Answer

(b) Given,

$$ \begin{aligned} & v=50 \mathrm{~Hz}, I_{\mathrm{rms}}=5 \mathrm{~A} \\ & t=\frac{1}{300} \mathrm{~s} \end{aligned} $$

We have to find $I(t)$

$$ \begin{aligned} I_{0} & =\text { Peak value }=\sqrt{2}, I_{\mathrm{rms}}=\sqrt{2} \times 5 \\ & =5 \sqrt{2} \mathrm{~A} \\ I & =I_{0} \sin \omega t=5 \sqrt{2} \sin 2 \pi v t=5 \sqrt{2} \sin 2 \pi \times 50 \times \frac{1}{300} \\ & =5 \sqrt{2} \sin \frac{\pi}{3}=5 \sqrt{2} \times \frac{\sqrt{3}}{2}=5 \sqrt{3 / 2} \mathrm{~A} \end{aligned} $$

Answer

(c) For delivering maximum power from the generator to the load, total internal reactance must be equal to conjugate of total external reactance.

Hence,

$$ \begin{aligned} X_{\text {int }} & ={ }^* X_{\text {ext }} \\ X_g & =\left(X_L\right)^*=-X_{L} \\ X_L & =-X_g \end{aligned} $$

Answer

(c) The voltmeter connected to $\mathrm{AC}$ mains reads mean value $\left(\left\langle v^{2}\right\rangle\right)$ and is calibrated in such a way that it gives value of $\left\langle v^{2}\right\rangle$, which is multiplied by form factor to give rms value.

Answer

(b) We know that resonant frequency in an $L-C-R$ circuit is given by

$$ v_{0}=\frac{1}{2 \pi \sqrt{L C}} $$

Now to reduce $v_{0}$ either we can increase $L$ or we can increase $C$.

To increase capacitance, we must connect another capacitor parallel to the first.

Thinking Process

For better tuning of an L-C-R circuit used for communication, quality factor of the circuit must be as high as possible.

Answer

(c) Quality factor $(Q)$ of an $L-C-R$ circuit is given by,

$$ Q=\frac{1}{R} \sqrt{\frac{L}{C}} $$

where $R$ is resistance, $L$ is inductance and $C$ is capacitance of the circuit. To make $Q$ high,

$R$ should be low, $L$ should be high and $C$ should be low.

These conditions are best satisfied by the values given in option (c).

Note We should be careful while writing formula for quality factor, because we are considering series L-C-R circuit.

Answer

(c) Given, $X_{L}=1 \Omega, R=2 \Omega$

$$ E_{\mathrm{rms}}=6 \mathrm{~V}, P_{\mathrm{av}}=? $$

Average power dissipated in the circuit

$$ \begin{aligned} P_{\mathrm{av}} & =E_{\mathrm{rms}} I_{\mathrm{rms}} \cos \phi \\ I_{\mathrm{rms}} & =\frac{I_{0}}{\sqrt{2}}=\frac{E_{\mathrm{rms}}}{Z} \\ Z & =\sqrt{R^{2}+X_{L}^{2}} \\ & =\sqrt{4+1}=\sqrt{5} \\ I_{\mathrm{rms}} & =\frac{6}{\sqrt{5}} \mathrm{~A} \\ \cos \phi & =\frac{R}{Z}=\frac{2}{\sqrt{5}} \\ P_{\mathrm{av}} & =6 \times \frac{6}{\sqrt{5}} \times \frac{2}{\sqrt{5}} \\ & =\frac{72}{\sqrt{5} \sqrt{5}}=\frac{72}{5}=14.4 \mathrm{~W} \end{aligned} $$

Answer

(a) Secondary voltage $V_{S}=24 \mathrm{~V}$

Power associated with secondary $P_{S}=12 \mathrm{~W}$

$$ \begin{aligned} I_{S} & =\frac{P_{S}}{V_{S}}=\frac{12}{24} \\ & =\frac{1}{2} \mathrm{~A}=0.5 \mathrm{~A} \end{aligned} $$

Peak value of the current in the secondary

$$ \begin{aligned} I_{0} & =I_{S} \sqrt{2} \\ & =(0.5)(1.414)=0.707=\frac{1}{\sqrt{2}} \mathrm{~A} \end{aligned} $$

Thinking Process

We can decide the elements, comprising the given circuit by predicting the variation in their reactances with frequency.

