Question

Match the following.

Column 1	Column 2
The phase difference between current and voltage in a purely resistive ac circuit	$\frac{𝜋}{2}$, the current leads the voltage
The phase difference between current and voltage in a pure inductive ac circuit	$0$
The phase difference between current and voltage in a pure capacitive ac circuit	$\frac{𝜋}{2}$, current lags voltage
The phase difference between current and voltage in an LCR series circuit	$ta{n}^{-1}\left(X_C–\frac{X_L}{R}\right)$

Answer 1

Part A:

1. In a purely resistive ac circuit there is no phase difference between current and voltage.
2. Voltage and current in the ac circuit are in the same phase, both reach their peak values simultaneously.
3. The phase difference is expressed as,
   $\begin{array}{rcl}& \Rightarrow & \tan \varphi =\frac{X_{L-}{X}_C}{R}\\ & \Rightarrow & \varphi ={\tan }^{-1}\left(\frac{X_{L-}{X}_C}{R}\right)\\ & \Rightarrow & \varphi ={\tan }^{-1}\left(\frac{0}{R}\right)\\ & =& 0^o\end{array}$ (where$X_L=$ Inductive reactance, $X_C=$ Capacitive reactance, $R=$Resistance)
4. As,
   $\begin{array}{rcl}& \Rightarrow & i=i_0\sin \omega t\\ & \Rightarrow & ∆\varphi =0^{\circ }\end{array}$ ($∆\varphi =$phase difference)

Hence, (A) - (ii)

Part B:

1. In a purely inductive ac circuit the current lags behind the emf or voltage by a phase difference of$\frac{\pi }{2}\left(or{90}^o\right)$.
2. The phase difference is expressed as,
   $\begin{array}{rcl}& \Rightarrow & \tan \varphi =\frac{X_{L-}{X}_C}{R}\\ & \Rightarrow & \varphi ={\tan }^{-1}\left(\frac{X_L}{0}\right)\\ & \Rightarrow & \varphi ={\tan }^{-1}\left(\infty \right)\\ & =& {90}^o\end{array}$ (where$X_L=$ Inductive reactance, $X_C=$ Capacitive reactance, $R=$Resistance)
3. As,
   $\begin{array}{rcl}& \Rightarrow & i=i_0\sin \left(\omega t-\frac{\mathrm{\pi }}{2}\right)\\ & \Rightarrow & ∆\varphi ={90}^{\circ }\end{array}$ ($∆\varphi =$phase difference)

Hence, (B) - (iii)

Part C:

1. In a purely capacitive ac circuit the current leads the emf or voltage by a phase angle$\frac{\pi }{2}\left(or{90}^o\right)$.
2. The phase of current is$\frac{\pi }{2}\left(or{90}^o\right)$ more than that of emf.
3. The phase difference is expressed as,
   $\begin{array}{rcl}& \Rightarrow & \tan \varphi =\frac{X_{L-}{X}_C}{R}\\ & \Rightarrow & \varphi ={\tan }^{-1}\left(\frac{X_C}{0}\right)\\ & \Rightarrow & \varphi ={\tan }^{-1}\left(\infty \right)\\ & =& {90}^o\end{array}$ (where$X_L=$ Inductive reactance, $X_C=$ Capacitive reactance, $R=$Resistance)
4. As,
   $\begin{array}{rcl}& \Rightarrow & i=i_0\sin \left(\omega t+\frac{\mathrm{\pi }}{2}\right)\\ & \Rightarrow & ∆\varphi ={90}^{\circ }\end{array}$ ($∆\varphi =$phase difference)

Hence, (C) - (i)

Part D:

1. In an LCR series ac circuit the phase difference between applied emf and circuit current is,
   $$\begin{gathered} \Rightarrow \tan \varphi =\frac{X_{L-}{X}_C}{R} \\ \Rightarrow \varphi ={\tan }^{-1}\left(\frac{X_{L-}{X}_C}{R}\right) \end{gathered}$$ (where$X_L=$ Inductive reactance, $X_C=$ Capacitive reactance, $R=$Resistance)

Thus (D) - (iv)

Hence, the correct answer is (A)-(ii), (B)-(iii), (C)-(i), (D)-(iv).