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A resistor R1 dissipate power P when connected with a ideal generator. When a resistance R2 is put in series, supply voltage remaining same,the power dissipated by R1 is  
Asked by sumsansam | 08 Jul, 2019, 09:08: PM
answered-by-expert Expert Answer
Let E be the emf ( electomotive force ) of generator.
 
When a resistance R is connected to generator, power dissipation P is given by,  
 
P = ( E2/R1 )   ..................... (1)
 
Hence emf of generator, E = ( P R1 )1/2  ...................(2)
 
when another resistance R2 is connected in series with R1 and this series combination is conneted to the generator,
then  current I drawn from generator is given by, 
 
I = E/( R1 + R2 )  = ( P R1 )1/2 / ( R1 + R2 ) ......................(3)
 
power dissipated by Resistor R1 = I2 R1 = [ ( P R1 ) / ( R1 + R2 )2 ]×R1 = P × [ R12 / ( R1 + R2 )2 ]
Answered by Thiyagarajan K | 08 Jul, 2019, 10:36: PM

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