Answer

$(a, d)$

Reactance of an inductor of inductance $L$ is, $X_{L}=2 \pi \nu L$ where $v$ is frequency of the AC circuit.

$$ \begin{aligned} X_{C} & =\text { Reactance of the capacitive circuit } \\ & =\frac{1}{2 \pi f C} \end{aligned} $$

On increasing frequency $v$, clearly $X_{L}$ increases and $X_{C}$ decreases.

For a $L-C-R$ circuit,

$$ \begin{aligned} Z & =\text { Impedance of the circuit } \\ & =\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}} \\ & =\sqrt{R^{2}+(2 \pi v L-\frac{1}{2 \pi v C}})^2 \end{aligned} $$

As frequency $(v)$ increases, $Z$ decreases and at certain value of frequency know as resonant frequency $\left(\nu_{0}\right)$, impedance $Z$ is minimum that is $Z_{\min }=R$ current varies inversely with impedance and at $Z_{\min }$ current is maximum.

Answer

$(c, d)$

According to the question, the current increases on increasing the frequency of supply. Hence, the reactance of the circuit must be decreases as increasing frequency.

For a capacitive circuit,

$$ X_{C}=\frac{1}{\omega C}=\frac{1}{2 \pi f C} $$

Clearly when frequency increases, $X_{C}$ decreases.

For $R$ - $C$ circuit,

$$ X=\sqrt{R^{2}+(\frac{1}{w C})^{2}} $$

when frequency increases $X$ decreases.

Thinking Process

Power loss due to transmission lines having resistance ( $R$ ) and rms current flowing $I_{r m s}$ is $I^{2}{ }_{r m s} R$

Answer

$(a, b, d)$

We have to transmit energy (power) over large distances at high alternating voltages, so current flowing through the wires will be low because for a given power $(P)$.

$$ \begin{aligned} P & =E_rms I_rms, I_rms \text { is low when } E_rms \text { is high. } \\ \text { Power loss } & =I^{2}{ }_{\text {rms }} R=\text { low } \quad\left(\because I_rms \text { is low }\right) \end{aligned} $$

Now at the receiving end high voltage is reduced by using step-down transformers.

Answer

$(a, b, c)$

According to question power transferred,

$$ P=I^{2} Z \cos \phi $$

where $I$ is the current, $Z=$ Impedance and $\cos \phi$ is power factor

As power factor,

where

$$ \begin{aligned} \cos \phi & =\frac{R}{Z} \\ R & >0 \text { and } Z>0 \\ \cos \phi & >0 \Rightarrow P>0 \end{aligned} $$

$$ \Rightarrow $$

Answer

(c, $d)$

When the AC voltage is applied to the capacitor, the plate connected to the positive terminal will be at higher potential and the plate connected to the negative terminal will be at lower potential.

The plate with positive charge will be at higher potential and the plate with negative charge will be at lower potential. So, we can say that the charge is in phase with the applied voltage.

$$ \begin{aligned} \text{Power applied to a circuit is} \quad P_av & =V_rms I_rms \cos \phi \\ \text{For capactive circuit,} \quad \phi & =90^{\circ} \\ \Rightarrow \quad \cos \phi & =0 \\ \Rightarrow \quad P_{\mathrm{av}} & =\text { Power delivered }=0 \end{aligned} $$

Answer

$(a, d)$

For house hold supplies, AC currents are used which are having zero average value over a cycle.

The line is having some resistance so power factor $\cos \phi=\frac{R}{Z} \neq 0$

so,

$$ \phi \neq \pi / 2 \Rightarrow \phi,<\pi / 2 $$

i.e., phase lies between 0 and $\pi / 2$.

Answer

If we consider a $L-C$ circuit analogous to a harmonically oscillating springblock system. The electrostatic energy $\frac{1}{2} C V^{2}$ is analogous to potential energy and energy associated with moving charges (current) that is magnetic energy $\frac{1}{2} L I^{2}$ is analogous to kinetic energy.

Thinking Process

The component with infinite resistance will be considered as open circuit and the component with zero resistance will be considered as short circuited.

Answer

We know that inductive reactance $X_{L}=2 \pi f L$

and capacitive reactance $X_{C}=\frac{1}{2 \pi f C}$

For very high frequencies $(f \rightarrow \infty), X_{L} \rightarrow \infty$ and $X_{C} \rightarrow 0$

When reactance of a circuit is infinite it will be considered as open circuit. When reactance of a circuit is zero it will be considered as short circuited.

