# NCERT Solutions for Class 12 Physics Chapter 7 Alternating Current

NCERT solutions for class 12 physics chapter 7 Alternating current- The supply that is received in our home is alternating in nature. Do you know what is the value of normal supply voltage that is coming in our home and what is the supply frequency? Solutions of NCERT class 12 physics chapter 7 alternating current will help you to clear such doubts. The alternating current in our daily application can be of a single phase which is utilised in our home or can be of three phases which are usually used in industries. Questions explained in the CBSE NCERT solutions for class 12 physics chapter 7 alternating current is based on single-phase alternating current. NCERT solutions are the basics for the CBSE board exam. In this chapter, you will study the concept of phasor diagrams. The knowledge of the mathematics chapter complex numbers will help you to understand the phasors and the problems in NCERT solutions for class 12 physics chapter 7 alternating current.

To solve the circuited related problem of NCERT class 12 chapter 7 alternating current you must refresh the Kirchhoff's voltage law and Kirchoff's current law that you have studied in current electricity chapter. The last topic of class 12 physics chapter 7 alternating current gives an introduction to single-phase transformers.

## NCERT solutions for class 12 physics chapter 7 alternating current exercises:

a)what is the RMS value of current?

Given,

RMS voltage in the circuit $V_{rms}=220V$

Resistance in the circuit $R=100\Omega$

Now,

RMS current in the circuit:

$I_{rms}=\frac{V_{rms}}{R}=\frac{220}{100}=2.2A$

Hence, the RMS value of current is 2.2A.

What is the rms value of current in the circuit?

Given,

RMS Voltage in the circuit $V_{rms}=200V$

Resistance in the circuit $R=100\Omega$

Now,

RMS Current in the circuit:

$I_{rms}=\frac{V_{rms}}{R}=\frac{220}{100}=2.2A$

Hence, the RMS value of current is 2.2A.

What is the net power consumed over a full cycle?

Given,

Supplied RMS Voltage $V_{rms}=220V$

Supplied RMS Current$I_{rms}=2.2A$

The net power consumed over a full cycle:

$P=V_{rms}I_{rms}=220*2.2=484W$

Hence net power consumed is 484W.

Given

Peak Value of ac supply:

$V_{peak}=300V$

Now as we know in any sinusoidal function

$RMSvalue=\frac{peakvalue}{\sqrt{2}}$

Since our ac voltage  supply is also sinusoidal

$V_{rms}=\frac{V_{peak}}{\sqrt{2}}=\frac{300}{\sqrt{2}}=212.13V$

Hence RMS value of voltage os 212.13V.

Given,

RMS value of current $I_{rms}=10A$

Since Current is also sinusoidal (because only resistance is present  in the circuit, not the capacitor and inductor)

$I_{rms}=\frac{I_{peak}}{\sqrt{2}}$

$I_{peak}=\sqrt{2}I_{rms}=\sqrt{2}*10=14.1A$

Hence the peak value of current is 14.1A.

Given

Supply Voltage $V=220V$

Supply Frequency $f=50Hz$

The inductance of the inductor connected $L=44mH=44*10^{-3}H$

Now

Inductive Reactance

$X_L=\omega L=2\pi fL=2\pi *50*44*10^{-3}$

RMS Value of the current :

$I_{rms}=\frac{V_{rms}}{X_L}= \frac{220}{2\pi *50*44*10^{-3}}=15.92A$

Hence the RMS Value of current is 15.92A.

Given,

Supply Voltage $V = 110V$

Supply Frequency $f=60Hz$

The capacitance of the connected capacitor   $C=60\mu F=60*10^{-6}F$

Now,

Capacitive Reactance

$X_C=\frac{1}{\omega C}=\frac{1}{2\pi fC}=\frac{1}{2\pi *60*60*10^{-6}}$

RMS Value of current

$I_{rms}=\frac{V_{rms}}{X_C}=V\omega C=V2\pi fC=110*2\pi *60*60*10^{-6}=2.49A$

Hence the RMS Value of current is 2.49A.

As we know,

Power absorbed $P=VIcos\phi$

Where $\phi$ is the phase difference between voltage and current.

$\phi$ for the inductive circuit is -90 degree and $\phi$ for the capacitive circuit is +90 degree.

In both cases (inductive and capacitive), the power absorbed by the circuit is zero because in both cases the phase difference between current and voltage is 90 degree.

