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#### Page No 40:

#### Question 1:

$\left(\mathrm{a}\right)5\sqrt{2}\mathrm{A}\phantom{\rule{0ex}{0ex}}\left(\mathrm{b}\right)5\sqrt{3/2}\mathrm{A}\phantom{\rule{0ex}{0ex}}\left(\mathrm{c}\right)5/6\mathrm{A}\phantom{\rule{0ex}{0ex}}\left(\mathrm{d}\right)5/\sqrt{2}\mathrm{A}$

#### Answer:

Frequency, *f* = 50 Hz

RMS current, *I*_{rms} = 5A

Time, *t* = $\frac{1}{300}s$

Instantaneous current, *I* = *I*_{0 }sinω*t*

$\Rightarrow I={I}_{o}\mathrm{sin}\left(\mathrm{\omega}\mathit{.}t\right)=\sqrt{2}{I}_{\mathrm{rms}}\mathrm{sin}\left(2\mathrm{\pi}f\mathit{.}t\right)\left({I}_{o}=\sqrt{2}{I}_{\mathrm{rms}}\mathrm{and}\mathrm{\omega}=2\mathrm{\pi}f\right)\phantom{\rule{0ex}{0ex}}\Rightarrow I=\sqrt{2}\times 5\mathrm{sin}\left(2\mathrm{\pi}\times 50\times \frac{1}{300}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow I=\sqrt{2}\times 5\times \frac{\sqrt{3}}{2}\phantom{\rule{0ex}{0ex}}=5\sqrt{\frac{3}{2}}A$

Hence, the correct answer is option (B).

#### Page No 40:

#### Question 2:

*Rg*and an internal reactance

*Xg*. It is used to supply power to a passive load consisting of a resistance

*Rg*and a reactance

*X*. For maximum power to be delivered from the generator to the load, the value of

_{L}*X*is equal to

_{L}*X*

_{g}.

*X*

_{g.}

*R*

_{g}.

#### Answer:

For maximum power to be delivered from generator to load, the total reactance must be zero.

i.e., *x _{L} + x_{g }*

_{= 0 ⇒ }

*x*=

_{L}*– x*

_{g}Hence, the correct answer is option (C).

#### Page No 41:

#### Question 3:

*v*

^{2}> and is calibrated to read $\sqrt{<{v}^{2}>}.$

(d) the pointer of the meter is stuck by some mechanical defect.

#### Answer:

The voltmeter connected to A-C mains reads mean value (<*v*^{2}>) and is caliberated in such a way that it gives value of <*v*^{2}>, which is multiplied by form factor to give 'rms' value.

Hence, the correct answer is option (C).^{}^{}

#### Page No 41:

#### Question 4:

#### Answer:

Resonant frequency for an LCR circuit,

${f}_{0}=\frac{1}{2\mathrm{\pi}\sqrt{LC}}$

To reduce the resonant frequency either inductance (*L*) or capacitance (*C*) should be increased.

In the given case, if we add another capacitor in parallel, the net capacitance will increase and hence the resonant frequency will decrease.

Hence, the correct answer is option (B).

#### Page No 41:

#### Question 5:

*LCR*circuit used for communication?

*R*= 20 â„¦,

*L*= 1.5 H,

*C*= 35μF.

*R*= 25 â„¦,

*L*= 2.5 H,

*C*= 45μF.

*R*= 15 â„¦,

*L*= 3.5 H,

*C*= 30μF.

*R*= 25 â„¦,

*L*= 1.5 H,

*C*= 45μF.

#### Answer:

For better tuning of *L-C-R* circuit. The quality factor should be as high as possible.

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

where, *R* is resistance*L* is the inductance*C* is the capacitance

For '*Q*' (quality factor) to be high:

'*R*' should be low

'*L*' should be high and '*C*' should be low.

Hence, the correct answer is option (C).

#### Page No 41:

#### Question 6:

#### Answer:

Given,*X*_{L }= 1Ω*R* = 2Ω*E*_{rms} = 6 V

Here,* I*_{rms }= $\frac{{E}_{\mathrm{rms}}}{Z}$

where, Z is the impedance.