So, $C_{1}, C_{2} \rightarrow$ shorted and $L_{1}, L_{2} \rightarrow$ opened.

So, effective impedance $=R_{\text {eq }}=R_{1}+R_{3}$

Answer

Let,

$$ \begin{aligned} & \left(I_{\mathrm{rms}}\right) a=\text { rms current in circuit }(\mathrm{a}) \\ & \left(I_{\mathrm{rms}}\right) b=\text { rms current in circuit }(\mathrm{b}) \\ & \left(I_{\mathrm{rms}}\right) a=\frac{V_{\mathrm{rms}}}{R}=\frac{V}{R} \\ & \left(I_{\mathrm{rms}}\right) b=\frac{V_{\mathrm{rms}}}{Z}=\frac{V}{\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}} \end{aligned} $$

(a) When

$$ \begin{aligned} \left(I_{\mathrm{rms}}\right) a & =\left(I_{\mathrm{rms}}\right) b \\ R & =\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}} \end{aligned} $$

$$ \Rightarrow \quad X_{L}=X_{C} \text {, resonance condition } $$

(b) As $Z \geq R$

$$ \begin{array}{rlrl} \Rightarrow \frac{\left(I_{\mathrm{rms}}\right) a}{\left(I_{\mathrm{rms}}\right) b} & =\frac{\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}}{R} \\ =\frac{Z}{R} \geq 1 \\ \Rightarrow \left(I_{\mathrm{rms}}\right) a & \geq\left(I_{\mathrm{rms}}\right) b \end{array} $$

No, the rms current in circuit (b), cannot be larger than that in (a).

Answer

Let the applied emf

and current developed is

$$ E=E_{0} \sin (\omega t) $$

and cure

$$ I=I_{0} \sin (\omega t \pm \phi) $$

Instantaneous power output of the $\mathrm{AC}$ source

$$ \begin{aligned} P & =E I=\left(E_{0} \sin \omega t\right) \\ & =E_{0} I_{0} \sin \omega t \cdot \sin (\omega t+\phi) \\ & =\frac{E_{0} I_{0}}{2}[\cos \phi-\cos (2 \omega t+\phi)] \end{aligned} $$

Average power

$$ \begin{aligned} P_{\mathrm{av}}= & \frac{V_{0}}{\sqrt{2}} \frac{I_{0}}{\sqrt{2}} \cos \phi \\ & =V_{\mathrm{rms}} I_{\mathrm{rms}} \cos \phi \end{aligned} $$

where $\phi$ is the phase difference.

Clearly, from Eq. (i)

when

$$ \begin{gathered} \cos \phi<\cos (2 \omega t+\phi) \\ P<0 \end{gathered} $$

Yes, the instantaneous power output of an $A C$ source can be negative From Eq. (ii)

$$ \begin{aligned} P_{\mathrm{av}} & >0 \\ \cos \phi & =\frac{R}{Z}>0 \end{aligned} $$

Because

No, the average power output of an $A C$ source cannot be negative.

Answer

Consider the diagram .

Bandwidth $=\omega_{2}-\omega_{1}$

where $\omega_{1}$ and $\omega_{2}$ corresponds to frequencies at which magnitude of current is $\frac{1}{\sqrt{2}}$ times of maximum value.

$$ I_{\mathrm{rms}}=\frac{I_{\max }}{\sqrt{2}}=\frac{1}{\sqrt{2}} \approx 0.7 \mathrm{~A} $$

Clearly from the diagram, the corresponding frequencies are $0.8 \mathrm{rad} / \mathrm{s}$ and $1.2 \mathrm{rad} / \mathrm{s}$.

$$ \Delta \omega=\text { Bandwidth }=1 \cdot 2-0.8=0.4 \mathrm{rad} / \mathrm{s} $$

Answer

$I_{\mathrm{rms}}=\mathrm{rms}$ current

$$ =\sqrt{\frac{1^{2}+2^{2}}{2}}=\sqrt{\frac{5}{2}}=1.58 \mathrm{~A} \approx 1.6 \mathrm{~A} $$

The rms value of the current $\left(I_{\mathrm{rms}}\right)=1.6 \mathrm{~A}$ is indicated in the graph.

