Match List - I with List - II List - I List - II (a) Phase difference between current and voltage in a purely resistive AC circuit (i) \pi \over 2 ; current leads voltage (b) Phase difference between current and voltage in a pure inductive AC circuit (ii) zero (c) Phase difference between current and voltage in a pure capacitive AC circuit (iii) \pi \over 2 ; current lags voltage (d) Phase difference between current and voltage in an LCR series circuit (iv) \tan ^ - 1( X_C - X_L \over R ) Choose the most appropriate answer from the options given below :

(a)-(ii), (b)-(iii), (c)-(i), (d)-(iv)

(a)-(i), (b)-(iii), (c)-(iv), (d)-(ii)

(a)-(ii), (b)-(iii), (c)-(iv), (d)-(i)

(a)-(ii), (b)-(iv), (c)-(iii), (d)-(i)

Alternating Current PYQ — JEE Main 2021 · 17 Mar · Shift 2 · Q45 with Solution

Match List - I with List - II

	List - I		List - II
(a)	Phase difference between current and voltage in a purely resistive AC circuit	(i)	${\pi \over 2}$ ; current leads voltage
(b)	Phase difference between current and voltage in a pure inductive AC circuit	(ii)	zero
(c)	Phase difference between current and voltage in a pure capacitive AC circuit	(iii)	${\pi \over 2}$ ; current lags voltage
(d)	Phase difference between current and voltage in an LCR series circuit	(iv)	${\tan ^{ - 1}}\left( {{{{X_C} - {X_L}} \over R}} \right)$

Choose the most appropriate answer from the options given below :

View written solutionFree

Correct answer: A

Purely resistive AC circuit
In a resistor, current and voltage are in phase. Hence phase difference is $\phi = 0$ So, $(a) \to (ii)$
Pure inductive AC circuit
In an inductor, current lags voltage by $\frac{\pi}{2}$ .
So phase difference is $\frac{\pi}{2}\text{; current lags voltage}$ Thus, $(b) \to (iii)$
Pure capacitive AC circuit
In a capacitor, current leads voltage by $\frac{\pi}{2}$ .
So, $(c) \to (i)$
LCR series circuit
For a series $LCR$ circuit, the phase angle is given by $\tan \phi = \frac{X_L - X_C}{R}$ If written for current with respect to voltage, equivalently the listed form is $\tan^{-1}\left(\frac{X_C-X_L}{R}\right)$ which matches the given option (iv). Thus, $(d) \to (iv)$
Final matching

(a)-(ii),\quad (b)-(iii),\quad (c)-(i),\quad (d)-(iv)

This corresponds to Option A.

Match List - I with List - II

	List - I		List - II
(a)	Phase difference between current and voltage in a purely resistive AC circuit	(i)	${\pi \over 2}$ ; current leads voltage
(b)	Phase difference between current and voltage in a pure inductive AC circuit	(ii)	zero
(c)	Phase difference between current and voltage in a pure capacitive AC circuit	(iii)	${\pi \over 2}$ ; current lags voltage
(d)	Phase difference between current and voltage in an LCR series circuit	(iv)	${\tan ^{ - 1}}\left( {{{{X_C} - {X_L}} \over R}} \right)$

Choose the most appropriate answer from the options given below :

View written solutionFree

Correct answer: A

Purely resistive AC circuit
In a resistor, current and voltage are in phase. Hence phase difference is $\phi = 0$ So, $(a) \to (ii)$
Pure inductive AC circuit
In an inductor, current lags voltage by $\frac{\pi}{2}$ .
So phase difference is $\frac{\pi}{2}\text{; current lags voltage}$ Thus, $(b) \to (iii)$
Pure capacitive AC circuit
In a capacitor, current leads voltage by $\frac{\pi}{2}$ .
So, $(c) \to (i)$
LCR series circuit
For a series $LCR$ circuit, the phase angle is given by $\tan \phi = \frac{X_L - X_C}{R}$ If written for current with respect to voltage, equivalently the listed form is $\tan^{-1}\left(\frac{X_C-X_L}{R}\right)$ which matches the given option (iv). Thus, $(d) \to (iv)$
Final matching

(a)-(ii),\quad (b)-(iii),\quad (c)-(i),\quad (d)-(iv)

This corresponds to Option A.