Match List-I with List-II: List-I List-II (a) L,>,1/ C (i) Current is in phase with emf (b) L = 1 C (ii) Current lags behind the applied emf (c) L, <, 1 C (iii) Maximum current occurs (d) resonant frequency (iv) Current leads the emf Choose the correct answer from the options given below:

(A) (a) - (iv) ; (b) - (iii) ; (c) - (ii) ; (d) - (i)

(B) (a) - (iii) ; (b) - (i) ; (c) - (iv) ; (d) - (ii)

(D) (a) - (ii) ; (b) - (i) ; (c) - (iii) ; (d) - (iv)

JEE Mains · Physics · STD 12 -Alternating current

Match List\(-I\) with List\(-II\) :

List\(-I\) List\(-II\)

\((a)\) \(\omega L\,>\,\frac{1}{\omega C}\) \((i)\) Current is in phase with \(emf\)

\((b)\) \(\omega {L}=\frac{1}{\omega {C}}\) \((ii)\) Current lags behind the applied \(emf\)

\((c)\) \(\omega {L}\, < \,\frac{1}{\omega {C}}\) \((iii)\) Maximum current occurs

\((d)\) resonant frequency \((iv)\) Current leads the \(emf\)

Choose the correct answer from the options given below :

Pick an option to check your answer.

Verified Solution

Correct Answer

(C) \((a) - (ii) ; (b) - (i) ; (c) - (iv) ; (d) - (iii)\)

Detailed explanation

\((a)\) For \({x}_{{L}}\,>\,{x}_{{C}}\), voltage leads the current

\((ii)\)

\((b)\) For \(x_{L}=x_{C}\), voltage \(\&\) current are in same phase

\((i)\)

\((c)\) For \({x}_{{L}}\, < \,{x}_{{C}}\), current leads the voltage

\((iv)\)

\((d)\) For resonant frequency \({x}_{{L}}={x}_{{C}}\), current is maximum

\((iii)\)

Same subject

From JEE Mains

List\(-I\)	List\(-II\)
\((a)\) \(\omega L\,>\,\frac{1}{\omega C}\)	\((i)\) Current is in phase with \(emf\)
\((b)\) \(\omega {L}=\frac{1}{\omega {C}}\)	\((ii)\) Current lags behind the applied \(emf\)
\((c)\) \(\omega {L}\, < \,\frac{1}{\omega {C}}\)	\((iii)\) Maximum current occurs
\((d)\) resonant frequency	\((iv)\) Current leads the \(emf\)