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A cuuent of 4.8 A is flowing in a computer.he number of e/second moving any cross section normal to direction of flow is

(ii)If current in bulb decreases by 20% then power decreases by
it is current sry
i did the answ for increase in current for 20%.
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Charge of an electron is  e = 1.602 * 10^-19 coulombs
1 ampere is equal to  1 coulomb of charge passing across any cross section normal to the direction of flow of current.

 number of electrons / second = (4.8  coulombs/sec) / (1.6 * 10^-19 Coulombs)
                   = 3 * 10^19 per second


  Power P =  I^2 R  ,    R is a constant for one bulb.

     P1 = I1^2 * R          and  P2 = I2^2 * R
         I2 = I1 -  I1 * 20/100  = 0.80 * I1
           P2 = (0.8* I1)^2 * R  = 0.64 * P1

     ( P2 - P1 ) / P1 = -0.36
             =>  36 % decrease in power.

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what can i do if you do not understand the simple solution. please ask your friend to explain this answer. i cannot explain in more detail. this is good enough.
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