$Z=\sqrt{{X}_{L}^{2}+{R}^{2}}=\sqrt{{1}^{2}+{2}^{2}}=\sqrt{5}\phantom{\rule{0ex}{0ex}}\Rightarrow {I}_{\mathrm{rms}}=\frac{6}{\sqrt{5}}\mathrm{A}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{cos\varphi}=\frac{R}{Z}=\frac{2}{\sqrt{5}}$

Average power,

${P}_{\mathrm{av}}={I}_{\mathrm{rms}}{E}_{\mathrm{rms}}\mathrm{cos\varphi}$

${P}_{\mathrm{av}}=\frac{6}{\sqrt{5}}\times 6\times \frac{2}{\sqrt{5}}=14.4\mathrm{W}$

Hence, the correct answer is option (C).

#### Page No 41:

#### Question 7:

#### Answer:

Given,

Secondary voltage, *V _{s }*= 24 V

Power associated with secondary,

*P*= 12 V

_{s}*P*

_{s }=

*I*

_{s }

*V*

_{s}

$\Rightarrow {I}_{\mathrm{s}}=\frac{{P}_{\mathrm{s}}}{{V}_{\mathrm{s}}}=\frac{12}{24}=\frac{1}{2}A$

Peak value of the current,

${I}_{0}={I}_{\mathrm{s}}\sqrt{2}=\frac{1}{2}\times \sqrt{2}=\frac{1}{\sqrt{2}}A$

Hence, the correct answer is option (A).

#### Page No 42:

#### Question 8:

#### Answer:

Impedance of a series *LCR* circuit containing resistor, inductor and capacitor is given by,

$Z=\sqrt{{R}^{2}+{\left({X}_{\mathrm{L}}-{X}_{\mathrm{C}}\right)}^{2}}$

$\Rightarrow Z=\sqrt{{R}^{2}+{\left(2\mathrm{\pi}fL-\frac{1}{2\mathrm{\pi}fC}\right)}^{2}}$

At resonance frequency, *Z* is minimum and current is maximum.

For any other value of frequency, the value of current will decrease (when compared to current at resonance).

Clearly the circuit must have capacitor and inductor. However, it may be possible that the circuit also has a resistor besides inductor & capacitor.

Hence, the correct options are (a) and (d).

#### Page No 42:

#### Question 9:

#### Answer:

According to question, the value of current is increasing with the increase in frequency. That is possible for a capacitive circuit.

${X}_{\mathrm{C}}=\frac{1}{\mathrm{\omega}C}=\frac{1}{2\mathrm{\pi}fC}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{So},\mathrm{current}I=\frac{E}{{\mathrm{X}}_{\mathrm{C}}}$

Here, $f\uparrow ,{X}_{C}\downarrow $ and $I\uparrow $

This is also true for an R-C circuit.

$Z=\sqrt{{R}^{2}+{X}_{\mathrm{C}}^{2}}\phantom{\rule{0ex}{0ex}}\mathrm{So},I=\frac{E}{Z}$

(Here, all symbols have their usual meaning.)

Hence, the correct options are (C) and (D).

#### Page No 42:

#### Question 10:

#### Answer:

Power, *P* = *E*_{rms}*I*_{rms}.

For a given power level, if '*E*_{rms}' is high then '*I*_{rms}' will be low.

Also, power loss = ${I}_{\mathrm{rms}}^{2}R$ ...(1)

If *I*_{rms }is low then power loss is also low. (from equation 1)

The high alternating voltage can be easily reduced by using step down transformer at the receiving end.

Hence, the correct options are (A), (B) and (D).

#### Page No 42:

#### Question 11:

*LCR*circuit, the power transferred from the driving source to the driven oscillator is

*P*=

*I*

^{2}

*Z*cos Ï•.

*P*≥ 0.

*P*= 0) in some cases.

#### Answer:

According to the question,*P* = *I*^{2 }*Z* cos Ï•

where, *I* = Current*Z *= Impedance

cos Ï• = Power factor

Power factor, $\mathrm{cos}\mathrm{\varphi}=\frac{R}{Z}$

where* R* > 0 and *Z* > 0

⇒ cos Ï• > 0

⇒ *P* > 0

Hence, the correct options are (A), (B) and (C).