This can be seen as The elements(Inductor and Capacitor)  are not absorbing the power, rather storing it. The capacitor is storing energy in electrostatic form and Inductor is storing the energy in magnetic form.

Given, in a circuit,

Inductance,  $L=2H$

Capacitance,    $C=32\mu F=32*10^{-6}F$

Resistance, $R=10\Omega$

Now,

Resonance frequency (frequency of maximum current OR minimum impedance OR frequency at which inductive reactance cancels out capacitive reactance )

$\omega _r=\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{2*32*10^{-6}}}=\frac{1}{8*10^{-3}}=125s^{-1}$

Hence Resonance frequency is 125 per second.

Q-Value:

$Q=\frac{1}{R}\sqrt{\frac{L}{C}}=\frac{1}{10}\sqrt{\frac{2}{32*10^{-6}}}=25$

Hence Q - value of the circuit is 25.

Given

Capacitance  $C=30\mu F=30*10^{-6}$

Inductance $L = 27mH = 27*10^{-3}H$

Now,

Angular Frequency

$\omega _r=\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{30*10^{-6}*27*10^{-3}}}=1.11*10^{3}rad/sec$

Hence Angular Frequency is $1.11*10^{3}rad/sec$

Given

Capacitance  $C=30\mu F=30*10^{-6}$

Inductance $L = 27mH = 27*10^{-3}H$

Charge on the capacitor  $Q=6mC=6*10^{-3}C$

Now,

The total energy stored in Capacitor :

$E=\frac{Q^2}{2C}=\frac{(6*10^{-3})^2}{2*30*10^{-6}}=\frac{6}{10}=0.6J$

Total energy later will be same because energy is being shared with capacitor and inductor and none of them loses the energy, they just store it and transfer it.

Given,

Resistance $R=20\Omega$

Inductance $L=1.5H$

Capacitance $C=35\mu F=35*10^{-6}F$

Voltage supply $V = 200V$

At resonance, supply frequency is equal to the natural frequency, and at the natural frequency, the total impedance of the circuit is equal to the resistance of the circuit

as inductive and capacitive reactance cancels each other. in other words,

$Z = \sqrt{\left ( \omega L-\frac{1}{\omega C} \right )^2+R^2}=\sqrt{0^2+R^2}=R=20\Omega$

As

$\omega L=\frac{1}{\omega C}$

Now,

Current in the circuit

$I=\frac{V}{Z}=\frac{200}{20}=10A$

Average Power transferred in the circuit :

$P=VI=200*10=2000W$

Hence average power transferred is 2000W.

Given,

Range of the frequency in which radio can be tune = $(800kHz\: to1200kHz)$

The effective inductance of the Circuit = $200\mu H$

Now, As we know,

$w^2=1/\sqrt{LC}$

$C=1/w^2L$

where $w$ is tuning frequency.

For getting the range of the value of a capacitor, let's calculate the two values of the capacitor, one maximum, and one minimum.

first, let's calculate the minimum value of capacitance which is the case when tuning frequency = 800KHz.

$C_{minimum}=\frac{1}{w_{minimum}^2L}=\frac{1}{(2\pi(800*10^3))^2*200*10^{-6}}=1.981*10^{-10}F$

Hence the minimum value of capacitance is 198pF.

Now, Let's calculate the maximum value of the capacitor.

in this case, tuning frequency = 1200KHz

$C_{maximum}=\frac{1}{w_{maximum}^2L}=\frac{1}{(2\pi(1200*10^3))^2*200*10^{-6}}=88.04*10^{-12}F$

Hence the maximum  value of the capacitor is 88.04pf

Hence the Range of the values of the capacitor is $88.04pF to 198.1pF$.

(a)Determine the source frequency which drives the circuit in resonance.

Given,

Variable frequency supply voltage $V$ = 230V

Inductance $L=5.0H$

Capacitance $C=80\mu F=80*10^{-6}F$

Resistance $R=40\Omega$

a) Resonance angular frequency in this circuit is given by :

$w_{resonance}=\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{5*80*10^{-6}}}=\frac{1000}{20}=50rad/sec$

Hence this circuit will be in resonance when supply frequency is 50 rad/sec.

$L=5.0H$$C=80\mu F$, $R=40\Omega$.

(b) Obtain the impedance of the circuit and the amplitude of current at the resonating frequency.

