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 | 30th 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 Syntax error from line 1 column 49 to line 1 column 72. Unexpected '<mstyle '. = 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 | 30th Jun, 2014, 08:58: PM

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