#### Page No 42:

#### Question 12:

*C*

#### Answer:

The maximum value of voltage, *${V}_{\mathrm{peak}}=\sqrt{2}{V}_{\mathrm{rms}}=\sqrt{2}\times 220\mathrm{V}=311.08\mathrm{V}$
V*

_{rms }

*I*

_{rms }As the circuit is pure capacitive, so the current developed leads the applied voltage by a phase angle of 90°.

Thus, the power delivered,

*P*

_{av}=

*V*

_{rms }

*I*

_{rms }cos90° = 0

When an AC voltage of 220 V is applied to a capacitor

*C*, the charge on the plates is in phase with the applied voltage.

Hence, the correct options are (C) and (D).

#### Page No 43:

#### Question 13:

#### Answer:

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

The line is having some resistance so power factor, $\mathrm{cos}\mathrm{\varphi}=\frac{R}{Z}\ne 0$

So, $\mathrm{\varphi}\ne \frac{\mathrm{\pi}}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \left|\mathrm{\varphi}\right|<\frac{\mathrm{\pi}}{2}$

i.e., phase lies between $0\frac{\mathrm{\pi}}{2}$.

Hence, the correct options are (A) and (D).

#### Page No 43:

#### Question 14:

*LC*circuit is considered analogous to a harmonically oscillating spring block system, which energy of the

*LC*circuit would be analogous to potential energy and which one analogous to kinetic energy?

#### Answer:

The electrostatic energy $\frac{1}{2}C{V}^{2}$ is analogous to potential energy.

Energy associated with moving charges (current) that is magnetic $\left(\frac{1}{2}L{I}^{2}\right)$ is analogous to kinetic energy.

#### Page No 43:

#### Question 15:

Draw the effective equivalent circuit of the circuit shown in the given figure, at very high frequencies and find the effective impedance.

#### Answer:

Inductive reactance, ${X}_{\mathrm{L}}=2\mathrm{\pi}fL$

Capacitive reactance, ${X}_{\mathrm{C}}=\frac{1}{2\mathrm{\pi}fC}$

For very high frequencies

$\Rightarrow {X}_{\mathrm{L}}\to \infty $ and ${X}_{\mathrm{C}}\to 0$

For infinite resistance, the circuit can be considered as open circuit.

If the reactance of circuit is zero, it will be considered as short circuited.

Effective impedance, Z = *R*_{1} + *R*_{3}

#### Page No 43:

#### Question 16:

#### Answer:

Let the supply voltage be *V*.

For pure resistive circuit:

$I=\frac{V}{R}$

For series LCR circuit:

$I\text{'}=\frac{V}{Z}=\frac{V}{\sqrt{{R}^{2}+{\left({X}_{\mathrm{L}}-{X}_{\mathrm{C}}\right)}^{2}}}\left(\mathrm{For},\mathrm{LCR}\mathrm{circuit},Z=\sqrt{{R}^{2}+{\left({X}_{\mathrm{L}}-{X}_{\mathrm{C}}\right)}^{2}}\right)$

(a).

If capacitive reactance and inductive reactance in circuit (b) becomes equal.

i.e., *X*_{L}* = X*_{C}

Then, $I\text{'}=\frac{V}{Z}=\frac{V}{\sqrt{{R}^{2}+{\left({X}_{\mathrm{L}}-{X}_{\mathrm{C}}\right)}^{2}}}=\frac{V}{R}=I$

Then, same current will from in both circuit (a) and (b)

(b).

No, the rms current in circuit (b) can not be larger than circuit (a).

#### Page No 43:

#### Question 17:

#### Answer:

Let the applied e.m.f., *E* = *E*_{0}â€‹sin(ω*t*)*I* = *I*_{0}â€‹sin(ω*t*−Ï•)

Then, instantaneous power, $P=EI=\frac{{E}_{0}{I}_{0}}{2}\left[\mathrm{cos\varphi}-\mathrm{cos}\left(2\mathrm{\omega}t+\mathrm{\varphi}\right)\right]$

where Ï• is the phase difference.

when, cos Ï• < cos (2ω*t* + Ï•)

Then, *P < 0 *

So, the instantaneous power output source can be negative.