Given,

Variable frequency supply voltage $V$ = 230V

Inductance $L=5.0H$

Capacitance $C=80\mu F=80*10^{-6}F$

Resistance $R=40\Omega$

Now,

The impedance of the circuit is

$Z=\sqrt{(wL-\frac{1}{wC})^2+R^2}$

at Resonance Condition

:$wL=\frac{1}{wC}$

$Z=R=40\Omega$

Hence, Impedance at resonance is 40$\Omega$.

Now, at resonance condition, impedance is minimum which means current is maximum which will happen when we have a peak voltage, so

Current in the Resonance circuit is Given by

$I_{resonance}=\frac{V_{peak}}{Z}=\frac{\sqrt{2}* 230}{40}=8.13A$

Hence amplitude of the current at resonance is 8.13A.

$L=5.0H$$C=80\mu F$, $R=40\Omega$.

Determine the rms potential drops across the three elements of the circuit. Show that the potential drop across the LC combination is zero at the resonating frequency.

Potential difference across any element = $I_{rms}*(impedance)$

$I_{rms}=\frac{I_{peak}}{\sqrt{2}}=\frac{8.13}{\sqrt{2}}=5.85A$

Now

The potential difference across the capacitor:

$V_{capacitor}=I_{rms}*\left (\frac{1}{w_{resonance}C} \right ) =5.85*\left ( \frac{1}{50*80*10^{-6}} \right )=1437.5V$

The potential difference across the inductor

$V_{inductor}=I_{rms}*(w_{resonance}L) =5.85* 50*5=1437.5V$

The potential difference across Resistor

$V_{resistor}=I_{rms}*\R =5.85* 40=230V$

Potential difference across LC combination

$V_{LC}=I_{rms}*\left ( wL-\frac{1}{wC} \right )=5.85*0=0$

Hence at resonating, frequency potential difference across LC combination is zero.

## NCERT solutions for class 12 physics chapter 7 alternating current additional exercises:

What is the total energy stored initially? Is it conserved during $LC$ oscillations?

Given,

The inductance of the inductor:

$L=20mH=20*10^{-3}H$

The capacitance of the capacitor :

$C=50\mu F=50*10^{-6}F$

The initial charge on the capacitor:

$Q=10mC=10*10^{-3}C$

Total energy present at the initial moment:

$E_{initial}=\frac{Q^2}{2C}=\frac{(10*10^{-3})^2}{2*50*10^{-6}}=1J$

Hence initial energy in the circuit is 1J. Since we don't have any power-consuming element like  resistance in the circuit, the energy will be conserved

Given,

The inductance of the inductor:

$L=20mH=20*10^{-3}H$

The capacitance of the capacitor :

$C=50\mu F=50*10^{-6}F$

The initial charge on the capacitor:

$Q=10mC=10*10^{-3}C$

The natural angular  frequency of the circuit:

$w_{natural}=\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{(20*10^{-3}*50*10^{-6})}}=10^3rad/sec$

Hence the natural angular frequency of the circuit is $10^3rad/sec$.

The natural frequency of the circuit:

$f_{natural}=\frac{w_{natural}}{2\pi}=\frac{10^3}{2\pi}=159Hz$

Hence the natural frequency of the circuit is 159Hz.

( c)  At what time is the energy stored

(i) completely electrical (i.e., stored in the capacitor)?

at any instant, the charge on the capacitor is:

$Q=Q_0cos(w_{natural}t)=Q_0cos(2\pi f_{natural}t)=Q_0cos\left ( \frac{2\pi t}{T} \right )$

Where time period :

$T=\frac{1}{f_{natural}}=\frac{1}{159}=6.28ms$

Now, when the total energy is purely electrical, we can say that

$Q=Q_0$

$Q_0=Q_0cos(\frac{2\pi}{T})$

$cos(\frac{2\pi t}{T})=1$

this is possible when

$t=0,\frac{T}{2},T,\frac{3T}{2}....$

Hence Total energy will be purely electrical(stored in a capacitor) at

$t=0,\frac{T}{2},T,\frac{3T}{2}....$.

(ii) completely magnetic (i.e., stored in the inductor)?

The stored energy will we purely magnetic when the pure electrical stored is zero. i.e. when the charge on the capacitor is zero, all energy will be stored in the inductor.

So, t for which charge on the capacitor is zero is

$t=\frac{T}{4},\frac{3T}{4},\frac{5T}{4}..$

Hence at these times, the total energy will be purely magnetic.

At what times is the total energy shared equally between the inductor and the capacitor?

