Physics Vidyapith ✍️

Mathematical Analysis of L-C Series Circuit : Let us consider, a circuit containing inductor $L$ capacitor $C$ and these are connected in series. If an alternating voltage source is applied across it then the resultant voltage of the L-C circuit $V=V_{L} - V_{C} \qquad(1)$ We know that: $V_{L} = iX_{L}$ $V_{C} = iX_{C}$ So from equation $(1)$ $V= iX_{L} - iX_{C} $ $V=i \left(X_{L} - X_{C} \right) $ $\frac{V}{i}=\left(X_{L} - X_{C} \right) $ $Z=\left(X_{L} - X_{C} \right) \qquad(2)$ Where $Z \rightarrow$ Impedance of L-C circuit. $X_{L} \rightarrow$ Inductive Reactance which has value $\omega L$ $X_{C} \rightarrow$ Capacitive Reactance which has value $\frac{1}{\omega C}$ So from equation $(2)$, we get $Z=\left( \omega L - \frac{1}{\omega C} \right) \qquad(3)$ The phase of resultant voltage: The phase of resultant voltage from current is $90^{\circ}$ as shown in the figure above. The Impedance and Phase at Re

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}

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

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}

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 su

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

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,

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} co

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-

Derivation of 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_{m