Average power, ${P}_{\mathrm{av}}=\frac{{V}_{0}}{\sqrt{2}}\frac{{I}_{0}}{\sqrt{2}}\mathrm{cos\varphi}$

$\Rightarrow {P}_{\mathrm{av}}={V}_{\mathrm{rms}}{I}_{\mathrm{rms}}\mathrm{cos\varphi}$

But, we know that,

$\mathrm{cos\varphi}=\frac{R}{Z}>0$

$\Rightarrow {P}_{\mathrm{av}}>0$

Hence, the average power output of an AC source cannot be negative.

#### Page No 43:

#### Question 18:

In series LCR circuit, the plot of *I*_{max} vs ω is shown in the given figure. Find the bandwidth and mark in the figure.

#### Answer:

Bandwidth, $\u2206\mathrm{\omega}={\mathrm{\omega}}_{2}-{\mathrm{\omega}}_{1}$

where, ${\mathrm{\omega}}_{1}{\mathrm{\omega}}_{2}$ corresponds to the frequencies at which magnitude of current $\frac{1}{\sqrt{2}}$ times of maximum value.

${I}_{\mathrm{rms}}=\frac{{I}_{0}}{\sqrt{2}}=\frac{1}{\sqrt{2}}=0.7A\left(\mathrm{maximum}\mathrm{current},{I}_{0}=1A\right)$

From the given diagram,

$\u2206\mathrm{\omega}={\mathrm{\omega}}_{2}-{\mathrm{\omega}}_{1}=1.2-0.8$

$\u2206\mathrm{\omega}=0.4\mathrm{rad}{s}^{-1}$

#### Page No 44:

#### Question 19:

The alternating current in a circuit is described by the graph shown in the given figure . Show rms current in this graph.

#### Answer:

Square root of mean of squared value of current is called rms current.

${I}_{\mathrm{rms}}=\sqrt{\frac{{1}^{2}+{2}^{2}}{2}}\approx 1.6A$

Desired graph is as given below (horizontal line parallel to time-axis):

#### Page No 44:

#### Question 20:

*Ï•*, by which the supply voltage leads the current in an

*LCR*series circuit, change as the supply frequency is gradually increased from very low to very high values.

#### Answer:

For an *L-C-R* circuit,

Let the phase angle be Ï• and* Z *be impedence.

And the voltage leads the current in LCR series circuit.

$\mathrm{tan\varphi}=\frac{{X}_{\mathrm{L}}-{X}_{\mathrm{C}}}{Z}$

$\Rightarrow \mathrm{tan\varphi}=\frac{\mathit{2}\mathrm{\pi}fL-{\displaystyle \frac{1}{\mathit{2}\mathrm{\pi}fC}}}{Z}$

tan Ï• < 0 (for *f* < *f*_{0})

tan Ï• > 0 (for *f* > *f*_{0})

tan Ï• < 0 (for *f* = *f*_{0})

#### Page No 44:

#### Question 21:

#### Answer:

(a) Power is given by product of voltage and current. So it will have maximum amplitude. Thus, the curve *A* represent the power consumption over a full cycle.

(b) Average power consumption over a complete cycle is zero.

(c) The device X may be L or C or LC.

#### Page No 44:

#### Question 22:

#### Answer:

In *DC*, the current is defined as the ratio between the charge flowing through a circuit in a particular time period.

An AC current changes direction with the source frequency and the attractive force would average to zero. Thus, the a.c 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 a.c

#### Page No 44:

#### Question 23:

#### Answer:

Given, *L* = 10^{–2} H*R* = 1 Ω*v* = 200 V*f* = 50 Hz

Impedance, $Z=\sqrt{{R}^{2}+{x}_{\mathrm{L}}^{2}}=\sqrt{{R}^{2}+{\left(2\mathrm{\pi}fL\right)}^{2}}$

⇒ *Z *= 3.3 Ω

$\mathrm{tan}\mathrm{\varphi}=\frac{\mathrm{\omega}L}{R}=3.14$

⇒ Ï• â‰ƒ 72°

Phase difference, $\mathrm{\varphi}=\frac{72\times \mathrm{\pi}}{180}\mathrm{rad}$

Time lag, $\u2206t=\frac{\mathrm{\varphi}}{\mathrm{\omega}}=\frac{72\mathrm{\pi}}{180\times 2\mathrm{\pi}\times 50}=\frac{1}{250}$

âˆ† *t* = 4 × 10^{–3} s

#### Page No 44:

#### Question 24:

#### Answer:

Given,* P*_{s} = 60 ω*I*_{s} = 0.54 A

⇒ *P*_{s} = *V*_{s}*I*_{s}

$\Rightarrow {V}_{\mathrm{S}}=\frac{{P}_{\mathrm{s}}}{{I}_{\mathrm{s}}}=\frac{60}{0.54}=11\mathrm{V}$

Voltage in secondary is less than voltage in primary.

