Circuit containing Capacitor and Resistor in Series (C-R Series Circuit )

Mathematical Analysis of C-R Series Circuit :

Let us consider, a circuit containing capacitor $C$ resistor $R$ and these are connected in series. If an alternating voltage source is applied across it then the resultant voltage of the C-R circuit

$V=\sqrt{ V_{C} ^{2} + V^{2}_{R}} \qquad(1)$

We know that:

$V_{R} = iR$
$V_{C} = iX_{C}$

So from equation $(1)$

$V=\sqrt{\left( iX_{C} \right)^{2} + \left(iR\right)^{2}} $

$V=i\sqrt{\left( X_{C} \right)^{2} + R^{2}} $

$\frac{V}{i}=\sqrt{\left( X_{C} \right)^{2} + R^{2}} $

$Z=\sqrt{\left( X_{C} \right)^{2} + R^{2}} \qquad(2)$

Where
$Z \rightarrow$ Impedance of C-R circuit.
$X_{C} \rightarrow$ Capacitive Reactance which has value $\frac{1}{\omega C}$

So from equation $(2)$, we get

$Z=\sqrt{\left( \frac{1}{\omega C} \right)^{2} + R^{2}} \qquad(3)$

The phase of resultant voltage:

If the phase of resultant voltage from from current is $\phi$ then

$tan \phi = \frac{X_{C} }{R} \qquad(4)$

$tan \phi = \frac{\frac{1}{\omega C}}{R} $

$tan \phi = \frac{1}{\omega C R} $

$\phi = tan^{1} \left(\frac{1}{\omega C R}\right) $

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Circuit containing Inductor and Resistor in Series (L-R Series Circuit )

Mathematical Analysis of L-R Series Circuit :

Let us consider, a circuit containing inductor $L$ resistor $R$ and these are connected in series. If an alternating voltage source is applied across it then the resultant voltage of the L-R circuit

$V=\sqrt{ V_{L} ^{2} + V^{2}_{R}} \qquad(1)$

We know that:

$V_{R} = iR$
$V_{L} = iX_{L}$

So from equation $(1)$

$V=\sqrt{\left( iX_{L} \right)^{2} + \left(iR\right)^{2}} $

$V=i\sqrt{\left( X_{L} \right)^{2} + R^{2}} $

$\frac{V}{i}=\sqrt{\left( X_{L} \right)^{2} + R^{2}} $

$Z=\sqrt{\left( X_{L} \right)^{2} + R^{2}} \qquad(2)$

Where
$Z \rightarrow$ Impedance of L-R circuit.
$X_{L} \rightarrow$ Inductive Reactance which has value $\omega L$

So from equation $(2)$, we get

$Z=\sqrt{\left( \omega L \right)^{2} + R^{2}} \qquad(3)$

The phase of resultant voltage:

If the phase of resultant voltage from from current is $\phi$ then

$tan \phi = \frac{X_{L} }{R} \qquad(4)$

$tan \phi = \frac{\omega L }{R} $

$\phi = tan^{1} \left(\frac{\omega L }{R}\right) $

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Circuit containing Inductor, Capacitor, and Resistor in Series (L-C-R Series Circuit )

Mathematical Analysis of L-C-R Series Circuit :

Let us consider, a circuit containing inductor $L$, capacitor $C$, and resistor $R$ and these are connected in series. If an alternating voltage source is applied across it then the resultant voltage of the L-C-R circuit

$V=\sqrt{\left( V_{L} -V_{C} \right)^{2} + V^{2}_{R}} \qquad(1)$

We know that:

$V_{R} = iR$
$V_{L} = iX_{L}$
$V_{C} = iX_{C}$

So from equation $(1)$

$V=\sqrt{\left( iX_{L} - iX_{C} \right)^{2} + (iR)^{2}} $

$V=i\sqrt{\left( X_{L} - X_{C} \right)^{2} + R^{2}} $

$\frac{V}{i}=\sqrt{\left( X_{L} - X_{C} \right)^{2} + R^{2}} $

$Z=\sqrt{\left( X_{L} - X_{C} \right)^{2} + R^{2}} \qquad(2)$

Where
$Z \rightarrow$ Impedance of L-C-R circuit.
$X_{L} \rightarrow$ Inductive Reactance which has value $\omega L$
$X_{C} \rightarrow$ Inductive Reactance which has value $\frac{1}{\omega C}$

