InterviewSolution
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InterviewSolution
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When an inductor coil is connected to an ideal battery of emf `10V`, a constant current `2.5 A` flows. When the same inductor coil is connected to an ac source of `10V` and `50 HZ` then the current is `2A`. Find out the inductance of the coil. |
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Answer» When the coil is connected to dc source, the final current is decided by the resitance of the coil. `therefore r=(10)/(2.5)=4Omega` When the coil is connected to ac, the final current is decided by the impedance of the coil. `therefore Z=(10)/(2)=5Omega` `"But" Z=sqrt((r)^(2)+(X_(L))^(2))=X_(L)^(2)=5^(2)-4^(2)=9` `X_(L)=3Omega` `therefore omegaL=2pifL=3` `therefore 2pi 50 L =3 " " therefore L=3//100pi"Henry"` |
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