â€‹i.e., *V*s < *V*p

The transformer is step down transformer.

⇒ Power impact = Power output

$\Rightarrow {V}_{\mathrm{p}}{I}_{\mathrm{p}}={V}_{\mathrm{s}}{I}_{\mathrm{s}}\phantom{\rule{0ex}{0ex}}\Rightarrow {I}_{\mathrm{p}}=\frac{{V}_{\mathrm{s}}{I}_{\mathrm{s}}}{{V}_{\mathrm{p}}}\phantom{\rule{0ex}{0ex}}\Rightarrow {I}_{\mathrm{p}}=\frac{111\times 0.54}{220}=0.27\mathrm{A}$

Hence, the current drawn in primary coil is 0.27 A

#### Page No 45:

#### Question 25:

#### Answer:

Reactance provided by a capacitor is given by, ${x}_{\mathrm{c}}=\frac{1}{\mathrm{\omega}c}=\frac{1}{2\mathrm{\pi}fC}$

For a given value of capacitance (C).

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

#### Page No 45:

#### Question 26:

#### Answer:

Inductive reactance, *X _{L}_{}*= ω

*L*

⇒

*X*= 2π

_{L}_{}*fL*

We know that, 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. Now if the current is decreasing, the polarity of the induced emf will be so as to increase the current and vice versa. But the induced emf is proportional to the rate of change of current, so it will provide greater reactance to the flow of current if the rate of change is faster, i.e., when frequency increases.

Hence, the reactance of an inductor, is proportional to the frequency, being given by ω

*L*.

#### Page No 45:

#### Question 27:

*R*, (ii)

*X*–

_{C}*X*, and (iii)

_{L}*I*. Another device has twice the values for

_{M}*R, X*and

_{C}*X*. How are the answers affected?

_{L }#### Answer:

*P* = 2 *k*ω = 2000 ω

tanÏ• = $-\frac{3}{4}$*V*_{rms} = *V* = 223 v

Power, $P=\frac{{V}^{2}}{Z}$

$Z=\frac{{V}^{2}}{P}=25\mathrm{\Omega}$

Impedance, $\Rightarrow 625={R}^{2}+{\left({X}_{\mathrm{L}}-{X}_{\mathrm{C}}\right)}^{2}.....\left(1\right)$

But, $\mathrm{tan\varphi}=\frac{{X}_{\mathrm{L}}-{X}_{\mathrm{C}}}{R}=\frac{3}{4}$

$\Rightarrow {X}_{\mathrm{L}}-{X}_{\mathrm{C}}=\frac{3R}{4}.....\left(2\right)$

Putting values in equation (1)

$625={R}^{2}+{\left(\frac{3R}{4}\right)}^{2}$

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

(b) ${X}_{\mathrm{L}}-{X}_{\mathrm{C}}=\frac{3R}{4}=15\mathrm{\Omega}$

(c) *I*_{rms }= $\sqrt{2}I$ = 12.6 A

If *R*, *X*_{C}, *X*_{L} are all doubled, tanÏ• does not change. *Z* is doubled, current is solved. So power is also solved.

#### Page No 45:

#### Question 28:

(ρ

_{Cu}= 1.7 × 10

^{–8}SI unit)

#### Answer:

(i) The distance between power station and tower is 10 km. Length of copper wire to used, *L* = 20 km = 20,000 m.

Resistance of copper wire, $R=\mathrm{\rho}\frac{l}{A}$

$\Rightarrow R=\frac{1.7\times {10}^{\u20138}.20,000}{3.14\times {\left(0.5\times {10}^{\u20132}\right)}^{2}}=4\mathrm{\Omega}$*p *= 10^{6} ω

If, *V* = 220 V

then, *I* = $\frac{P}{V}=\frac{{10}^{6}}{220}$ = 0.45 × 10^{4} A

Power loss = *I*^{2}*R* = 0.4 × (0.45 × 10^{4})^{2 }> 10^{6} ω

Thus, this method cannot be used for transmission.

