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(a). Consider circuit in figure. How much energy is absorbed by electrons from the initial state of no current (ignore thermal motion) to the state of drift velocity? (b). Electrons give up energy at the rate of RI^(2) per second to the thermal energy. what time scale would number associate with energy in problem (a)? n=number f electron/volume=10^(29)//m^(3). Length of circuit =10cm, cross-section=A=(1mm)^(2). |
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Answer» Solution :(a). By ohm's law, current I is given by `I=6V//6Omega=1A` But, `I "net"Av_(d)` or `v_(d)=(i)/("neA")` On substituting the values For, `n=`number of electron/volume `=10^(29)//m^(3)` Length of cirucit=10cm, cross-section `=A=(1mm)^(2)` `v_(d)=(1)/(10^(29)xx1.6xx10^(-19)xx10^(-6))` `=(1)/(1.6)xx10^(-4)m//s` THEREFORE, the energy absorbed int he form of KE is given by `KE=(1)/(2)m_(e)v_(d)^(2)xxnAI` `=(1)/(2)xx9.1xx10^(31)xx(1)/(2.56)xx10^(20)xx10^(8)xx10^(6)xx10^(1)` `=2xx10^(-17)J` (b). POWER loss is given by `P=I^2R=6xx1^(2)=6W=6J//s` Since, `P=(E)/(t)` Therefore, `E=Pxxt` or `t=(E)/(P)=(2xx10^(-17))/(6)=10^(-17)s` |
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