Answer

The phase angle $(\phi)$ by which voltage leads the current in $L-C-R$ series circuit is given by

$$ \begin{array}{ll} \tan \phi=\frac{X_{L}-X_{C}}{R}=\frac{2 \pi \nu L-\frac{1}{2 \pi \nu C}}{R} & \\ \tan \phi<0\left(\text { for } \nu<\nu_{0}\right) & \text { for } \nu=\nu_{0}=\frac{1}{2 \pi \sqrt{2 C}} \end{array} $$

Answer

(a) We know that Power $=P=V I$

that is curve of power will be having maximum amplitude, equals to multiplication of amplitudes of voltage $(V)$ and current $(I)$ curve. So, the curve will be represented by $A$.

(b) As shown by shaded area in the diagram, the full cycle of the graph consists of one positive and one negative symmetrical area.

Hence, average power over a cycle is zero.

(c) As the average power is zero, hence the device may be inductor $(L)$ or capacitor $(C)$ or the series combination of $L$ and $C$.

Answer

For a Direct Current (DC),

$$ 1 \text { ampere }=1 \text { coulomb/sec } $$

An $A C$ current changes direction with the source frequency and the attractive force would average to zero. Thus, the AC ampere must be defined in terms of some property that is independent of the direction of current.

Joule’s heating effect is such property and hence it is used to define rms value of AC.

Answer

Given, inductance $L=0.01 \mathrm{H}$ resistance $R=1 \Omega$, voltage $(V)=200 \mathrm{~V}$

and frequency $(f)=50 \mathrm{~Hz}$.

Impedance of the circuit $\quad Z=\sqrt{R^{2}+X_{L}^{2}}=\sqrt{R^{2}+(2 \pi f L)^{2}}$ $$=\sqrt{1^{2}+(2 \times 3.14 \times 50 \times 0.01)^{2}}$$

$$ or \quad Z=\sqrt{10.86}=3.3 \Omega $$

$$ \begin{aligned} \tan \phi & =\frac{\omega L}{R}=\frac{2 \pi f L}{R}=\frac{2 \times 3.14 \times 50 \times 0.01}{1}=3.14 \\ \phi & =\tan ^{-1}(3.14) \approx 72^{\circ} \\ \text{Phase difference} \qquad \phi & =\frac{72 \times \pi}{180} \mathrm{rad} . \end{aligned} $$

Time lag between alternating voltage and current

$$ \Delta t=\frac{\phi}{\omega}=\frac{72 \pi}{180 \times 2 \pi \times 50}=\frac{1}{250} \mathrm{~s} $$

Answer

Given, $P_{S}=60 \mathrm{~W}, I_{S}=0.54 \mathrm{~A}$

Current in the primary $I_{D}=$ ?

Taking line voltage as $220 \mathrm{~V}$.

We can write Since,

$$ \begin{array}{ll} \Rightarrow & P_{L}=60 \mathrm{~W}, I_{L}=0.54 \mathrm{~A} \\ \Rightarrow & V_{L}=\frac{60}{0.54}=110 \mathrm{~V} . \end{array} $$

Voltage in the secondary $\left(E_{S}\right)$ is less than voltage in the primary $\left(E_{p}\right)$.

Hence, the transformer is step down transformer.

Since, the transformation ratio

Substituting the values,

On solving

$$ \begin{aligned} r & =\frac{V_{s}}{V_{p}}=\frac{I_{p}}{I_{s}} \\ \frac{110 \mathrm{~V}}{220 \mathrm{~V}} & =\frac{I_{p}}{0.54 \mathrm{~A}} \\ I_{p} & =0.27 \mathrm{~A} \end{aligned} $$

Answer

A capacitor does not allow flow of direct current through it as the resistance across the gap is infinite. When an alternating voltage is applied across the capacitor plates, the plates are alternately charged and discharged. The current through the capacitor is a result of this changing voltage (or charge).

Thus, a capacitor will pass more current through it if the voltage is changing at a faster rate, i.e. if the frequency of supply is higher. This implies that the reactance offered by a capacitor is less with increasing frequency.

Mathematically, the reactance can be written as $X_{C}=\frac{1}{\omega C}$.

Answer

An inductor opposes flow of current through it by developing a back emf according to Lenz’s law. The induced voltage has a polarity so as to maintain the current at its present value. If the current is decreasing, the polarity of the induced emf will be so as to increase the current and vice -versa.

