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Initally switch `S_(1)` is closed, after a long time, `S_(1)` is opened and `S_(2)` is closed. Find the heat produced in coil after long time. |
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Answer» When `S_(1)` is closed, in steady-state current in the coil `i_(0)=E//R` . The energy stored in inductor `=(1)/(2)Li_(0)^(2)` . After closing off `S_(2)` , after long time, heat generated in the coil i.e. in `R` `=((R)/(R+R_(0)))((1)/(2)Li_(0)^(2))` `=((R)/(R+R_(0)))((1)/(2)L(E^(2))/(R^(2)))=(LE^(2))/(2R(R+R_(0))` OR After closing of `S_(2)` , current in the coil i.e. in `R` `i=i_(0)e^-((R+R_(0))/(L))t` Heat produced in coil i.e. in `R` `H=int_(0)^(oo)i^(2)Rdt=i_(0)^(2)int_(0)^(00)e^(2(R+R_(0)t)/(L)dt` `=(E^(2))/(R^(2))R(|e^(-(2(R+R_(0)))/(L)t)|_(0)^(oo))/((2R+R_(0))/(L))` `=-(LE^(2))/(2R(R+R_(0)))|e^(-e^(oo))-e^(0)=(1)/(e^(oo))-1=(1)/(oo-1)=0-1|` `=(LE)^(2)/(2R(R+R_(0)))` |
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