flows in it more than its rated capacity in the circuit. We went for comparing the currents everywhere because the definition of fusing is closely related to current flow, so that is the quantity to be compared. One should mind here that with the given ratings of power and voltage, we can only tell the limits for the quantities and not the exact quantities. Power rating though gives the exact power consumed by the bulb in unit time.
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