Since, the induced emf is proportional to the rate of change of current, it will provide greater reactance to the flow of current if the rate of change is faster, i.e., if the frequency is higher. The reactance of an inductor, therefore, is proportional to the frequency. Mathematically, the reactance offered by the inductor is given by $X_{L}=\omega L$.

Thinking Process

We have to apply the formula for phase relation, net reactance as well as instantaneous power associate with the circuit in terms of voltage and current.

Answer

Given, power drawn $=P=2 \mathrm{~kW}=2000 \mathrm{~W}$

$$ \begin{array}{rlrl} \tan \phi =-\frac{3}{4}, I_{M}=I_{0}=?, R=?, X_{C}-X_{L}=? \\ V_{\mathrm{rms}} =V=223 \mathrm{~V} \\ \text { Power } P =\frac{V^{2}}{Z} \\ \Rightarrow Z =\frac{V^{2}}{P}=\frac{223 \times 223}{2 \times 10^{3}}=25 \\ \text { Impedance } Z =25 \Omega \\ \text { Impedance } Z =\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}} \\ \Rightarrow \qquad 25 =\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}} \\ or \qquad 625 =R^{2}+\left(X_{L}-X_{C}\right)^{2} \\ \text{Again,}\qquad \tan \phi =\frac{X_{L}-X_{C}}{R}=\frac{3}{4} \\ or \qquad X_{L}-X_{C} =\frac{3 R}{4} \end{array} $$

From Eq. (ii), we put $X_{L}-X_{C}=\frac{3 R}{4}$ in Eq. (i), we get

$$ \begin{aligned} & 625=R^{2}+\frac{3 R}{4}^{2}=R^{2}+\frac{9 R^{2}}{16} \\ or \qquad & 625=\frac{25 R^{2}}{16} \end{aligned} $$

(a) Resistance $R=\sqrt{25 \times 16}=\sqrt{400}=20 \Omega$

(b) $X_{L}-X_{C}=\frac{3 R}{4}=\frac{3}{4} \times 20=15 \Omega$

(c) Main current $I_{M}=\sqrt{2} I=\sqrt{2} \frac{V}{Z}=\frac{223}{25} \times \sqrt{2}=12.6 \mathrm{~A}$

As $R, X_{C}, X_{L}$ are all doubled, $\tan \phi$ does not change. $Z$ is doubled, current is halved. So, power is also halved.

Answer

(i) The town is $10 \mathrm{~km}$ away, length of pair of Cu wires used, $L=20 \mathrm{~km}=20000 \mathrm{~m}$.

Resistance of Cu wires,

$$ \begin{aligned} R & =\frac{l}{A}=\frac{l}{\pi(r)^{2}} \\ & =\frac{1.7 \times 10^{-8} \times 20000}{3.14\left(0.5 \times 10^{-2}\right)^{2}}=4 \Omega \end{aligned} $$

$I$ at $220 \mathrm{~V}$

$$ \begin{aligned} V I & =10^{6} \mathrm{~W} ; I=\frac{10^{6}}{220}=0.45 \times 10^{4} \mathrm{~A} \\ R I^{2} & =\text { power loss } \\ & =4 \times(0.45)^{2} \times 10^{8} \mathrm{~W} \\ & >10^{6} \mathrm{~W} \end{aligned} $$

Therefore, this method cannot be used for transmission.

(ii) When power $P=10^{6} \mathrm{~W}$ is transmitted at $11000 \mathrm{~V}$.

$$ \begin{aligned} V^{\prime} I^{\prime} & =10^{6} \mathrm{~W}=11000 I^{\prime} \\ \text { Current drawn, } I^{\prime} & =\frac{1}{1.1} \times 10^{2} \\ \text { Power loss } & =R I^{2}=\frac{1}{1.21} \times 4 \times 10^{4} \\ & =3.3 \times 10^{4} \mathrm{~W} \\ \text { Fraction of power loss } & =\frac{3.3 \times 10^{4}}{10^{6}}=3.3 % \end{aligned} $$

Thinking Process

The circuit consists of inductor (L) and capacitor (C) connected in series and the combination is connected parallel with a resistance $R$. Due to this combination there is oscillation of electromagnetic energy.

Answer

In the given figure $i$ is the total current from the source. It is divided into two parts $i_{1}$ through $R$ and $i_{2}$ through series combination of $C$ and $L$.

