Question 14 Marks
For the $L - C - R$ series AC circuit derive the phase relationship between instantaneous current and voltage, using the phasor diagram.
Answer
→As shown in the fig., a resistor, an inductor and a capacitor are connected in series in an AC circuit.
→Voltage of the AC source $v=v_m \sin \omega t$
→As the components are in series, ac current in each element is same, having the same amplitude and phase.
→Let current $i$ be :
$i=i_m \sin (\omega t+\phi)$
where, $\phi$ - is the phase difference between the voltage across the source and the current in the circuit.
→In the fig. (a), the phasor representing the current in the circuit as given by eq. (1), is shown by $\vec{I}$.
→Further, $\vec{V}_L, \vec{V}_R, \vec{V}_C$ and $\vec{V}$ represent the voltage across inductor $L$, resistor $R$, capacitor C and the source, respectively.
→All the phasors are shown in the fig. with their corresponding phase difference.
→The length of these phasors or the amplitudes of $\overrightarrow{ V }_{ R }, \overrightarrow{ V }_{ C }$ and $\overrightarrow{ V }_{ L }$ are :
→$v_{ R m}=i_m R , v_{ C m}=i_m X _{ C }, v _{ L m}=i_m X _{ L }$
→From the phasor diagram, the equation of resultant voltage is as follows :
$\overrightarrow{ V }_{ L }+\overrightarrow{ V }_{ R }+\overrightarrow{ V }_{ C }=\overrightarrow{ V }$
(Note : voltage equation from the vertical components of above phasor relation can be written as : $v_L+v_R+v_C=v$ )
→Since $\vec{V}_C$ and $\vec{V}_L$ are always along the same line and in opposite directions, they can be combined into a single phasor $\left(\overrightarrow{ V }_{ C }+\overrightarrow{ V }_{ L }\right)$ which has a magnitude $\left|v_{ Cm }-v_{ Lm }\right|$.
→Since $\vec{V}$ is represented as the hypotenuse of a right-angle whose sides are $\vec{V}_R$ and $\vec{V}_C+\vec{V}_L$ the phythagorean theorem gives :
$\begin{aligned}
→v_m^2 & =v_{ R m}^2+\left(v_{ C m}-v_{ L m}\right)^2 \\
v_m^2 & =\left(i_m R \right)^2+\left(i_m X _{ C }-i_m X _{ L }\right)^2 \\
\therefore v_m^2 & =i_m^2 R ^2+i_m^2\left( X _{ C }- X _{ L }\right)^2 \\
\therefore v_m^2 & =i_m^2\left[ R ^2+\left( X _{ C }- X _{ L }\right)^2\right] \\
\therefore i_m^2 & =\frac{v_m^2}{ R ^2+\left( X _{ C }- X _{ L }\right)^2} \\
\therefore i_m & =\frac{v_m}{\sqrt{ R ^2+\left( X _{ C }- X _{ L }\right)^2}}
\end{aligned}$
→The equation can also be written as follows :
$\therefore i_m=\frac{U_m}{Z}$
where, $Z=\sqrt{R^2+\left(X_C-X_L\right)^2}$
→Z is known as impedence of the given AC circuit, which is analogous to resistance in a DC circuit. Its unit is ohm $(\Omega)$ (And it has the same dimension as resistance).
→Since phasor $\vec{I}$ is always parallel to phasor $\overrightarrow{V_R}$, the phase angle $\phi$ is the angle between $\overrightarrow{V_R}$ and $\overrightarrow{ V }$ and it is shown in fig. (c).
→$\begin{array}{l}
\tan \phi=\frac{v_{ Cm }-v_{ Lm }}{v_{ Rn }} \\
\tan \phi=\frac{i_m X _{ C }-i_m X _{ L }}{i_{m R }} \\
\therefore \tan \phi=\frac{ X _{ C }- X _{ L }}{ R }
\end{array}$
→The diagram shown in fig. (c) is known as Impedence diagram, which is a right-triangle with $Z$ as its hypotenuse.
