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Q.

If two bulbs of 25W and 100W rated at 200V are connected in 440V supply, then


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a

100 watt bulb will fuse      

b

25 watt bulb will fuse

c

none of the bulb will fuse

d

both the bulbs will fuse 

answer is B.

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Detailed Solution

Given two bulbs having power, P1=25W and P2=100Wboth are rated at voltage, V=200volts and connected in series with 440 volts supply.
The expression for power is given as,
P=V2/ R
Where, V is the voltage and R is the resistance.
From the above expression,
R=V2/ P
Hence, we can find the resistance of each bulb using this equation for the given power and voltage rating.
⇒R1=V2/ P1⇒(200V)2/25W⇒1600Ω
The resistance of 25W bulb is 1600Ω. And,
⇒R2=V2/ P2⇒(200V)2/100W⇒400Ω
The resistance of 100W is 400Ω.
Since the two bulbs are connected in series, the total resistance will be R1+R2.
The voltage across each bulb will be different. They are connected to 440 volts supply also.
Hence, the voltage across the 25W bulb is given as,
⇒V1=440V ((R1/(R1+R2))
Substituting the values in the above expression,
⇒V1=440V (1600Ω/(1600Ω+400Ω)) = 352V
The voltage across 25W is 352V. This is higher than the rated voltage 200volts. Therefore, the bulb will fuse.
The voltage across 100W bulb is given as,
⇒V2=440V ((R2/(R1+R2))
Substituting the values in the above expression,
⇒V2=440V ((400Ω/(1600Ω+400Ω)) = 88V
The voltage across 100W is 88V. This is lower than the rated voltage 200volts. Therefore, the bulb will not fuse.
Therefore, only the 25W bulb will fuse.
 
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