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The voltage applied to a supply inductive coil of self inductance 15.9 mH is given by the equation V=100 sin314t+75 sin 942 t+50sin 1570t. Find the equation for current wave. |
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Answer» Solution :The standard equation of the voltage can be written as `V=V_(01)sin omega_(1)t+V_(02)sinomega_(2)t+V_(03) sin omega_(3)t` On comparing this with the GIVEN equation, we have `omega_(1)=313("rad")/(s),X_(L1)=omega_(1)L=314xx15.9xx10^(-3)=5Omega` `Omega_(2)=942("rad")/(S),X_(L2)=omega_(2)L=942xx15.9xx10^(-3)=15Omega` `and omega_(3)=1570("rad")/(S),X_(L3)=omega_(3)L=1570xx15.9xx10^(-3)=25Omega` Hence : `i_(01)=(V_(01))/(X_(L1))=(100)/(5)=20A`, `i_(02)=(V_(02))/(X_(L2))=(75)/(15)=5A, and i_(03)=(V_(03))/(X_(L3))=(50)/(25)=2A` THUS `i=i_(01)sin(omega_(1)t-phi)+i_(02)sin (omega_(2)t-phi)+i_(03)sin (omega_(3)t-phi)` `=20sin(314t-(pi)/(2))+5SIN(942t-(pi)/(2))+2sin (1570t-(pi)/(2))` |
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