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 $
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 $