The energy will be shared equally when the energy in the capacitor is half of the maximum energy it can store.i.e.

$\frac{Q^2}{2C}=\frac{1}{2}\frac{Q_0^2}{2C}$

From here, we got

$Q=\frac{Q_0}{\sqrt{2}}$

So now, we know the charge on the capacitor, we can calculate the time for which

$\frac{Q_0}{\sqrt{2}}=Q_0cos\left ( \frac{2\pi t}{T} \right )$

$\frac{1}{\sqrt{2}}=cos\left ( \frac{2\pi t}{T} \right )$

From here,

$t=\frac{T}{8},\frac{3T}{8},\frac{5T}{8}..$

Hence for these times, the total energy will be shared equally between capacitor and inductor.

If a resistor is inserted in the circuit, how much energy is eventually dissipated as heat?

If the resistance is added to the circuit, the whole energy will dissipate as heat eventually. energy will keep moving between the capacitor and inductor with reducing in magnitude in each cycle and eventually all energy will be dissipated.

Given,

The inductance of the coil $L=0.50H$

the resistance of the coil $R=100\Omega$

Supply voltage $V=240V$

Supply voltage frequency$f=50Hz$

Now, as we know peak voltage = $\sqrt2$(RMS Voltage)

Peak voltage

$V_{peak}=\sqrt2*240=339.4V$

The impedance of the circuit :

$Z=\sqrt{R^2+(wL)^2}=\sqrt{100^2+(2\pi (.5) *50)^2}$

Now peak current in the circuit :

$I_{peak}=\frac{V_{peak}}{Z}=\frac{339}{\sqrt{100^2+(2\pi (.5) *50)^2}}=1.82A$

Hence peak current is 1.82A in the circuit.

Let the voltage  in the circuit be

$V = V_0coswt$ and

Current in the circuit be

$I = I_0cos(wt-\phi )$

Where $\phi$ is the phase difference between voltage and current.

V is maximum At

t = 0

$I$ is maximum At

$t=\frac{w}{\phi }$

Hence, the time lag between voltage maximum and the current maximum is

$\frac{w}{\phi }$.

For phase difference $\phi$  we have

$tan\phi =\frac{wL}{R}=\frac{2\pi *50*0.5}{100}=1.57$

$\phi =57.5^0$

$t=\frac{\phi}{w}=\frac{57.5*\pi}{180*2\pi *50}=3.2ms$

Hence time lag between the maximum voltage and the maximum current is $3.2ms$

## Q7.14 Obtain the answers (a) to (b) in Exercise 7.13 if the circuit is connected to a high-frequency supply $(240 V\: ,10 kHz)$. Hence, explain the statement that at very high frequency, an inductor in a circuit nearly amounts to an open circuit. How does an inductor behave in a dc circuit after the steady state?

Given,

The inductance of the coil $L=50H$

the resistance of the coil $R=100\Omega$

Supply voltage $V=240V$

Supply voltage frequency$f=10kHz$

a)

Now, as we know peak voltage = $\sqrt2$(RMS Voltage)

Peak voltage $V_{peak}=\sqrt2*240=339.4V$

Now,

The impedance of the circuit :

$Z=\sqrt{R^2+(wL)^2}=\sqrt{100^2+(2\pi 10*10^3 *50)^2}$

Now peak current in the circuit :

$I_{peak}=\frac{V_{peak}}{Z}=\frac{339}{\sqrt{100^2+(2\pi 10*10^3 *50)^2}}=1.1*10^{-2}A$

Hence peak current is $1.1*10^{-2}A$ in the circuit.

The current in the circuit is very small, which is one of the indications of inductor working as a nearly open circuit in the case of high frequency.

b)

For phase difference $\phi$  we have

$tan\phi =\frac{wL}{R}=\frac{2\pi *10*10^3*0.5}{100}=100\pi$

$\phi =89.82^0$

Now

$t=\frac{\phi}{w}=\frac{89.82*\pi}{180*2\pi *10^3}=25\mu s$

Hence time lag between the maximum voltage and the maximum current is $25\mu s$.

In the DC circuit, after attaining the steady state, inductor behaves line short circuit as $w$ is Zero.

## Q7.15 (a)   A $100\mu F$ capacitor in series with a $40\Omega$ resistance is connected to a $110\: V$, $60\: Hz$supply.  What is the maximum current in the circuit?

