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In a series LCR circuit, voltage drop ax...

In a series LCR circuit, voltage drop axross L can be more than applied voltage (True/false).

Text Solution

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True, voltage across inductor is given by
`V_(Lv)=I_(v)X_(L)=(E_(v)X_(L))/(sqrt(R^(2)+(X_(L)-X_(C)^(2)))`
for certain values of R, `X_(L) and X_(C) , X_(L)` can be grater than `Z=sqrt(R^(2)+(X_(L)-X_(C)^(2))`. In that case `V_(Lv)` can be greater than `E_(v)`.
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