Question

Match List-I with List-II :

	List–I		List–II
$\left(a\right)$	$\omega L>\frac{1}{\omega C}$	$\left(i\right)$	Current is in same phase with emf.
$\left(b\right)$	$\omega L=\frac{1}{\omega C}$	$\left(ii\right)$	Current lags behind the applied emf.
$\left(c\right)$	$\omega L<\frac{1}{\omega C}$	$\left(iii\right)$	Maximum current occurs.
$\left(d\right)$	Resonant frequency	$\left(iv\right)$	Current leads the emf.

Choose the correct answer from the options given below :

• A. $\left(a\right)–\left(ii\right);\left(b\right)–\left(i\right);\left(c\right)-\left(iv\right);\left(d\right)–\left(iii\right)$
• B. $\left(a\right)–\left(iii\right);\left(b\right)–\left(i\right);\left(c\right)-\left(iv\right);\left(d\right)–\left(ii\right)$
• C. $\left(a\right)–\left(ii\right);\left(b\right)–\left(i\right);\left(c\right)-\left(iii\right);\left(d\right)–\left(iv\right)$
• D. $\left(a\right)–\left(iv\right);\left(b\right)–\left(iii\right);\left(c\right)-\left(ii\right);\left(d\right)–\left(i\right)$

Answer 1

The correct option is A $\left(a\right)–\left(ii\right);\left(b\right)–\left(i\right);\left(c\right)-\left(iv\right);\left(d\right)–\left(iii\right)$

We know that,

$X_L=\omega L$ and $X_C=\frac{1}{\omega C}$

$\text{Option (a):}$ For $X_L>X_{{}_C,}$, the circuit is inductive, so voltage leads the current.
So, $\left(a\right)-\left(ii\right)$.

$\text{Option (b):}$ For $X_L=X_{{}_C,}$, resonance condition, voltage & current are in same phase.
So, $\left(b\right)-\left(i\right)$.

$\text{Option (c):}$ For $X_L<X_{{}_C,}$, the circuit is capacitive, so current leads the voltage.
So, $\left(c\right)-\left(iv\right)$.

$\text{Option (d):}$ For resonant frequency, current is maximum.
So, $\left(d\right)-\left(iii\right)$.

Hence, $\left(A\right)$ is the correct answer.