So from equation $(2)$, we get

$Z=\sqrt{\left( \omega L - \frac{1}{\omega C} \right)^{2} + R^{2}} \qquad(3)$

The phase of resultant voltage:

If the phase of resultant voltage from from current is $\phi$ then

$tan \phi = \frac{X_{L} - X_{C}}{R} \qquad(4)$

$tan \phi = \frac{\omega L - \frac{1}{\omega C}}{R} $

$ \phi = tan^{-1} \left( \frac{\omega L - \frac{1}{\omega C}}{R} \right) $

The Impedance and Phase at Resonance Condition:($X_{L} = X_{C}$):

At resonance $X_{L} = X_{C} \qquad(5)$

$\omega L = \frac{1}{\omega C}$

$\omega^{2} = \frac{1}{L C}$

$\omega = \sqrt{\frac{1}{L C}}$

$2 \pi f = \sqrt{\frac{1}{L C}}$

$ f = \frac{1}{2 \pi}\sqrt{\frac{1}{L C}}$

Where $f \rightarrow$ Natural frequency of the circuit

1.) The Impedance of the circuit at resonance condition:

Substitute the resonance condition i.e. $X_{L} = X_{C}$ in equation $(2)$ then the impedance of the L-C-R Circuit

$Z=R$

The impedance of the L-C-R circuit at resonance condition is equal to the resistance of the resistor applied in a circuit.

2.) The Phase of resultant voltage at resonance condition:

$tan\phi =0$

$tan\phi = tan 0^{\circ}$

$\phi=0^{\circ}$

The phase of resultant voltage at resonance condition is zero. i.e. the direction of resultant voltage in the direction of current in the circuit.

Note: There are following cases arise in the L-C-R circuit at resonance condition

Case -1:

If $X_{L} \gt X_{C}$, the $tan \phi$ is positive, i.e. $\phi$ is positive. In this case, the voltage leads to the current. Therefore, the circuit is more inductive rather than capacitive or resistive.

Case -2:

If $X_{L} \lt X_{C}$, the $tan \phi$ is negative, i.e. $\phi$ is negative. In this case, the voltage lags behind the current. Therefore, the circuit has a more capacitance-dominated circuit.

Case -3:

If $X_{L} = X_{C}$, the $tan \phi$ is zero, i.e. $\phi$ is negative. In this case, the voltage and the current are in phase. Therefore, the circuit is purely resistive.

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Merits and Demerits of Alternating Current in Comparison to Direct Current

Merits and Demerits of AC in Comparison to DC

(a) Merits

(i) Alternating current can be increased or decreased by using a transformer.

This is the reason that Alternating current can be transmitted from one place to other place at relatively lower expenditure and minimum loss of energy. In Direct current, it is not possible.

(ii) Alternating current can be controlled by choke coil or capacitor at very small loss of energy. To control Direct current resistance is required in which energy loss is very high.

(ii) Alternating current can easily be converted into Direct current by using a rectifier but converting Direct current into Alternating current is not easy.

(iv) Alternating current is cheaper than DC. (life of a cell or battery is very limited).

(b) Demerits

(i) Alternating current is more dangerous as compared to Direct current.

(ii) Alternating current cannot be used in electrolysis.

(iii) Most of the Alternating current of high frequency flows on the surface of the wire, therefore, a thick wire by joining number of thin insulated wires in parallel are to be used.

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Power in Alternating Current Circuit

Definition of Power in Alternating Current Circuit:

The rate of power consumption in an alternating current circuit is known as power in the circuit.

Mathematical Analysis:

Let us consider an alternating current circuit in which the voltage and current at any instant are given by

$V=V_{\circ} sin \omega t \qquad(1)$

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

Where $\phi$ $\rightarrow$ Phase difference between voltage and current

So instantaneous Power

$P=Vi$

$P=\left\{V_{\circ} sin \omega t \right\} \left\{ i_{\circ} sin \left( \omega t - \phi \right) \right\}$

$P=V_{\circ} i_{\circ} \: sin \omega t \: sin \left( \omega t - \phi \right)$

$P=\frac{V_{\circ} i_{\circ}}{2} \left[2\: sin \: \omega t \: sin \left( \omega t - \phi \right)\right]$

$P=\frac{V_{\circ} i_{\circ}}{2} \left[ cos \left( \omega t -\omega t + \phi \right) \\ \qquad - cos \left( \omega t + \omega t - \phi \right) \right]$

$P=\frac{V_{\circ} i_{\circ}}{2} \left[ cos \left(\phi \right) - cos \left( 2\omega t - \phi \right) \right]$