So, we can write $i=i_{1}+i_{2}$ As, $V_{m} \sin \omega t=R i_{1}$ $\Rightarrow$

$$ i_{1}=\frac{V_{m} \sin \omega t}{R} $$

[from the circuit diagram]

If $q_{2}$ is charge on the capacitor at any time $t$, then for series combination of $C$ and $L$.

Applying KVL in the Lower circuit as shown,

$$ \begin{array}{rlrl} \frac{q_{2}}{C}+\frac{L d i_{2}}{d t}-V_{m} \sin \omega t=0 \\ \Rightarrow \frac{q_{2}}{C}+\frac{L d_{2} q_{2}}{d t^{2}} =V_{m} \sin \omega t \\ \text { Let } \therefore \frac{d q_{2}}{d t} =q_{m} \omega \cos (\omega t+\phi) \\ \Rightarrow \frac{d^{2} q_{2}}{d t^{2}} =-q_{m} \omega^{2} \sin (\omega t+\phi) \end{array} $$

$$ \Rightarrow \quad \frac{q_{2}}{C}+\frac{L d_{2} q_{2}}{d t^{2}}=V_{m} \sin \omega t \quad \because i_{2}=\frac{d q_{2}}{d t} $$

Now putting these values in Eq. (ii), we get

$$ q_{m} \frac{1}{C}+L\left(-\omega^{2}\right) \sin (\omega t+\phi)=V_{m} \sin \omega t $$

If $\phi=0$ and $\frac{1}{C}-L \omega^{2}>0$,

then

$$ q_{m}=\frac{V_{m}}{\frac{1}{C}-L \omega^{2}} $$

From Eq. (iii),

$$ \begin{aligned} i_{2} =\frac{d q_{2}}{d t}=\omega q_{m} \cos (\omega t+\phi) \\ i_{2} =\frac{\omega V_{m} \cos (\omega t+\phi)}{\frac{1}{C}-L \omega^{2}} \\ \text { Taking } \phi =0 ; i_{2}=\frac{V_{m} \cos (\omega t)}{\frac{1}{\omega C}-L \omega} \end{aligned} $$

using Eq. (iv),

From Eqs. (i) and (v), we find that $i_{1}$ and $i_{2}$ are out of phase by $\frac{\pi}{2}$.

Now,

$$ i_{1}+i_{2}=\frac{V_{m} \sin \omega t}{R}+\frac{V_{m} \cos \omega t}{\frac{1}{\omega C}-L \omega} $$

Put

$$ \frac{V_{m}}{R}=A=C \cos \phi \text { and } \frac{V_{m}}{\frac{1}{\omega C}-L \omega}=B=C \sin \phi $$

$\therefore \quad i_{1}+i_{2}=C \cos \phi \sin \omega t+C \sin \phi \cos \omega t$

$$ =C \sin (\omega t+\phi) $$

where

$$ C=\sqrt{A^{2}+B^{2}} $$

and

$$ \phi=\tan ^{-1} \frac{B}{A} C=\frac{V_{m}^{2}}{R^{2}}+\frac{V_{m}^{2}}{\frac{1}{\omega C}-L \omega} $$

and

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

Hence,

$$ i=i_{1}+i_{2}=\frac{V_{m}^{2}}{R^{2}}+\frac{V_{m}^{2}}{\frac{1}{\omega C}-L \omega} \sin (\omega t+\phi) $$

or

$$ \frac{i}{V_{m}}=\frac{1}{Z}=\frac{1}{R^{2}}+\frac{1}{\frac{1}{\omega C}-L \omega^{2}} $$

This is the expression for impedance $Z$ of the circuit.

Note In this problem, we should not apply the formulae of L-C-R series circuit directly.

Thinking Process

Apply KVL for the given L-C-R series circuit and find the required relations. Also find energy loss through the resistors to know net loss of energy through the circuit.

Answer

Consider the $L-C-R$ circuit. Applying KVL for the loop, we can write

$\begin{gathered}V=V_m \sin \omega t \\ L \frac{d i}{d t}+\frac{q}{C}+i R=V_m \sin \omega t\end{gathered}$

Multiplying both sides by $i$, we get

$$ L i \frac{d i}{d t}+\frac{q}{C} i+i^{2} R=\left(V_{m} i\right) \sin \omega t=V i $$

where

$$ \begin{aligned} L i \frac{d i}{d t}=\frac{d}{d t} & \frac{1}{2} L i^{2}=\text { rate of change of energy stored in an inductor. } \\ R i^{2} & =\text { joule heating loss } \\ \frac{q}{C} i=\frac{d}{d t} & \frac{q^{2}}{2 C}=\text { rate of change of energy stored in the capacitor. } \end{aligned} $$

$V i=$ rate at which driving force pours in energy. It goes into (i) ohmic loss and (ii) increase of stored energy.

