CBSE Class 10 Answered
this is a question
An electric bulb is rated 220 V and 100 W. when it is operated on 110 V, the poweer consumed will be
1. 100 W 2. 75 W 3. 50 W 4. 25 W
And here is a solution you just tell out of both which is correct solution or if any one is wrong then tell its region OR BOTH CORRECT THEN TELL THEIR REGION
Solution 1
power , p=100 W potential difference , V=220 V
and resistance ,R = to be calculated
Now, P= square of V by R
so, R= 220multiply220 and by 100
so R= 484 ohms
IN SECOND CASE:
power, P= ?
P.D. , V= 110V
And resistance, R = 484 ohms
NOW, P=square of V by R
P= square of (110) by 484
P= 25 W
SOLUTION 2
in first case
Given
Power, P= 100 W P.D. ,V= 220 V
so P= V multiply I or it can be I = P by V
so I= 100 by 220
I= 5 by 11 AMPERE
in second case
power, P= ? P.D. , V= 110V CERRENT ,I= 5 by 11 so P= V multiply I
so P= 110 multiply 5 by 11
so P= 50 W
NOTES:- SOLUTION 1 IS CORRECT ACCORDIND TO NCERT BOOK . and also see in both solution the answer is differ
Asked by Akshay Rajput | 30 Jun, 2014, 05:41: PM
Expert Answer
The Solution no. -1 is correct way to solve the above question.
In the solution no.- 2
The value P = V x I = 50 W gives the total power supplied by the source.
This power has two components.
One of the component is the power consumed by the Bulb i.e 
= 25 W (Which is your solution 1)
The other component is the power loss through the conductor in the form of heat.(which is 50 - 25 = 25 watts)
Remember that when a load consumes some power W1 , then along with that power, there is always a heat loss component W2
Hence total power supplied by the source(or Battery) is W1 + W2.
Answered by Ravindra Kapal | 30 Jun, 2014, 08:58: PM
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