When an inductor of inductance ( (6)/(π) H ), a capacitor of...

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. (

	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

)

Options

A( $\begin{array}{cc}\text{ A } & \text{ B } & \text{ C } & \text{ D } \\ \text{(iv) } & \text{ (ii) } & \text{ (i) } & \text{ (iii) }\end{array}$ )
B( $\begin{array}{cc}\text{ A } & \text{ B } & \text{ C } & \text{ D } \\ \text{ (iv) } & \text{ (iii) } & \text{ (i) } & \text{ (ii) } \end{array}$ )
C( $\begin{array}{cc}\text{ A } & \text{ B } & \text{ C } & \text{ D } \\ \text{ (iv) } & \text{ (i) } & \text{ (ii) } & \text{ (iii) } \end{array}$ )
D( $\begin{array}{cc}\text{ A } & \text{ B } & \text{ C } & \text{ D } \\ \text{ (i) } & \text{ (iv) } & \text{ (iii) } & \text{ (ii) }\end{array}$ )

Correct answer

C. ( $\begin{array}{cc}\text{ A } & \text{ B } & \text{ C } & \text{ D } \\ \text{ (iv) } & \text{ (i) } & \text{ (ii) } & \text{ (iii) } \end{array}$ )

Step-by-step 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)$

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