The average power for one cycle is:

$P_{avg}$ = The average value of $\left\{ \frac{V_{\circ} i_{\circ}}{2} \left[ cos \left(\phi \right) - cos \left( 2\omega t - \phi \right) \right] \right\}$

The term $cos \left( 2\omega t - \phi \right)$ is time-dependent and the average value of term $cos \left( 2\omega t - \phi \right)$ for complete one cycle is zero. So the average power dissipation in one cycle is

$P_{avg}= \frac{V_{\circ} i_{\circ}}{2} cos \phi $

$P_{avg}= \frac{V_{\circ}}{\sqrt{2}} \frac{i_{\circ}}{\sqrt{2}} cos \phi $

$P_{avg}= V_{rms} \: i_{rms} cos \phi $

Case -1 The circuit containing Pure Resistor only:

If the circuit contains a pure resistance only then voltage $V$ and current $i$ are always in the same phase. i.e. $\phi=0^{\circ}$ and $cos \phi = +1$, then average power

$P_{avg}= V_{rms} \: i_{rms} $

Case -2 The circuit containing Pure Inductor only:

If the circuit containing pure inductor only then voltage $V$ leads over current $i$ by $90^{\circ}$ i.e. $\phi=90^{\circ}$ and $cos 90^{\circ} = 0$, then average power

$P_{avg}= 0 $

Case -3 The circuit containing Pure Capacitor only:

If the circuit containing pure capacitor only then voltage $V$ lags behind current $i$ by $90^{\circ}$ i.e. $\phi=90^{\circ}$ and $cos 90^{\circ} = 0$, then average power

$P_{avg}= 0 $

Case -4 The circuit containing inductor and resistor only (L-R Circuit):

We know that the phase in the L-R circuit is

$tan \phi = \frac{X_{L}}{R}$

From the above equation, Draw the triangle as shown in the figure below:

From the above figure, the value of $cos\phi$ is

$cos \phi = \frac{R}{\sqrt{R^{2} + X^{2}_{L}}}$

So average power

$P_{avg}= V_{rms} \: i_{rms} \frac{R}{\sqrt{R^{2} + X^{2}_{L}}} $

$\left(P_{avg}\right)_{L-R} \lt \left(P_{avg}\right)_{R}$

Thus, the average power consumption in an L-R circuit is less than the power consumed in a purely resistive circuit

Case -5 The circuit containing capacitor and resistor only (C-R Circuit):

We know that the phase in the C-R circuit is

$tan \phi = \frac{X_{C}}{R}$

From the above equation, Draw the triangle as shown in the figure below:

From the above figure, the value of $cos\phi$ is

$cos \phi = \frac{R}{\sqrt{R^{2} + X^{2}_{C}}}$

So average power

$P_{avg}= V_{rms} \: i_{rms} \frac{R}{\sqrt{R^{2} + X^{2}_{C}}} $

$\left(P_{avg}\right)_{C-R} \lt \left(P_{avg}\right)_{R}$

Thus, the average power consumption in a C-R circuit is less than the power consumed in a purely resistive circuit

Case -6 The circuit containing capacitor and inductor only (L-C Circuit):

We know that the phase in the L-R circuit is

$\phi = 90^{\circ}$ and $cos 90^{\circ} = 0$

So average power

$P_{avg}= 0 $

Case -7 The circuit containing inductor, capacitor, and resistor only (L-C-R Circuit):

We know that the phase in the L-C-R circuit is

$tan \phi = \frac{X_{L} - X_{C}}{R}$

From the above equation, Draw the triangle as shown in the figure below:

From the above figure, the value of $cos\phi$ is

$cos \phi = \frac{R^{2}}{\sqrt{R^{2} + \left( X^{2}_{L} - X^{2}_{C} \right) }}$

So average power

$P_{avg}= V_{rms} \: i_{rms} \frac{R^{2}}{\sqrt{R^{2} + \left( X^{2}_{L} - X^{2}_{C} \right) }} $

$\left(P_{avg}\right)_{L-C-R} \lt \left(P_{avg}\right)_{R}$

Thus, the average power consumption in an L-C-R circuit is less than the power consumed in a purely resistive circuit

Wattless Current:

If the circuit contains only resistor or only inductor or only inductor and capacitor then power consumption in the circuit will be zero (i.e. there will be no energy dissipation in the circuit) due to the phase difference $90^{\circ}$ between voltage(or impedance) and current. So the current in the circuit is called the "Wattless current".

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