1.

Calculate the back e.m.f of a 10H, 200 Omega coil 100 ms after a 100Vdc supply is connected to it.

Answer»

Solution :The VALUE of current at 100 ms after the switch is CLOSED is
`I=I_(0)[1-e^((-t)/(T_(0)))]`, Here, `I_(0) =(100)/(200)` =0.5 amp,
`tau_(0)=(L)/(R)=(10)/(200)=0.05` sec, t=0.1 sec
`I=0.5(1-e^(-0.1//0.05))=0.5(1-e^(-2))=0.4325A`
Now, `E=IR+L(dI)/(dt)`, or `100= 0.4325 xx 200+L(dI)/(dt)`
Back `e.m.f=L(dI)/(dt) = 100 -0.4325 xx 200 = 13.5V`


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