Given,

The capacitance of the capacitor $C=100\mu F$

The resistance of the circuit $R=40\Omega$

Voltage supply $V = 100V$

Frequency of voltage supply $f=60Hz$

The maximum current in the circuit

$I_{max}=\frac{V_{max}}{Z}=\frac{\sqrt{2}V}{\sqrt{R^2+\left ( \frac{1}{wC} \right )^2}}=\frac{\sqrt{2}*110}{\sqrt{40^2+\left ( \frac{1}{2\pi *60*100*10^{-6}} \right )^2}}=3.24A$

Hence maximum current in the circuit is 3.24A.

In the case of a capacitor, we have

$tan\phi=\frac{\frac{1}{wC}}{R}=\frac{1}{wCR}$

So,

$tan\phi=\frac{1}{wCR}=\frac{1}{2\pi 60 *100*10^{-6}*40}=0.6635$

$\phi=33.56^0$

So the time lag between max voltage and the max current is :

$t=\frac{\phi }{w}=\frac{33.56\pi}{180*2\pi*60}=1.55ms$

Given,

The capacitance of the capacitor $C=100\mu F$

The resistance of the circuit $R=40\Omega$

Voltage supply $V = 100V$

Frequency of voltage supply $f=12kHz$

The maximum current in the circuit

$I_{max}=\frac{V_{max}}{Z}=\frac{\sqrt{2}V}{\sqrt{R^2+\left ( \frac{1}{wC} \right )^2}}=\frac{\sqrt{2}*110}{\sqrt{40^2+\left ( \frac{1}{2\pi *12*10^3*100*10^{-6}} \right )^2}}=3.9A$

Hence maximum current in the circuit is 3.9A.

b)

In the case of capacitor, we have

$tan\phi=\frac{\frac{1}{wC}}{R}=\frac{1}{wCR}$

So,

$tan\phi=\frac{1}{wCR}=\frac{1}{2\pi 10*10^3 *100*10^{-6}*40}=\frac{1}{96\pi}$

$\phi=0.2^0$

So the time lag between max voltage and max current is :

$t=\frac{\phi }{w}=\frac{0.2\pi}{180*2\pi*60}=0.04\mu s$

At high frequencies, $\phi$ tends to zero. which indicates capacitor acts as a conductor at high frequencies.

In the DC circuit, after a steady state is achieved, Capacitor acts like an open circuit.

As we know, in the case of a parallel RLC circuit:

$\frac{1}{Z}=\sqrt{\frac{1}{R^2}+(wC-\frac{1}{wL})^2}$

$I=\frac{V}{Z}={V}{\sqrt{\frac{1}{R^2}+\left (wC- \frac{1}{wL} \right )^2}}$

The current will be minimum when

$wC=\frac{1}{wL}$

Which is also the condition of natural frequency. Hence the total current is minimum when source frequency is equal to the natural frequency.

RMS value of current in R

$I_{rms}=\frac{V_{rms}}{R}=\frac{230}{40}=5.75A$

RMS value in Inductor

$I_{inductor}=\frac{V_{rms}}{wL}=\frac{230}{5*50}=0.92A$

RMS value in capacitor

$I_{capacitor}=\frac{V_{rms}}{1/wL}={230*50*80*10^{-6}}=0.92A$

Capacitor current and inductor current will cancel out each other so the current flowing in the circuit is 5.75A.

The inductance of the inductor $L=80mH=80*10^3H$

The capacitance of the capacitor $C=60\mu F$

Voltage supply $V = 230V$

Frequency of voltage supply $f=50Hz$.

Here, we have

$V=V_{max}sinwt=V_{max}sin2\pi ft$

Impedance

$Z=\sqrt{R^2+\left ( wL-\frac{1}{wC} \right )^2}$

$Z=\sqrt{0^2+\left ( 2\pi*50*80*10^{-3}-\frac{1}{2\pi 50*60*10^{-6}} \right )^2}=8\pi-\frac{1000}{6\pi }$

Now,

Current in the circuit will be

$I=\frac{V}{Z}=\frac{V_{max}sinwt}{Z\angle \phi }=I_{max}sin(wt-\phi )$

where,

$I_{max}=\frac{V_{max}}{Z}=\frac{\sqrt{2}*230}{8\pi-\frac{1000}{6\pi}}=-11.63A$

The negative sign is just a matter of the direction of current.so,

$I=11.63sin(wt-\phi )$

here

$tan\phi=\frac{wL-\frac{1}{wC}}{R}$

But, since the value of R is zero(since our circuit have only L and C)

$\phi=90^0$

Hence

$I=11.63sin(wt-\frac{\pi}{2} )$

Now,

RMS value of this current:

$I_{rms}=\frac{I_{max}}{\sqrt{2}}=\frac{11.63}{\sqrt{2}}=8.22A$.

