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
rajusetu:
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.
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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