Question 54: AC fundamentals

Question

A series LCR circuit is connected to a $45 \sin (\omega t)$ Volt source. The resonant angular frequency of the circuit is $10^5 ~\mathrm{rad}~ \mathrm{s}^{-1}$ and current amplitude at resonance is $I_0$. When the angular frequency of the source is $\omega=8 	imes 10^4 ~\mathrm{rad} ~\mathrm{s}^{-1}$, the current amplitude in the circuit is $0.05 I_0$. If $L=50 ~\mathrm{mH}$, match each entry in List-I with an appropriate value from List-II and choose the correct option.

List - I
List - II

(P) $I_0$ in $\mathrm{mA}$
(1) 44.4

(Q) The quality factor of the circuit
(2) 18

(R) The bandwidth of the circuit in $\mathrm{rad}~ \mathrm{s}^{-1}$
(3) 400

(S) The peak power dissipated at resonance in Watt
(4) 2250

(5) 500

Accepted Answer

$\begin{aligned} & \omega_0=10^5 \mathrm{rad} \mathrm{s}^{-1}, L=50 	imes 10^{-3} \mathrm{H} \\ & \Rightarrow X_L=5000 \Omega=X_C 	ext { (at resonance) } \\ & \frac{1}{\sqrt{L C}}=\omega_0 \\ & \Rightarrow L_C=\frac{1}{\omega_0^2} \\ & \Rightarrow C=2 	imes 10^{-9} \mathrm{~F} \\ & I_0=\frac{\varepsilon_0}{\sqrt{R^2+\left(X_L-X_Cight)^2}} \\ & I_0=\frac{\varepsilon_0}{R} ..........(i) \end{aligned}$
$$
\begin{aligned}
 0.05 I_0 & =\frac{\varepsilon_0}{\sqrt{R^2+\left(X_L-X_Cight)^2}}  ..........(ii) \\
	ext { At } \omega & =8 	imes 10^4 \mathrm{rad} \mathrm{s}^{-1} \\
X_L & =8 	imes 10^4 	imes 50 	imes 10^{-3}=4000 \Omega \\
X_C & =\frac{1}{\omega_C}=6250 \Omega
\end{aligned}
$$
From (i) and (ii), we get
$$
\begin{aligned}
& \sqrt{R^2+\left(X_L-X_Cight)^2}  =\frac{R}{0.05} \\
& R^2+(6250-4000)^2  =400 R^2 \\
& \Rightarrow R^2 \simeq \frac{(2250)^2}{400}  \\
& R  =\frac{2250}{20}=112.5 \Omega \\
&I_0  =\frac{\varepsilon_0}{R}=\frac{45}{112.5} \\
& =400 \mathrm{~m} \mathrm{~A}
\end{aligned}
$$
$(\mathrm{P} ightarrow 3)$
$\begin{aligned} ightarrow Q & =\frac{1}{R} \sqrt{\frac{L}{C}} \\ \Rightarrow Q & =\frac{2}{2250} 	imes \sqrt{\frac{50 	imes 10^{-3}}{2 	imes 10^{-7}}}=\frac{10^4}{225} \\ & =44.4\\ (\mathrm{Q} ightarrow 1) \\ Q & =\frac{\omega_0}{B W}\end{aligned}$
$\Rightarrow B W=\frac{\omega_0}{Q}=\frac{10^5}{\frac{10^4}{225}}=2250$
$(\mathrm{R} ightarrow 4)$
$\begin{array}{rrr}P  =\frac{\varepsilon_0^2}{R}  	ext { (peak power) } \\
=\frac{45^2}{112.5}=18 \mathrm{~W}  \end{array}$
$(S ightarrow 2)$

Answer

$P \rightarrow 2, Q \rightarrow 3, R \rightarrow 5, S \rightarrow 1$

Answer

$P \rightarrow 3, Q \rightarrow 1, R \rightarrow 4, S \rightarrow 2$

Answer

$P \rightarrow 4, Q \rightarrow 5, R \rightarrow 3, S \rightarrow 1$

Answer

$P \rightarrow 4, Q \rightarrow 2, R \rightarrow 1, S \rightarrow 5$

AC fundamentals

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