As we know,

RMS potential drop across an element with impedance Z:

$V_{element}=I_{rms}Z_{element}$

SO,

RMS potential difference across inductor:

$V_{inductor}=I_{rms}*wL=8.22*2\pi *60*80*10^{-3}=206.61V$

RMS potential drop across capacitor

$V_{capacitor}=I_{rms}*\frac{1}{wC}=8.22*\frac{1}{2\pi*60*60*10^{-6}}=436.3V$

(c) What is the average power transferred to the inductor?

Since

$I=I_{max}sin(wt-\phi )$

Current flowing in the circuit is sinusoidal and hence average power will be zero as the average of sin function is zero.in other words, the inductor will store energy in the positive half cycle of the sin (0 degrees to 180 degrees) and will release that energy in the negative half cycle(180 degrees to 360 degrees), and hence average power is zero.

As we know,

Average power $P=VIcos\theta$  where $\theta$ is the phase difference between voltage and current.

Since in the circuit, phase difference $\theta$ is $\pi/2$, the average power is zero.

Since the phase difference between voltage and current is 90 degree, even the total power absorbed by the circuit is zero. This is an ideal circuit, we can not have any circuit in practical that consumes no power, that is because practically resistance of any circuit is never zero. Here only inductor and capacitor are present and none of them consumes energy, they just store it and transfer it like they are doing it in this case.

The inductance of the inductor $L=80mH=80*10^3H$

The capacitance of the capacitor $C=60\mu F$

The resistance of a $15\Omega$resistor

Voltage supply $V = 230V$

Frequency of voltage supply $f=50Hz$

As we know,

Impedance

$Z=\sqrt{R^2+\left ( wL-\frac{1}{wC} \right )^2}$

$Z=\sqrt{15^2+\left ( 2\pi*50*80*10^{-3}-\frac{1}{2\pi 50*60*10^{-6}} \right )^2}=31.728$

Current flowing in the circuit :

$I=\frac{V}{Z}=\frac{230}{31.72}=7.25A$

Now,

Average power transferred to the resistor:

$P_{resistor}=I^2R=(7.25)^2*15=788.44W$

Average power transferred to the inductor = 0

Average power transferred to the capacitor = 0:

Total power absorbed by circuit :

$P_{resistor}+p_{inductor}+P_{capacitor}=788.44+0+0=788.44W$

Hence circuit absorbs 788.44W.

What is the source frequency for which current amplitude is maximum. Obtain this maximum value.

The inductance of the inductor $L=0.12H$

The capacitance of the capacitor $C=480n F$

The resistance of the resistor $R=23\Omega$

Voltage supply $V = 230V$

Frequency of voltage supply $f=50Hz$

As we know,

the current amplitude is maximum at the natural frequency of oscillation, which is

$w_{natural}=\sqrt\frac{1}{LC}=\frac{1}{\sqrt{0.12*480*10^{-9}})}=4166.67rad/sec$

Also, at this frequency,

$Z=R=23$

SO,

The maximum current in the circuit :

$I_{max}=\frac{V_{max}}{Z}=\frac{V_{max}}{R}=\frac{\sqrt{2*}230}{23}=14.14A$

Hence maximum current is 14.14A.

Since the resistor is the only element in the circuit which consumes the power, the maximum absorbed power by circuit will be maximum when power absorbed by the resistor will be maximum. power absorbed by the resistor will be maximum at when current is maximum which is the natural frequency case,

Hence when source frequency will be equal to the natural frequency, the power absorbed will be maximum.

Hence frequency

$f=\frac{w_r}{2\pi}=\frac{4166.67}{2\pi}=663.48Hz$

Maximum Power Absorbed

$P=I^2R=(14.14)^2*23=2299.3W$.

The value of maximum angular frequency is calculated in the first part of the question and whose magnitude is 4166.67

Q-factor of any circuit is given by

$Q=\frac{w_rL}{R}=\frac{4166.67*0.12}{23}=21.74$

Hence Q-factor for the circuit is 21.74.

As

Power $P=I^2R$

Power $P$ will be half when the current $I$ is  $1/\sqrt{2}$ times the maximum current.

