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:

Q7.1 (a) 100\Omega resistor is connected to a 220\: V50\: Hz ac supply.

a)what is the RMS value of current?

Answer:

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.

Q7.1 (b)  100W resistor is connected to a 220\: V50Hz ac supply.

 What is the rms value of current in the circuit?

Answer:

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.

Q7.1 (c) 100\Omega resistor is connected to a 220\: V50\: Hz ac supply.

What is the net power consumed over a full cycle?

Answer:

Given,

Supplied RMS Voltage V_{rms}=220V

Supplied RMS CurrentI_{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.

Q7.2 (a) The peak voltage of an ac supply is 300 V. What is the RMS voltage?

Answer:

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.

Q7.2 (b) The RMS value of current in an ac circuit is 10\: A . What is the peak current?

Answer:

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.

Q7.3  44\: mH  inductor is connected to 220\: V50\: Hz  ac supply. Determine the RMS value of the current in the circuit.

Answer:

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.

Q7.4   60\mu F capacitor is connected to a 100\: V60\; Hz  ac supply. Determine the rms value of the current in the circuit.

Answer:

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.

Q7.5  In Exercises 7.3 and 7.4, what is the net power absorbed by each circuit over a complete cycle. Explain your answer.

Answer:

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.

Q7.6  Obtain the resonant frequency \omega _{r} of a series LCR circuit with L=2.0HC=32\mu F and  R=10\Omega . What is the Q -value of this circuit?

Answer:

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.

Q7.7  A charged 30\mu F capacitor is connected to a 27mH  inductor. What is the angular frequency of free oscillations of the circuit?

Answer:

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

Q7.8 Suppose the initial charge on the capacitor in Exercise 7.7 is  6mC . What is the total energy stored in the circuit initially? What is the total energy at later time?

Answer:

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.

Q7.9  A series LCR circuit with R=20\OmegaL=1.5H  and c=35\mu F is connected to a variable-frequency 200\; V ac supply. When the frequency of the supply equals the natural frequency of the circuit, what is the average power transferred to the circuit in one complete cycle?

Answer:

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.

Q7.10  A radio can tune over the frequency range of a portion of MW broadcast band:  (800kHz\: to1200kHz)  . If its LCcircuit has an effective inductance of 200\mu H, what must be the range of its variable capacitor?

Answer:

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.

Q7.11 (a) Figure shows a series LCR circuit connected to a variable frequency 230\: V source.  L=5.0HC=80\mu F, R=40\Omega.

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

        

Answer:

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.

Q7.11 (b) Figure  shows a series LCR circuit connected to a variable frequency 230\: V source.

L=5.0HC=80\mu F, R=40\Omega.

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

            

Answer:

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.

Q7.11 Figure shows a series LCR circuit connected to a variable frequency 230\: V source.

L=5.0HC=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.

            

Answer:

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:

Q7.12  (a) An LC circuit contains a 20mH inductor and a 50\mu F capacitor with an initial charge of 10mC. The resistance of the circuit is negligible. Let the instant the circuit is closed be t=0.

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

Answer:

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

Q7.12 (b)  An LC circuit contains a 20mH inductor and a 50\mu F capacitor with an initial charge of 10mC. The resistance of the circuit is negligible. Let the instant the circuit is closed be t=0.

 What is the natural frequency of the circuit?            

Answer:

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.

Q7.12 (c-i)  An LC circuit contains a 20mH inductor and a 50\mu F capacitor withan initial charge of 10mC. The resistance of the circuit is negligible. Let the instant the circuit is closed be t=0.

 ( c)  At what time is the energy stored

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

Answer:

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}.....

Q7.12 (c-ii)  An LC circuit contains a 20mH inductor and a 50\mu F capacitor with an initial charge of 10mC. The resistance of the circuit is negligible.
(C) Let the instant the circuit is closed be t=0.

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

Answer:

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.

Q7.12 (d)    An LC circuit contains a 20mH inductor and a 50\mu F capacitor with an initial charge of 10mC. The resistance of the circuit is negligible. Let the instant the circuit is closed be t=0.

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

Answer:

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.

Q7.12 (e)    An LC circuit contains a 20mH inductor and a 50\mu F capacitor with an initial charge of 10mC. The resistance of the circuit is negligible. Let the instant the circuit is closed be t=0.

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

Answer:

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.

Q 7.13 (a)  A coil of inductance 0.50Hand resistance 100\Omega is connected to a  240V50Hz  ac supply. What is the maximum current in the coil?

Answer:

Given,

The inductance of the coil L=0.50H

the resistance of the coil R=100\Omega

Supply voltage V=240V

Supply voltage frequencyf=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.

Q 7.13 (b) A coil of inductance 0.50Hand resistance 100\Omega is connected to a 200V50Hz  ac supply. What is the time lag between the voltage maximum and the current maximum?

Answer:

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?

Answer:

Given,

The inductance of the coil L=50H

the resistance of the coil R=100\Omega

Supply voltage V=240V

Supply voltage frequencyf=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)   100\mu F capacitor in series with a 40\Omega resistance is connected to a 110\: V, 60\: Hzsupply.
 What is the maximum current in the circuit?