(ii) When power is transmitted at voltage, *V *= 1100 V

Current drawn, *I*^{1} = 91 A

Power loss = (*I*')^{2 }*R* = (91)^{2} × 4 = 3.3 × 10^{4} ω

Fraction of power loss $=\frac{3.3\times {10}^{4}}{{10}^{6}}=3.3\%$

#### Page No 45:

#### Question 29:

Consider the *LCR *circuit shown in the given figure. Find the net current *i* and the phase of *i*. Show that $i=\frac{v}{Z}.$ Find the impendence Z for this circuit.

#### Answer:

In the given figure '*L*' & '*C*' are connected in series and resistor '*R*' is connected in parallel to both.

From the circuit, *i* = *i*_{1} + *i*_{2}

where '*i*'* *is the total current flowing through the circuit.

Applying KVL in lower circuit.

$\frac{{q}_{2}}{C}+Ld{i}_{2}$

⇒ *V*_{C} + *V*_{L} = *V*_{m} sinω*t$\Rightarrow \frac{{q}_{1}}{C}+\frac{Ld{i}_{1}}{dt}={V}_{\mathrm{m}}\mathrm{sin\omega}t\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{q}_{1}}{C}+\frac{L{d}^{2}{q}_{1}}{d{t}^{2}}={V}_{\mathrm{m}}\mathrm{sin\omega}t.....\left(1\right)$*

Let

*q*

_{1}=

*q*

_{m}sin (ω

*t*+ Ï•) .....(2)

*i*

_{1 }= $\frac{d{q}_{1}}{dt}$=

*q*

_{m}ω

^{2}sin(ω

*t*+ Ï•) .....(3)

$\frac{{d}^{2}{q}_{1}}{d{t}^{2}}$= –

*qm*ω

^{2}sin(ω

*t*+ Ï•) .....(4)

Substitute the values of equation (2) & (4) in equation (1)

$\frac{{q}_{\mathrm{m}}\mathrm{sin}(\mathrm{\omega}t+\mathrm{\varphi})}{c}$–

*Lq*

_{m}ω

^{2}sin(ω

*t*+ Ï•) =

*v*

_{m}sinω

*t*

q

q

_{m}

*sin(*ω

*t +*Ï•) $\left[\frac{1}{C}-{\mathrm{\omega}}^{2}L\right]$ =

*v*

_{m}sinω

*t*

at Ï• = 0

${q}_{\mathrm{m}}=\frac{{V}_{\mathrm{m}}}{\mathrm{\omega}\left[{\displaystyle \frac{1}{\mathrm{\omega}C-\mathrm{\omega}L}}\right]}.....\left(5\right)$

Applying KCL at junction,

*i*=

*i*

_{1}+

*i*

_{2}

*i*= $\frac{{v}_{\mathrm{m}}\mathrm{sin\omega}t}{R}$ +

*q*

_{m}ωcos (ω

*t*+ Ï•)

Using relation (5) for

*q*

_{m }at Ï• = 0

$i=\frac{{v}_{\mathrm{m}}\mathrm{sin\omega}t}{R}+\frac{{v}_{\mathrm{m}}\mathrm{\omega cos\omega}t}{\mathrm{\omega}\left({\displaystyle \frac{1}{\mathrm{\omega}c}}-\mathrm{\omega}L\right)}$

$i=\frac{{v}_{\mathrm{m}}}{R}\mathrm{sin\omega}t+\frac{{v}_{\mathrm{m}}}{\left({\displaystyle \frac{1}{\mathrm{\omega}c}}-\mathrm{\omega}L\right)}\mathrm{cos\omega}t$

$\mathrm{Let},A=\frac{{v}_{\mathrm{m}}}{R}=C\mathrm{cos\varphi}.....\left(6\right)\phantom{\rule{0ex}{0ex}}B=\frac{{v}_{\mathrm{m}}}{{\displaystyle \frac{1}{\mathrm{\omega}C}}-\mathrm{\omega}L}=C\mathrm{sin\varphi}.....\left(7\right)$