As,

$I =I_{max}Sin(wt-\phi)$

At half powerpoint :

$\frac{i_{max}}{\sqrt{2}} =I_{max}Sin(wt-\phi)$

$\frac{1}{\sqrt{2}} =Sin(wt-\phi)$

$wt=\phi+\frac{\pi}{4}$

here,

$\phi=tan^{-1}(\frac{wL-\frac{1}{wC}}{R})$

On putting values, we get, two values of $w$ for which

$wt=\phi+\frac{\pi}{4}$

And they are:

$w_1=678.75Hz$

$w_2=648.22Hz$

Also,

The current amplitude at these frequencies

$I_{halfpowerpoint}=\frac{I_{max}}{\sqrt{2}}=\frac{14.14}{1.414}=10A$

The inductance of the inductor $L=0.3H$

The capacitance of the capacitor $C=27\mu F$

The resistance of the resistor $R=7.4\Omega$

Now,

Resonant frequency

$w_r=\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{0.3*27*10^{-6}}}=111.11rad/sec$

Q-Factor of the circuit

$Q=\frac{w_rL}{R}=\frac{111.11*0.3}{7.4}=45.0446$

Now, to improve the sharpness of resonance by reducing its full width at half maximum, by a factor of 2 without changing$w_r$,

we have to change the resistance of the resistor to half of its value, that is

$R_{new}=\frac{R}{2}=\frac{7.4}{2}=3.7\Omega$

Yes, at any instant the applied voltage will be distributed among all element and the sum of the instantaneous voltage of all elements will be equal to the applied. But this is not the case in RMS because all elements are varying differently and they may not be in the phase.

Yes, we use capacitors in the primary circuit of an induction coil to avoid sparking. when the circuit breaks, a large emf is induced and the capacitor gets charged from this avoiding the case of sparking and short circuit.

Q 7.22 (c)  Answer the following questions:

An applied voltage signal consists of a superposition of a $dc$  voltage and an ac voltage of high frequency. The circuit consists of an inductor and a capacitor in series. Show that the dc signal will appear across $C$ and the ac signal across $L$.

For a high frequency, the inductive reactance and capacitive reactance:

$X_L=wL=Large \:value\: And \:X_C = \frac{1}{wC}=Very\:small$

Hence the capacitor does not offer resistance to a higher frequency, so the ac voltage appears across L.

Similarly

For DC, the inductive reactance and capacitive reactance:

$X_L=wL=Very\:small\: And \:X_C = \frac{1}{wC}=Large \:value$

Hence DC signal appears across Capacitor only.

Q 7.22 (d) Answer the following questions:

A choke coil in series with a lamp is connected to a dc line. The lamp is seen to shine brightly. Insertion of an iron core in the choke causes no change in the lamp’s brightness. Predict the corresponding observations if the connection is to an ac line.

For a steady state DC, the increasing inductance value by inserting iron core in the choke, have no effect in the brightness of the connected lamp, whereas, for ac when the iron core is inserted, the light of the lamp will shine less brightly because of increase in inductive impedance.

Q7.22 (e) Answer the following questions:

Why is choke coil needed in the use of fluorescent tubes with ac mains? Why can we not use an ordinary resistor instead of the choke coil?

We need choke coil in the use of fluorescent tubes with ac mains to reduce the voltage across the tube without wasting much power. If we use simply resistor for this purpose, there will be more power loss, hence we do not prefer it.

Given,

Input voltage:

$V_{input}=2300V$

Number of turns in the primary coil

$N_{primary}= 4000$

Output voltage:

$V_{output}=230V$

Now,

Let the number of turns in secondary be

$N=N_{secondary}$

Now as we know, in a transformer,

$\frac{V_{primary}}{V_{secondary}}=\frac{N_{primary}}{N_{secondary}}$

${N_{secondary}} =\frac{V_{secondary}}{V_{primary}}*N_{primary}=\frac{230}{2300}*4000=400$

Hence the number of turns in secondary winding id 400.

## Q7.24   At a hydroelectric power plant, the water pressure head is at a height  of $300m$ and the water flow available is $100m^{3}s^{-1}$. If the turbine generator efficiency is $60^{0}/_{0}$, estimate the electric power availablefrom the plant $(g=9.8ms^{-2})$.

