Alternating Current Circuit containing Capacitance only (C-Circuit)

Alternating Current Circuit Containing Capacitance Only (C-Circuit): Let us consider, A circuit containing a capacitor of capacitance $C$ only which is connected with an alternating EMF i.e electromotive force source i.e.

Let us consider, A circuit containing a coil of inductance $L$ only which is connected with an alternating EMF i.e electromotive force source i.e.

$E=E_{\circ}sin\omega t\qquad(1)$

When alternating emf is applied across the capacitor plates then the charge on capacitor plates varies continuously and correspondingly current flows in the connecting leads. Let the charge on the capacitor plates is $q$ and the current in the circuit at any instant is $i$. Since there is no resistance in the circuit then the instantaneous potential difference is $\frac{q}{C}$ across the capacitor plates must be equal to the applied emf i.e.

$\frac{q}{C} = E_{\circ} sin \omega t$

$q = CE_{\circ} sin \omega t \qquad(2)$

The instantaneous current $i$ in the circuit is, therefore

$i=\frac{dq}{dt} \qquad(3)$

Now substitute the value of $q$ from equation $(2)$ to equation $(3)$

$i=\frac{d}{dt} CE_{\circ} sin \omega t $

$i=CE_{\circ} \omega cos\omega t$

$i=\omega CE_{\circ} cos\omega t$

$i=\frac{E_{\circ}}{\frac{1}{\omega C}} cos\omega t$

$i=\frac{E_{\circ}}{X_{C}} cos\omega t$

Where $X_{C}= \frac{1}{\omega C}$ is known as capacitive reactance.

$i=i_{\circ} cos\omega t \qquad(4)$

Where the $i_{\circ}=\frac{E_{\circ}}{X_{C}}$ is the maximum current in the circuit. Now apply Ohm's law in this equation and we find that the term $X_{C}=\frac{1}{\omega C}$ has the dimension of resistance. It represents the 'effective opposition' of the capacitor to the flow of alternating current. It is known as the 'reactance of the capacitor' or 'capacitive reactance' and is denoted by $X_{C}$. The capacitive reactance is infinite for DC for which $f=0$.

$i=i_{\circ} sin \left( \omega t + \frac{\pi}{2} \right) \qquad(4)$

Now compare equation $(1)$ and equation $(4)$ which shows that an alternating circuit containing a capacitor only, the current leads the emf by a phase angle of $\frac{\pi}{2}$ or $90^\circ$ (or the emf lags behind the current by a phase angle of $\frac{\pi}{2}$). The phase diagram between EMF and the current of a capacitor is shown below-

The phasor diagram between the EMF and current of a capacitor is also shown in the given figure below-

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Alternating Current Circuit containing Inductance only (L-Circuit)

Alternating Current Circuit Containing Inductance only (L-Circuit): Let us consider, An alternating current circuit containing a coil of inductance $L$ only. This inductor is connected with an alternating EMF i.e electromotive force source i.e.

$E=E_{\circ}sin\omega t\qquad(1)$

The current $i$ in coil varies continuously then an opposite emf is induced in the coil whose magnitude is $L\frac{di}{dt}$ So the net instantaneous of the circuit:

$E_{\circ}sin\omega t -L\frac{di}{dt}=0$

$E_{\circ}sin\omega t =L\frac{di}{dt}$

$di=\frac{E_{\circ}}{L}sin\omega dt$

Now integrate the above equation then the above equation can be written as

$\int di=\int \frac{E_{\circ}}{L}sin\omega dt$

$\int di= \frac{E_{\circ}}{L} \int sin\omega dt$

$i= \frac{E_{\circ}}{L} \frac{-cos\omega t}{\omega}$

$i= -\frac{E_{\circ}}{\omega L} cos\omega t$

$i= -\frac{E_{\circ}}{X_{L}} cos\omega t$

Where $X_{L}= \omega L$ is known as inductive reactance.

$i= -i_{\circ} cos\omega t$

Where $i_{\circ}=\frac{E_{\circ}}{X_{L}}$ is known as the maximum value of current in the circuit. Now compare this equation to Ohm's law then we find that the term $X_{L}=\omega L$ has the dimensions of resistance. It defines the 'effective opposition' of the coil to the flow of alternating current. it is known as the 'reactance of the coil' or 'inductive reactance' and it is denoted by $X_{L}$. The inductive reactance $X_{L}$ is zero for DC at which frequency is zero.

$i= -i_{\circ} sin \left(\frac{\pi}{2}- \omega t \right)$

$i= i_{\circ} sin \left(\omega t - \frac{\pi}{2} \right) \qquad(2)$

Now compare equation $(1)$ and equation $(2)$ which shows that an alternating circuit containing an inductor only, the current lags behind the emf by a phase angle of $\frac{\pi}{2}$ or $90^\circ$ (or the emf leads the current by a phase angle of $\frac{\pi}{2}$). The phase diagram between EMF and the current of an inductor is shown below-

The phasor diagram between the EMF and current of an inductor is also shown in the given figure below-

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Alternating Current Circuit containing Resistance only (R-Circuit)

Alternating Current Circuit containing Resistance (R-Circuit): Let us consider, An alternating current circuit containing resistance $R$ only. This resistance $R$ is connected with an alternating EMF i.e electromotive force source i.e.