*i*=

*C*sin(ω

*t*+ Ï•)

Squaring and adding (6), (7)

*A*

^{2}+

*B*

^{2}=

*C*

^{2 }cos

^{2}Ï• +

*C*

^{2}sin

^{2}Ï•

$\Rightarrow C=\sqrt{{A}^{2}+{B}^{2}}\phantom{\rule{0ex}{0ex}}\mathrm{\varphi}={\mathrm{tan}}^{-1}\left(\frac{B}{A}\right)={\mathrm{tan}}^{-1}\left(\frac{{v}_{\mathrm{m}}}{{\displaystyle \frac{{\displaystyle \frac{1}{\mathrm{\omega}C-\mathrm{\omega}L}}}{{\displaystyle \frac{{V}_{\mathrm{m}}}{R}}}}}\right)\phantom{\rule{0ex}{0ex}}\therefore \mathrm{tan\varphi}=\frac{R}{{\displaystyle \frac{1}{\mathrm{\omega}C}}-\mathrm{\omega}L}\phantom{\rule{0ex}{0ex}}\Rightarrow C=\sqrt{\frac{{V}_{\mathrm{m}}^{2}}{{R}^{2}}+\frac{{V}_{\mathrm{m}}^{2}}{{\left({\displaystyle \frac{1}{\mathrm{\omega}{C}_{2}}}-\mathrm{\omega}L\right)}^{2}}}\phantom{\rule{0ex}{0ex}}\therefore i=\left[\frac{{V}_{\mathrm{m}}^{2}}{R}+\frac{{V}_{\mathrm{m}}}{{\left({\displaystyle \frac{1}{\mathrm{\omega}C}}-\mathrm{\omega}L\right)}^{2}}\right]\mathrm{sin}\left(\mathrm{\omega}t+\mathrm{\varphi}\right).....\left(8\right)\phantom{\rule{0ex}{0ex}}\mathrm{And}\mathrm{\varphi}={\mathrm{tan}}^{\u20131}\frac{R}{\left({\displaystyle \frac{1}{\mathrm{\omega}C}}-\mathrm{\omega}L\right)}\phantom{\rule{0ex}{0ex}}\therefore i=\frac{V}{Z}$

For AC

$i=\frac{V}{Z}\mathrm{sin}\left(\mathrm{\omega}t+\mathrm{\varphi}\right).....\left(9\right)$

Comparing (8) & (9)

So, $\frac{1}{Z}=\sqrt{\frac{1}{{R}^{2}}+\frac{1}{{\left({\displaystyle \frac{1}{\mathrm{\omega}C}}-\mathrm{\omega}L\right)}^{2}}}$

#### Page No 45:

#### Question 30:

*LCR*circuit driven at frequency ω, the equation reads $L\frac{di}{dt}+Ri+\frac{q}{C}={v}_{i}={v}_{m}\mathrm{sin}\omega t$

*i*and simplify where possible.

*v*and

*i*must be acute.

#### Answer:

$L\frac{di}{dt}+{R}_{\mathrm{i}}+\frac{q}{c}={V}_{\mathrm{i}}={V}_{\mathrm{m}}\mathrm{sin\omega}t.....\left(1\right)$

Multiplying both the sides by *i*, we get

$Li\frac{di}{dt}+\frac{q}{c}i+{i}^{2}R=\left({V}_{\mathrm{m}}i\right)\mathrm{sin\omega}t={V}_{\mathrm{i}}.....\left(2\right)$

where, ${L}_{\mathrm{i}}\frac{di}{dt}=\frac{d}{dt}\left(\frac{1}{2}L{i}^{2}\right)$ = rate of change of energy stored in an inductor

${i}^{2}R$ = Joule heating loss,

$\frac{q}{C}i=\frac{d}{dt}\left(\frac{{q}^{2}}{2C}\right)$ = rate of change of energy stored in the capacitor.*V*_{i }= Rate at which driving force pours in energy.

It goes into (i) ohmic loss and (ii) increase of stored energy.

Hence, equation (2) in the form of conservation of energy statement. Integrating both the sides of equation (2) with respect to time over one full cycle (0 → T) we may write.

