Question

When an inductor of inductance $\frac{6}{\pi} H , a$ capacitor of capacitance $\frac{50}{\pi} \mu F$ and resistor of resistance $R$ are connected in series with an AC supply of rms voltage $220\, V$ and frequency $50\, Hz$, the rms current through the circuit is $440\, mA$. 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.

List I		List II
(A)	$X_L$	(I)	200 $\Omega$
(B)	$X_C$	(II)	300 $\Omega$
(C)	$R$	(III)	500 $\Omega$
(D)	$Z$	(IV)	600 $\Omega$

• A. A $\to$ IV , B $\to$ II, C $\to$ I , D $\to$ III
• B. A $\to$ IV , B $\to$ III, C $\to$ I , D $\to$ II
• C. A $\to$ IV , B $\to$ I, C $\to$ II , D $\to$ III
• D. A $\to$ I , B $\to$ IV, C $\to$ III , D $\to$ II

Answer 1

Answer: (C)

Given, inductance of inductor, $L=\frac{6}{\pi} \,H$
capacitance of capacitor, $C=\frac{50}{\pi} \mu F$
supply voltage, $V_{ rms }=220 \,V$,
supply frequency, $f=50 \,Hz$
and supply current, $I_{ rms }=440 \,mA$
Now, (A) inductive reactance, $X_{L}=\omega L$
$X_{L} =2 \pi f \times L$
$ (\because \omega=2 \pi f) $
$=2 \pi \times 50 \times \frac{6}{\pi}=600 \Omega $
$X_{L} = 600 \,\Omega $
(B) Capacitance reactance, $X_{C}=\frac{1}{\omega C}=\frac{1}{2 \pi f C}$
$X_{C} =\frac{\pi \times 10^{6}}{2 \pi \times 50 \times 50} $
$\Rightarrow X_{C} =\frac{10^{6}}{2500 \times 2}$
$ \Rightarrow X_{C}=200 \,\Omega$
(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
$(500)^{2} =R^{2}+(600-200)^{2} $
$250000 =R^{2}+(400)^{2} $
$R^{2} =250000-160000 $
$R^{2} =90000$
$ \Rightarrow R=300\, \Omega$
(D) Impedance,
$Z=\frac{V_{ rms }}{I_{ rms }}=\frac{220}{440} \times 10^{3}$
$ \Rightarrow Z=500\, \Omega$