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A `750 Hz, 20 V` source is connected to as resistance of `100 Omega` an inductance of `0.1803 H` and a capacitance of `10 muF` all in sereis.Calculate the time in which the resistance (thermalcapacity `2J//.^(@)C`) wil get heated by `10^(@)C`. |
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Answer» Here, `v=750 Hz, E_(v)=20V, R=100 Omega` `L=0.1803 H, C=10 muF=10^(-5) F, t=?` `Delta theta=10^(@)C`, thermal capacity `=2J//.^(@)C` `X_(L)= omegaL=2pi vL=2xx3.14xx750xx0.1803` `X_(C)=1/(omega C)=(1)/(2pi xx750xx10^(-5))=21.2 Omega` `Z=sqrt(R^(2)+(X_(L)-X_(C))^(2))` `=sqrt(100^(2)+(850-21.2)^(2))=835 Omega` Power dissipated `=E_(v)I_(v) cos phi` `=E_(v) (E_(v)/Z)(R/Z)=(20^(2)xx100)/((835)^(2))=0.0574 W` Heat produced in resistance `=2xx10=20 J` If t is the required time, then `Pxxt=20 rArr t=20/P=20/0.0574=348 s` |
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