${\int}_{0}^{\mathrm{T}}\frac{d}{dt}\left(\frac{1}{2}L{i}^{2}+\frac{{q}^{2}}{2C}\right)dt+{\int}_{0}^{\mathrm{T}}R{i}^{2}dt={\int}_{0}^{\mathrm{T}}vidt\phantom{\rule{0ex}{0ex}}\Rightarrow 0+\left(+ve\right)={\int}_{0}^{\mathrm{T}}vidt\phantom{\rule{0ex}{0ex}}\Rightarrow {\int}_{0}^{\mathrm{T}}vidt0$

If phase difference between *v *& *i *is a constant and acute angle.

#### Page No 46:

#### Question 31:

*LCR*circuit shown in the given figure, the ac driving voltage is

*v = v*sin ω

_{m}*t*.

*q*(

*t*).

*t = t*

_{0}, the voltage source stops and

*R*

*is short circuited. Now write down how much energy is stored in each of*

*L*and

*C*.

#### Answer:

(i) The ac driving voltage in the given LCR circuit,

$v={v}_{m}\mathrm{sin}\omega t$

At a time *t,*

$\mathrm{Sum}\mathrm{of}\mathrm{potential}\mathrm{drop}\mathrm{across}\mathrm{inductor},\mathrm{resistor}\mathrm{and}\mathrm{the}\mathrm{capacitor},\phantom{\rule{0ex}{0ex}}L\frac{{d}^{2}q}{d{t}^{2}}+R\frac{dq}{dt}+\frac{q}{C}={v}_{m}\mathit{sin}\omega t\phantom{\rule{0ex}{0ex}}Let\mathit{,}\mathit{}q\mathit{=}{q}_{m}\mathrm{cos}\left(\omega t+\varphi \right)\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}(\mathrm{let},\varphi =\mathrm{phase})\phantom{\rule{0ex}{0ex}}\mathrm{Hence},\mathrm{the}\mathrm{current}\mathrm{at}\mathrm{time}t,\phantom{\rule{0ex}{0ex}}\left|i\right|=\left|\frac{dq}{dt}\right|=\left|\frac{d}{dt}\left[{q}_{m}\mathrm{cos}\left(\omega t+\varphi \right)\right]\right|={q}_{m}\omega \mathrm{sin}\mathit{(}\omega t\mathit{+}\varphi \mathit{)}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

$i={i}_{m}\mathrm{sin}\left(\omega t+\varphi \right)\left(\mathrm{Let},{i}_{m}={q}_{m}\omega \right)$

Now for given LCR circuit,

${i}_{m}=\frac{{v}_{m}}{Z}=\frac{{V}_{m}}{\sqrt{{R}^{2}+{\left({X}_{L}-{X}_{C}\right)}^{2}}}\phantom{\rule{0ex}{0ex}}\mathrm{Hence},\phantom{\rule{0ex}{0ex}}\mathrm{tan}\varphi =\frac{\left({X}_{L}-{X}_{C}\right)}{R}$

(ii) At, *t* =* t*_{o,}* *resistance R is short circuited.

Due to which energy stored in inductor L:

${U}_{L}=\frac{1}{2}L{i}^{2}=\frac{1}{2}L{\left[\frac{{v}_{m}}{Z}\right]}^{2}=\frac{1}{2}L{\left[\frac{{v}_{m}}{\sqrt{{R}^{2}+{\left({X}_{L}-{X}_{C}\right)}^{2}}}\right]}^{2}{\mathrm{sin}}^{2}\left(\omega {t}_{o}+\varphi \right)\phantom{\rule{0ex}{0ex}}\mathrm{Energy}\mathrm{stored}\mathrm{in}\mathrm{capacitor}:\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{U}_{L}=\frac{1}{2}\frac{{q}^{2}}{C}=\frac{1}{2C}{\left[\frac{{v}_{m}}{\sqrt{{R}^{2}+{\left({X}_{L}-{X}_{C}\right)}^{2}}}\right]}^{2}\left[\frac{1}{{\omega}^{2}}{\mathrm{cos}}^{2}\left(\omega {t}_{o}+\varphi \right)\right]\left(\mathrm{As},\omega {q}_{m}={i}_{m}\right)$

(iii) Now further the circuit behaves as an LC oscillator. The capacitor will go on discharge further and all energy will go to inductor L and back and forth continuously.

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