Alternating Current

In an AC resistive circuit, current and voltage are in phase.

So, $I = {V \over R} \Rightarrow I = {{220} \over {50}}\sin (100\pi t)$ ..... (i)

$\therefore$ Time period of one complete cycle of current is

$T = {{2\pi } \over \omega } = {{2\pi } \over {100\pi }} = {1 \over {50}}s$

So, current reaches its maximum value at

${t_1} = {T \over 4} = {1 \over {200}}s$

When current is half of its maximum value, then from Eq.(i), we have

$I = {{{I_{\max }}} \over 2} = {I_{\max }}\sin (100\pi {t_2})$

$ \Rightarrow \sin (100\pi {t_2}) = {1 \over 2} \Rightarrow 100\pi {t_2} = {{5\pi } \over 6}$

So, instantaneous time at which current is half of maximum value is ${t_2} = {1 \over {120}}s$

Hence, time duration in which current reaches half of its maximum value after reaching maximum value is

$\Delta t = {t_2} - {t_1} = {1 \over {120}} - {1 \over {200}} = {1 \over {300}}s = 3.3$ ms

Phase difference between I₂ and V, i.e. C $-$ R₂ circuit is given by

$\tan \phi = {{{X_C}} \over {{R_2}}} \Rightarrow \tan \phi = {1 \over {C\omega {R_2}}}$

Substituting the given values, we get

$\tan \phi = {1 \over {{{\sqrt 3 } \over 2} \times {{10}^{ - 6}} \times 100 \times 20}} = {{{{10}^3}} \over {\sqrt 3 }}$

$\therefore$ $\phi$₁ is nearly 90$^\circ$.

Phase difference between I₁ and V, i.e. in L $-$ R₁ circuit is given by

$\tan {\phi _2} = - {{{X_L}} \over {{R_1}}} = - {{L\omega } \over R}$

Substituting the given values, we get

$\tan {\phi _2} = - {{{{\sqrt 3 } \over {10}} \times 100} \over {10}} = - \sqrt 3 $

As, $\tan {\phi _2} = - \sqrt 3 $

$\therefore$ ${\phi _2} = 120^\circ $

Now, phase difference between I₁ and I₂ is

$\Delta \phi = {\phi _2} - {\phi _1} = 120^\circ - 90^\circ = 30^\circ $

It is given that e₀ = 283 V; $\omega$ = 320.

The inductor reactance is X_L = 320 $\times$ 25 $\times$ 10^$-$3 = 8 $\Omega$

The capacitor reactance is

${X_C} = {1 \over {\omega C}} = {1 \over {320 \times 1000 \times {{10}^{ - 6}}}} = {{1000} \over {320}} = 3.1\,\Omega $

It is given that R = 5 $\Omega$. Therefore, the total impedance is

$Z = \sqrt {{R^2} + {{({X_L} - {X_C})}^2}} = \sqrt {50} = 7\,\Omega $

and the phase difference between the voltage across the source and the current is

$\tan \phi = {{{X_L} - {X_C}} \over R} = {{8 - 3.1} \over 5} \approx 1 \Rightarrow \theta = 45^\circ $

From the circuit, we have

$R = {{80} \over {10}} = 8\,\Omega $

$10 = {{220} \over {\sqrt {{R^2} + X_L^2} }} \Rightarrow \sqrt {64 + X_L^2} = 22$

$ \Rightarrow X_L^2 = 484 - 64 = 420$

$ \Rightarrow {X_L} = \sqrt {420} = 20.5\,\Omega $

$ \Rightarrow \omega L = 20.5$

$L = {{20.5} \over {2\pi \times 50}} = {{20.5} \over {314}} = 0.065\,H$

The phase difference between the alternating current and emf in an AC circuit depends on the components in the circuit:

• In a purely resistive ($R$) circuit, the current and emf are in phase, meaning the phase difference is $0$.


• In a purely inductive ($L$) circuit, the current lags behind the emf by $\frac{\pi}{2}$, meaning the phase difference is $\frac{\pi}{2}$.


• In a purely capacitive ($C$) circuit, the current leads the emf by $\frac{\pi}{2}$, again meaning the phase difference is $\frac{\pi}{2}$.


• In an $L$-$R$ or $L$-$C$ circuit, the phase difference depends on the relative values of $L$, $R$, and $C$ and can be anywhere between $0$ and $\frac{\pi}{2}$.

Therefore, if the phase difference between the alternating current and emf is $\frac{\pi}{2}$, then the circuit cannot contain only a resistor ($R$) since that would give a phase difference of $0$. So, the answer is Option A: $R,L$. The phase difference would not be $\frac{\pi}{2}$ if the circuit contains both a resistor and an inductor.