When an inductor of inductance \(\frac{6}{\pi} \mathrm{H}\), a capacitor of capacitance \(\frac{50}{\pi} \mu…

When an inductor of inductance \(\frac{6}{\pi} \mathrm{H}\), a capacitor of capacitance \(\frac{50}{\pi} \mu \mathrm{F}\) and resistor of resistance \(R\) are connected in series with an AC supply of rms voltage \(220 \mathrm{~V}\) and frequency \(50 \mathrm{~Hz}\), the rms current through the circuit is \(440 \mathrm{~mA}\). Match the inductive reactance, \(X_L\) the capacitive reactance, \(X_C\) the resistance \(R\) and the impedance \(Z\) of the circuit given in List-I with the corresponding values given in List-II. \(\begin{array}{llll} \hline & \text { List- I } & & \text { List- II } \\ \hline \text {(A) } & X_L & \text { (i) } & 200 \Omega \\ \hline \text {(B) } & X_C & \text { (ii) } & 300 \Omega \\ \hline \text {(C) } & R & \text { (iii) } & 500 \Omega \\ \hline \text {(D) } & Z & \text { (iv) } & 600 \Omega \\ \hline \end{array}\)
  1. \(\begin{array}{cc}\text { A } & \text { B } & \text { C } & \text { D } \\ \text {(iv) } & \text { (ii) } & \text { (i) } & \text { (iii) }\end{array}\)
  2. \(\begin{array}{cc}\text { A } & \text { B } & \text { C } & \text { D } \\ \text { (iv) } & \text { (iii) } & \text { (i) } & \text { (ii) } \end{array}\)
  3. \(\begin{array}{cc}\text { A } & \text { B } & \text { C } & \text { D } \\ \text { (iv) } & \text { (i) } & \text { (ii) } & \text { (iii) } \end{array}\)
  4. \(\begin{array}{cc}\text { A } & \text { B } & \text { C } & \text { D } \\ \text { (i) } & \text { (iv) } & \text { (iii) } & \text { (ii) }\end{array}\)

Solution

Given, inductance of inductor, \(L=\frac{6}{\pi} \mathrm{H}\), capacitance of capacitor, \(C=\frac{50}{\pi} \mu \mathrm{F}\) supply voltage, \(V_{\text {rms }}=220 \mathrm{~V}\), supply frequency, \(f=50 \mathrm{~Hz}\) and supply current, \(I_{\mathrm{rms}}=440 \mathrm{~mA}\) Now, (A) inductive reactance, \(X_L=\omega L\) \(\begin{aligned} X_L & =2 \pi f \times L \quad(\because \omega=2 \pi f) \\ & =2 \pi \times 50 \times \frac{6}{\pi}=600 \Omega \\ X_L & =600 \Omega \end{aligned}\) (B) Capacitance reactance, \(X_C=\frac{1}{\omega C}=\frac{1}{2 \pi f C}\) \(\begin{aligned} X_C & =\frac{\pi \times 10^6}{2 \pi \times 50 \times 50} \\ \Rightarrow \quad X_C & =\frac{10^6}{2500 \times 2} \Rightarrow X_C=200 \Omega \end{aligned}\) (C) Resistance, \(R\) \(\therefore\) Impedance of LCR circuit, \(Z^2=R^2+\left(X_L-X_C\right)^2\) Putting the given values, we get \(\begin{aligned} (500)^2 & =R^2+(600-200)^2 \\ 250000 & =R^2+(400)^2 \\ R^2 & =250000-160000 \\ R^2 & =90000 \Rightarrow R=300 \Omega \end{aligned}\) (D) Impedance, \(Z=\frac{V_{\text {rms }}}{I_{\text {rms }}}=\frac{220}{440} \times 10^3 \Rightarrow Z=500 \Omega\)

Asked in: AP EAMCET 2019 (20 Apr Shift 1)

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