Hence Eq. (ii) is in the form of conservation of energy statement. Integrating both sides of Eq. (ii) with respect to time over one full cycle $(0 \rightarrow T)$ we may write

$$ \begin{array}{r} \int_{0}^{T} \frac{d}{d t} \frac{1}{2} L i^{2}+\frac{q^{2}}{2 C} d t+\int_{0}^{T} R i^{2} d t=\int_{0}^{T} V i d t \\ \Rightarrow 0+(+v e)=\int_{0}^{T} V i d t \\ \Rightarrow \int_{0}^{T} V i d t>0 \text { if phase difference between } V \text { and } i \text { is a constant and acute angle. } \end{array} $$

Answer

(a) Consider the R-L-C circuit shown in the adjacent diagram.

Given Let current at any instant be $i$

Applying KVL in the given circuit

$$ i R+L \frac{d i}{d t}+\frac{q}{C}-V_{m} \sin \omega t=0 $$

Now, we can write

$$ i=\frac{d q}{d t} \Rightarrow \frac{d i}{d t}=\frac{d^{2} q}{d t^{2}} $$

From Eq. (i)

$$ \frac{d q}{d t} R+L \frac{d^{2} q}{d t^{2}}+\frac{q}{C}=V_{m} \sin \omega t $$

$$ \Rightarrow \quad L \frac{d^{2} q}{d t^{2}}+R \frac{d q}{d t}+\frac{q}{C}=V_{m} \sin \omega t $$

This is the required equation of variation (motion) of charge. (b) Let $\quad q=q_{m} \sin (\omega t+\phi)=-q_{m} \cos (\omega t+\phi)$

$$ \begin{aligned} i & =i_{m} \sin (\omega t+\phi)=q_{m} \omega \sin (\omega t+\phi) \\ i_{m} & =\frac{V_{m}}{Z}=\frac{V_{m}}{\sqrt{R^{2}+\left(X_{C}-X_{L}\right)^{2}}} \\ \phi & =\tan ^{-1} \frac{X_{C}-X_{L}}{R} \end{aligned} $$

When $R$ is short circuited at $t=t_{0}$, energy is stored in $L$ and $C$.

and

$$ \begin{aligned} U_{L} & =\frac{1}{2} L i^{2}=\frac{1}{2} L \frac{V_{m}}{\sqrt{\left(R^{2}+X_{C}-X_{L}\right)^{2}}} \sin ^{2}\left(\omega t_{0}+\phi\right) \\ U_{C} & =\frac{1}{2} \times \frac{q^{2}}{C}=\frac{1}{2 C}\left[q^{2} m \cos ^{2}\left(\omega t_{0}+\phi\right)\right] \\ & =\frac{1}{2 C} \frac{V_{m}}{\sqrt{R^{2}+\left(X_{C}-X_{L}\right)^{2}}} \\ & =\frac{1}{2 C} \times \frac{i_{m}}{\omega} \cos ^{2}\left(\omega t_{0}+\phi\right) \\ & =\frac{i^{2} m}{2 C \omega^{2}} \cos ^{2}\left(\omega t_{0}+\phi\right) \\ & =\frac{1}{2 C} \frac{V_{m}}{\sqrt{R^{2}+\left(X_{C}-X_{L}\right)^{2}}} \frac{\cos ^{2}\left(\omega t_{0}+\phi\right)}{\omega^{2}} \quad\left[\because i_{m}=q_{m} \omega\right] \\ & =\frac{1}{2 C \omega^{2}} \frac{V_{m}}{\sqrt{R^{2}+\left(X_{C}-X_{L}\right)^{2}}} \cos ^{2}\left(\omega t_{0}+\phi\right) \end{aligned} $$

(c) When $R$ is short circuited, the circuit becomes an $L-C$ oscillator. The capacitor will go on discharging and all energy will go to $L$ and back and forth. Hence, there is oscillation of energy from electrostatic to magnetic and magnetic to electrostatic.