$E=E_{\circ}sin\omega t\qquad(1)$

The potential difference across the circuit

$E=iR$

Then from equation $(1)$

$iR=E_{\circ}sin\omega t $

$i=\frac{E_{\circ}}{R}sin\omega t $

$i=i_{\circ}sin\omega t \qquad(2) $

Where $i_{\circ}$ is the peak value or amplitude of the current in the circuit which has value $i_{\circ}=\frac{E_{\circ}}{R}$.

Now compare the equation $(1)$ and equation $(2)$ which shows that if a circuit is containing a resistor only then the current is always in phase with the applied EMF i.e electromotive force. The phase diagram between EMF and the current of resistance is shown below-

The phasor diagram between the EMF and current of resistance is also shown in the given figure below-

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Mean Value and Root Mean Square Value of Alternating Current

Mean Or Average Value of Alternating Current:

Let us consider alternating current $i$ propagating in a circuit then the average value of current.

$ i_{mean}=\frac{1}{\left ( \frac{T}{2} \right )}\int_{0}^{\frac{T}{2}}i \:dt \qquad (1)$

$ where \quad i = i_{0}sin \omega t\quad(2)$

Now substitute the value of current $i$ in above equation $(1)$ 

$ i_{mean}= \frac{2}{T}\int_{0}^{\frac{T}{2}}i_0. sin \omega t.dt$

$ i_{mean}= \frac{2.i_{0}}{T}\int_{0}^{\frac{T}{2}}\sin \omega t.dt$

$ i_{mean}= \frac{2 i_{0}}{T}[\frac{-cos\:\omega t}{\omega} ]_{0}^{\frac{T}{2}}$ The value of $\omega$ is $\frac{2 \pi}{T}$ i.e $\omega=\frac{2\pi}{T}$

$ i_{mean}= \frac{2 i_{0}}{T \left (\frac{2\pi}{T} \right )}\left [ -cos \left (\frac{2 \pi}{T} \right ) \left ( \frac{T}{2} \right ) \\ \qquad \qquad \qquad +cos0^\circ \right ] $

$ i_{mean}= \frac{i_{0}}{\pi}\left [ - cos\pi+cos0^{\circ} \right ]$

$ i_{mean}=\frac{i_{0}}{\pi}\left [1+1 \right ]$

$ i_{mean} =\frac{2 i_{0}}{\pi}$

$ i_{mean} = 0.637\: i_{0}$

Thus, The mean (or average) value of alternating current for the cycle is $0.637$ times or $63.7 \%$ of the peak value.

Root mean square value of Alternating Current:

Let us consider current $i$ propagating in a circuit then the mean square value of alternating current.

$ \left (i_{mean} \right )^{2} = \frac{1}{T}\int_{0}^{T}i^{2}.dt \qquad(1)$

$ where\quad i= i_{0} \dot sin\omega t\qquad (2)$

Now substitute the value of current $i$ in above equation $(1)$

$ \left ( i_{mean} \right )^{2} = \frac{1}{T}\int_{0}^{T} \left ( i_{0} \:sin\omega t\right )^{2} dt $

$ \left (i_{mean} \right )^{2} = \frac{i_{0}^2}{T}\quad\int_{0}^{T} sin^{2}\omega t.dt$

$ \left (i_{mean} \right )^{2} = \frac{i_0^{2}}{T}\int_{0}^{T} \frac{1-cos2\omega t}{2}\ dt$

$ \left (i_{mean} \right )^{2} = \frac{i_o^{2}}{2T}\int_{0}^{T}\left ( 1-cos2\omega t \right )dt$

$ \left (i_{mean} \right )^{2} = \frac{i_{0}^{2}}{2T}\left [ \left ( t \right )_{0}^{T} - \left ( \frac{sin2 \omega t}{2 \omega} \right )_{0}^{T} \right ] $

$ \left (i_{mean} \right )^{2} = \frac{i_0^{2}}{2T}\left [ \left ( T-0 \right ) \\ \qquad\qquad\qquad -\frac{1}{2\omega}\left ( sin2\omega T-sin 0^{\circ} \right ) \right ]$

$ \left (i_{mean} \right )^{2} = \frac{i_0^{2}}{2T}\left [ \left ( T-0 \right ) \\ \qquad\qquad\qquad -\frac{1}{2\omega}\left ( sin2\omega T-sin0^{\circ} \right ) \right ]$

$ \left (i_{mean} \right )^{2} = \frac{i_0^{2}}{2T}\left [ T-\frac{1}{2\omega}\left ( sin4\pi \\ \qquad \qquad\qquad  -sin0^{\circ} \right ) \right ]$

$ \left (i_{mean} \right )^{2} = \frac{i_0^{2}}{2T}\left [ T-\frac{1}{2\omega}\left ( 0-0 \right ) \right ]$

$ \left ( i_{mean} \right )^{2}=\frac{i_0^{2}}{2}$

$ \left (i_{mean} \right )^{2} = \frac{i_0^{2}}{2}$

So root mean square value of above equation:

$ i_{rms} = \sqrt{i_{mean}^{2}}$

$i_{rms} = \frac{i_{0}}{\sqrt{2}}$

$i_{rms} = 0.707\:i_{0}$

Thus, the root mean square value of an alternating current is $0.707$ times or $70.7 \%$ of the peak value.

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