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Question: Answered & Verified by Expert
An AC generator $10 \mathrm{~V}(\mathrm{rms})$ at $200 \mathrm{rad} / \mathrm{s}$ is connected in series with a $50 \Omega$ resistor, a $400 \mathrm{mH}$ inductor and a $200 \mu \mathrm{F}$ capacitor. The rms voltage across the inductor is
PhysicsAlternating CurrentTS EAMCETTS EAMCET 2017
Options:
  • A $2.5 \mathrm{~V}$
  • B $3.4 \mathrm{~V}$
  • C $6.7 \mathrm{~V}$
  • D $10.8 \mathrm{~V}$
Solution:
1706 Upvotes Verified Answer
The correct answer is: $10.8 \mathrm{~V}$
Given,
$$
\begin{aligned}
& E=10 \mathrm{~V} \\
& \omega=200 \mathrm{rad} / \mathrm{s} \\
& R=50 \Omega \\
& L=400 \mathrm{mH} \\
& C=200 \mu \mathrm{F}
\end{aligned}
$$
We know that,
$$
\begin{aligned}
Z & =\sqrt{R^2+\left(X_L-X_C\right)^2} \\
& =\sqrt{(50)^2+(80-25)^2}=\sqrt{(50)^2-(55)^2} \\
& =\sqrt{2500+3025} \\
Z & =\sqrt{5525}=74.3 \Omega \\
I & =\frac{E}{Z}=\frac{10}{74.3}=0.13459 \mathrm{~A} \\
E_L & =I X_L=0.13459 \times 80 \\
& =10.76 \mathrm{~V} \text { or } 10.8 \mathrm{~V}
\end{aligned}
$$

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