# 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

1
by rajusetu
it is current sry
i did the answ for increase in current for 20%.
create one query - one problem per question.

2014-12-24T15:38:17+05:30

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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

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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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