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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

TS EAMCET 2017

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 $