## In This post you will get answer of Q.

An AC generator producing 10V(rms) at 200 rad/s is connected in series with a $50 Omega$ resistor, a 400 mH inductor and a $200 mu F$ capacitor. The rms voltage across the inductor is

Solution:

## Given $E =10 V$

$omega =200 ,rad / s $

$R =50 ,Omega$

$L =400 ,mH $

$C =200 ,mu F$

We know that,

$Z=sqrt{R^{2}+left(X_{L}-X_{C}right)^{2}}$

$=sqrt{(50)^{2}+(80-25)^{2}}=sqrt{(50)^{2}-(55)^{2}} $

$=sqrt{2500+3025} $

$Z =sqrt{5525}=74.3 ,Omega $

$I =frac{E}{Z}=frac{10}{74.3}=0.13459 A$

$E_{L} =I X_{L}=0.13459 times 80 $

$=10.76 V text { or } 10.8 V $