4 Marks Questions

Question 14 Marks

Write the equation of average power for $L - C - R$ series AC circuit and discuss its special cases.

Answer

→Average power for $L - C - R$ series AC circuit,
$\begin{aligned}
P & = VI \cos \phi \\
\therefore \quad P & =\frac{v_m i_m}{2} \cos \phi
\end{aligned}$
• Special Cases :
(i) Case I : Resistive Circtuit :
→ For an AC circuit containing only resistance (i.e. for purely resistive circuits), so $\phi=0$
$\therefore \quad$ Average power dissipation,
$\begin{aligned}
& P & = VI \cos \phi \\
\therefore \quad & P & = VI \cos 0 \\
\therefore \quad & P & = VI \text { (Maximum) }
\end{aligned}$
→ Thus, in the AC circuit having only resistance, power dissipation is maximum.
(ii) Case II : Purely Inductive Or Purely Capacitive Circuit :
→If the circuit is purely inductive or purely capacitive, phase difference between voltage and current is $\frac{\pi}{2}$.
→Therefore, average power dissipated in circuit is :
$\begin{aligned}
P & = VI \cos \phi \\
\therefore P & = VI \cos \frac{\pi}{2} \\
\therefore P & =0
\end{aligned}$
→ Thus, eventhough a current is flowing in the circuit, power dissipated is zero.
→ Such a current is referred to as Wattless Current .
(iii) Case IIII: $L - C - R$ Series Circuit:
→In an $L - C - R$ series circuit, power dissipated is as per,
$P = VI \cos \phi$

Where, $\phi=\tan ^{-1}\left(\frac{ X _{ C }- X _{ L }}{ R }\right)$
→So, $\phi$ may be non zero in a RL or RC or RCL circuit, Even in such cases, power is dissipated only in the resistor.
(iv) Case IV : Power dissipated at Resonance in $L-C-R$ Circuit :
→At resonance, $X _{ C }= X _{ L }$ and $\phi=0$ so $\cos \phi=1$
→Therefore, power dissipated in circuit.
$\begin{array}{l}
P = VI \cos \phi \\
P = VI \text { (Maximum) }
\end{array}$
→ Thus, maximum power is dissipated in a circuit (through R) at Resonance.

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Question 24 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.

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Physics 4 Marks Questions

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