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A 20 V, 750 Hz source us connected to a series combination of `R = 100 Omega, C = 10 mu F` and `L = 0.1803 H`. Calculate the time in which resistance will get heated by `10^(@) C`, if thermal capacity of the material `= 2 J//.^(@)C` |
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Answer» Here, `E_(v) = 20 V, v = 750 Hz, R = 100 Omega, C mu F = 10^(-5) F, L = 0.1803 H, Delta theta = 10^(@)C, ms = 2 J//.^(@)C` `X_(L) = 2 pi v L = 2 xx (22)/(7) xx 750 xx 0.1803 Omega = 850 Omega, X_(C ) = (1)/(2 pi v C) = (1)/(2 xx (22)/(7) xx 750 xx 10 xx 10^(-6)) Omega = 21.2 Omega` `Z = sqrt(R^(2) + (X_(L) - X_(C))^(2)) = sqrt(100^(2) + (850 - 21.2)^(2)) = 834.8 Omega` `I_(v) = (E_(v))/(Z) = (20)/(834.8) = 0.02395` As `I_(v)^(2) Rt = (ms) Delta theta = 2 xx 10 = 20 :. t = 20//I_(v)^(2) R = (20)/((0.02395)^(2) xx 100) = 348.7 s` |
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