View full question & answer→→As shown in the fig., a resistor, an inductor and a capacitor are connected in series in an AC circuit.
→Voltage of the AC source $v=v_m \sin \omega t$
→As the components are in series, ac current in each element is same, having the same amplitude and phase.
→Let current $i$ be :
$i=i_m \sin (\omega t+\phi)$
where, $\phi$ - is the phase difference between the voltage across the source and the current in the circuit.
→In the fig. (a), the phasor representing the current in the circuit as given by eq. (1), is shown by $\vec{I}$.
→Further, $\vec{V}_L, \vec{V}_R, \vec{V}_C$ and $\vec{V}$ represent the voltage across inductor $L$, resistor $R$, capacitor C and the source, respectively.
→All the phasors are shown in the fig. with their corresponding phase difference.
→The length of these phasors or the amplitudes of $\overrightarrow{ V }_{ R }, \overrightarrow{ V }_{ C }$ and $\overrightarrow{ V }_{ L }$ are :
→$v_{ R m}=i_m R , v_{ C m}=i_m X _{ C }, v _{ L m}=i_m X _{ L }$
→From the phasor diagram, the equation of resultant voltage is as follows :
$\overrightarrow{ V }_{ L }+\overrightarrow{ V }_{ R }+\overrightarrow{ V }_{ C }=\overrightarrow{ V }$
(Note : voltage equation from the vertical components of above phasor relation can be written as : $v_L+v_R+v_C=v$ )
→Since $\vec{V}_C$ and $\vec{V}_L$ are always along the same line and in opposite directions, they can be combined into a single phasor $\left(\overrightarrow{ V }_{ C }+\overrightarrow{ V }_{ L }\right)$ which has a magnitude $\left|v_{ Cm }-v_{ Lm }\right|$.
→Since $\vec{V}$ is represented as the hypotenuse of a right-angle whose sides are $\vec{V}_R$ and $\vec{V}_C+\vec{V}_L$ the phythagorean theorem gives :
$\begin{aligned}
→v_m^2 & =v_{ R m}^2+\left(v_{ C m}-v_{ L m}\right)^2 \\
v_m^2 & =\left(i_m R \right)^2+\left(i_m X _{ C }-i_m X _{ L }\right)^2 \\
\therefore v_m^2 & =i_m^2 R ^2+i_m^2\left( X _{ C }- X _{ L }\right)^2 \\
\therefore v_m^2 & =i_m^2\left[ R ^2+\left( X _{ C }- X _{ L }\right)^2\right] \\
\therefore i_m^2 & =\frac{v_m^2}{ R ^2+\left( X _{ C }- X _{ L }\right)^2} \\
\therefore i_m & =\frac{v_m}{\sqrt{ R ^2+\left( X _{ C }- X _{ L }\right)^2}}
\end{aligned}$
→The equation can also be written as follows :
$\therefore i_m=\frac{U_m}{Z}$
where, $Z=\sqrt{R^2+\left(X_C-X_L\right)^2}$
→Z is known as impedence of the given AC circuit, which is analogous to resistance in a DC circuit. Its unit is ohm $(\Omega)$ (And it has the same dimension as resistance).
→Since phasor $\vec{I}$ is always parallel to phasor $\overrightarrow{V_R}$, the phase angle $\phi$ is the angle between $\overrightarrow{V_R}$ and $\overrightarrow{ V }$ and it is shown in fig. (c).
→$\begin{array}{l}
\tan \phi=\frac{v_{ Cm }-v_{ Lm }}{v_{ Rn }} \\
\tan \phi=\frac{i_m X _{ C }-i_m X _{ L }}{i_{m R }} \\
\therefore \tan \phi=\frac{ X _{ C }- X _{ L }}{ R }
\end{array}$
→The diagram shown in fig. (c) is known as Impedence diagram, which is a right-triangle with $Z$ as its hypotenuse.