Given,

Height of the water pressure head

$h=300m$

The volume of the water flow per second

$V=100m^3s^{-1}$

Turbine generator efficiency

$\eta =0.6$

Mass of water flowing per second

$M=100*10^3=10^5kg$

The potential energy stored in the fall for 1 second

$P=Mgh=10^5*9.8*300=294*10^6J$

Hence input power

$P_{input}=294*10^6J/s$

Now as we know,

$\eta =\frac{P_{output}}{P_{input}}$

$P_{output}=\eta *P_{input}=0.6*294*10^6=176.4*10^6W$

Hence output power is 176.4 MW.

Power required

$P=800kW=800*10^3kW$

The total resistance of the two-wire line

$R=2*15*0.5=15\Omega$

Input Voltage

$V_{input}=4000V$

Output Voltage:

$V_{output}=220V$

RMS Current in the wireline

$I=\frac{P}{V_{input}}=\frac{800*10^3}{4000}=200A$

Now,

Power loss in the line

$P_{loss}=I^2R=200^2*15=600*10^3=600kW$

Hence, power loss in line is 600kW.

How much power must the plant supply, assuming there is negligible power loss due to leakage?

Power required

$P=800kW=800*10^3kW$

The total resistance of the two-wire line

$R=2*15*0.5=15\Omega$

Input voltage

$V_{input}=4000V$

Output voltage:

$V_{output}=220V$

RMS current in the wireline

$I=\frac{P}{V_{input}}=\frac{800*10^3}{4000}=200A$

Now,

Total power delivered by plant = line power loss + required electric power = 800 + 600 = 1400kW.

Power required

$P=800kW=800*10^3kW$

The total resistance of the two-wire line

$R=2*15*0.5=15\Omega$

Input Voltage

$V_{input}=4000V$

Output Voltage:

$V_{output}=220V$

RMS Current in the wireline

$I=\frac{P}{V_{input}}=\frac{800*10^3}{4000}=200A$

Now,

Voltage drop in the power line = $IR=200*15=3000V$

Total voltage transmitted from the plant = 3000+4000=7000

as power is generated at 440V, The rating of the power plant is 440V-7000V.

Power required

$P=800kW=800*10^3kW$

The total resistance of the two-wire line

$R=2*15*0.5=15\Omega$

Input Voltage

$V_{input}=40000V$

Output Voltage:

$V_{output}=220V$

RMS current in the wireline

$I=\frac{P}{V_{input}}=\frac{800*10^3}{40000}=20A$

Now,

a) power loss in the line

$P_{loss}=I^2R=20^2*15=6kW$

b)

Power supplied by plant = 800 kW + 6 kW = 806kW.

c)

Voltage drop in the power line = $IR=20*15=300V$

Total voltage transmitted from the plant = 300+40000=40300

as power is generated at 440V, The rating of the power plant is 440V-40300V.

We prefer high voltage transmission because power loss is a lot lesser than low voltage transmission.

## NCERT solutions for class 12 physics- chapter wise

 NCERT solutions for class 12 physics chapter 1 Electric Charges and Fields Solutions of NCERT class 12 physics chapter 2 Electrostatic Potential and Capacitance CBSE NCERT solutions for class 12 physics chapter 3 Current Electricity NCERT solutions for class 12 physics chapter 4 Moving Charges and Magnetism Solutions of NCERT class 12 physics chapter 5 Magnetism and Matter CBSE NCERT solutions for class 12 physics chapter 6 Electromagnetic Induction NCERT solutions for class 12 physics chapter 7 Alternating Current Solutions of NCERT class 12 physics chapter 8 Electromagnetic Waves CBSE NCERT solutions for class 12 physics chapter 9 Ray Optics and Optical Instruments NCERT solutions for class 12 physics chapter 10 Wave Optics Solutions Solutions of NCERT class 12 physics chapter 11 Dual nature of radiation and matter CBSE NCERT solutions for class 12 physics chapter 12 Atoms NCERT solutions for class 12 physics chapter 13 Nuclei Solutions of NCERT class 12 physics chapter 14 Semiconductor Electronics Materials Devices and Simple Circuits

## Significance of NCERT solutions for class 12 physics chapter 7 alternating current in board exam:

From the unit electromagnetic induction and alternating current around 10% questions are asked in the CBSE board exam. If all the solutions of NCERT class 12 physics chapter 7 alternating current are covered then it is easy to answer questions asked in board exam related to the chapter. For JEE Main, NEET and other state board and competitive exams also this chapter is important. Prepare the chapter well with the help of CBSE NCERT solutions for class 12 physics chapter 7 alternating current.

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