Answer:

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.

Q 7.15 (b)  100\mu F capacitor in series with a 40\Omega resistance is connected to a 110\; V,  60\: Hz  supply. What is the time lag between the current maximum and the voltage maximum?

Answer:

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

Q7.16  Obtain the answers to (a) and (b) in Exercise 7.15 if the circuit is connected to a 110\: V10kHz  supply? Hence, explain the statement that a capacitor is a conductor at very high frequencies. Compare this behaviour with that of a capacitor in a dc circuit after the steady state.

Answer:

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.

Q7.17  Keeping the source frequency equal to the resonating frequency of the series LCR circuit, if the three elements, LC and R are arranged in parallel, show that the total current in the parallel LCR circuit is minimum at this frequency. Obtain the current rms value in each branch of the circuit for the elements and source specified in Exercise 7.11 for this frequency.

Answer:

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.

Q7.18 (a) A circuit containing a 80mH inductor and a 60\mu F  capacitor in series is connected to a 230\: V50\: Hz  supply. The resistance of the circuit is negligible. Obtain the current amplitude and rms values.

Answer:

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.

Q7.18 (b) A circuit containing a 80mH inductor and a 60\mu F  capacitor in series is connected to a 230\: V50\: Hz  supply. The resistance of the circuit is negligible. Obtain the rms values of potential drops across each element.

Answer:

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

Q7.18 (c)  A circuit containing a 80mH inductor and a 60\mu F  capacitor in series is connected to a 230\: V50\: Hz  supply. The resistance of the circuit is negligible

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

Answer:

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.

Q7.18 (d) A circuit containing a 80mH inductor and a 60\mu F  capacitor in series is connected to a 230\: V50\: Hz  supply. The resistance of the circuit is negligible. What is the average power transferred to the capacitor?

Answer:

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.

Q7.18 (e)  A circuit containing a 80mH inductor and a 60\mu F  capacitor in series is connected to a 230\: V50\: Hz  supply. The resistance of the circuit is negligible. What is the total average power absorbed by the circuit? [‘Average’ implies ‘averaged over one cycle’.]

Answer:

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.

Q7.19  Suppose the circuit in Exercise 7.18 has a resistance of 15\Omega. Obtain the average power transferred to each element of the circuit, and the total power absorbed.

Answer:

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

The capacitance of the capacitor C=60\mu F

The resistance of a 15\Omegaresistor 

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.

Q7.20 (a)  A series LCR circuit with L=0.12H, C=480nFR=23\Omega  is connected to a 230V variable frequency supply.

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

Answer:

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.

Q7.20 (b)  A series LCR circuit with L=0.12H, C=480nFR=23\Omega  is connected to a 230V variable frequency supply.

What is the source frequency for which average power absorbed by the circuit is maximum. Obtain the value of this maximum power.      

Answer:

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.

Q7.20 (c)  A series LCR circuit with L=0.12H, C=480nFR=23\Omega  is connected to a 230Vvariable frequency supplyWhat is the Q factor of the given circuit?

Answer:

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.

Q7.20 (d)  A series LCR circuit with L=0.12H, C=480nFR=23\Omega  is connected to a 230V variable frequency supply. For which frequencies of the source is the power transferred to the circuit half the power at resonant frequency? What is the current amplitude at these frequencies?

Answer:

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

Q7.21 Obtain the resonant frequency and \varrho-factor of a series LCR  circuit  with L=0.3H, C=27\mu F, and R=7.4\Omega . It is desired to improve the sharpness of the resonance of the circuit by reducing its ‘full width at half maximum’ by a factor of 2. Suggest a suitable way.

Answer:

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 changingw_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

Q7.22 (a) In any ac circuit, is the applied instantaneous voltage equal to the algebraic sum of the instantaneous voltages across the series elements of the circuit? Is the same true for rms voltage?

Answer:

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.

7.22  (b)  Answer the following questions:  A capacitor is used in the primary circuit of an induction coil.                

Answer:

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.

Answer:

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. 

Answer:

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?

Answer:

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.

 

Q7.23  A power transmission line feeds input power at 2300 Vto a stepdown transformer with its primary windings having 40000 turns. What should be the number of turns in the secondary in order to get output power at 230 V?

Answer:

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}).

Answer:

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.

Q7.25 (c) A small town with a demand of 800 kW of electric power at 220V is situated 15km away from an electric plant generating power at 440VThe resistance of the two wire line carrying power is 0.5\Omega  per km. The town gets power from the line through a 4000-220V step-down transformer at a sub-station in the town.  Characterise the step up transformer at the plant.

Answer:

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.

Q7.26 Do the same exercise as above with the replacement of the earlier transformer by a 40,000-220V  step-down transformer (Neglect, as before, leakage losses though this may not be a good assumption any longer because of the very high voltage transmission involved). Hence, explain why high voltage transmission is preferred?

Answer:

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

Subject wise solutions

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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