Wednesday, June 1, 2022

H C Verma solutions, ALTERNATING CURRENT, Chapter-39, EXERCISES, Q11 to Q19, Concepts of Physics, Part-II

Alternating Current


EXERCISES, Q11 to Q19


     11.  A resistor of resistance 100 Ω is connected to an AC source Ɛ =(12 V) sin (250 π s⁻¹)t. Find the energy dissipated as heat during t =0 to t =1.0 ms. 


ANSWER: ω =250π s⁻¹

Time period T =2π/ω s =2π/250π s

→T =0.008 s =8 ms.

Since we have to calculate the energy dissipated in 1.0 ms which is less than the time taken to complete one cycle, we can not calculate the average heat dissipated based on rms voltage. Thus we integrate.

H =∫(Ɛ²/R)dt

Limits of integration is t =0 to t =0.001 s

→H =∫{(Ɛₒsinωt)²/R} dt

  =(Ɛₒ²/R)∫sin²ωt dt

  =½(Ɛₒ²/R)∫(1-cos2ωt) dt

 =½(Ɛₒ²/R)[t -(sin2ωt)/2ω]

Putting the limits

H =½(12²/100)[0.001-sin(2*250π*0.001)/500π] J

=0.72[0.001-(sin π/2)/500π] J

=0.72[0.001 -1/500π] J

=2.61x10⁻⁴ J.

 

 



 


     12.  In a series RC circuit with an AC source, R =300 Ω, C =25 µF, Ɛₒ =50 V, and v =50/π Hz. Find the peak current and the average power dissipated in the circuit. 


ANSWER: R =300 Ω, C =25x10⁻⁶ F,

ω =2π𝝂 =2π(50/π) s⁻¹

   =100 s⁻¹

Ɛₒ =50 V. 

Reactance of the circuit X =1/ωC

→X =1/25x10⁻⁴ Ω

     =400 Ω.

R = 300 Ω.

Hence impedance of the circuit,

Z =√(R²+X²)

  =√(300²+400²)

  =500 Ω.

Hence peak current, iₒ =Ɛₒ/Z

     =50/500 A

     =0.10 A.


The average power dissipated in the circuit is,

P =Ɛᵣₘₛiᵣₘₛcos φ

   =(Ɛₒ/√2)(iₒ/√2)(R/Z)

   =(50*0.10/2)*(300/500) W

   =1.5 W.

 




 

     13.  An electric bulb is designed to consume 55 W when operated at 110 volts. It is connected to a 220 V, 50 Hz line through a choke coil in series. What should be the inductance of the coil for which the bulb gets the correct voltage?  


ANSWER: Let resistance of the bulb =R.

Power P =E²/R 

→R =E²/P

   =110²/55 Ω 

   =220 Ω. 

Now let the inductance of the added choke coil in the circuit =L.

The reactance of the circuit, 

X =ωL

   =2πfL

   =100πL

The impedance of the circuit,

Z=√(R²+X²)

  =√{220²+(100πL)²} 

So rms current through the circuit, 

I =220/Z 

The bulb is rated for 110 V, so the voltage drop through it in the new circuit should be 110 V. i.e. 

IR =110 

→(220/Z)*220 =110 

→Z =220²/110 

→√{220²+(100πL)²}=220²/110 

→(100πL)² =(220²*220²/110²)-220²

                =220²(4 -1)

                =220²*3

→100πL =220√3 

→L=220√3/100π =1.2 H   

 




 

     14.  In a series LCR circuit with an AC source, R =300 Ω, C =20 µF, L =1.0 Henry, Ɛᵣₘₛ =50 V and 𝝂 =50/π Hz. Find (a) the rms current in the circuit and (b) the rms potential difference across the capacitor, the resistor, and the inductor. Note that the sum of the rms potential difference across the three elements is greater than the rms voltage of the source. 


ANSWER: R =300 Ω, C =20x10⁻⁶ F, 

L =1.0 H, Ɛᵣₘₛ =50 V, 𝝂 =50/π. 

Capacitive reactance, 

Xc =1/ωC 

  =1/(2π𝝂C) 

  =1/{2π*(50/π)*20x10⁻⁶} 

  =500 Ω 

Inductive reactance,

Xi =ωL 

   =2π*(50/π)*1.0 Ω

   =100 Ω.      

Impedance of the circuit,

Z =√{R²+(Xc-Xi)²}

  =√{300²+(500 -100)²} Ω

  =√{300²+400²} Ω

  =500 Ω 

Given Ɛᵣₘₛ =50 V

(i) rms current =Ɛᵣₘₛ/Z

             =50/500 A

             =0.10 A 


(ii) rms potential difference across capacitor

=iᵣₘₛ*Xc 

=0.10*500 V

=50 V.

rms potential difference across the resistor =iᵣₘₛR

    =0.10*300 V

   =30 V.

rms potential difference across the inductor =iᵣₘₛXi

     =0.10*100 V

    =10 V.

Sum of the rms potential differences across the three elements,

=50 V +30 V +10 V

=90 V

It is greater than the rms voltage of the source which is 50 V. 

    




 

     15.  Consider the situation of the previous problem. Find the average electric field energy stored in the capacitor and the average magnetic field energy stored in the coil.  


ANSWER: Average electric field energy stored in a capacitor, 

=½CV² 

=½*20x10⁻⁶*50² J 

=0.025 J

=25 mJ. 

Average magnetic field energy stored in the coil, 

=½Liᵣₘₛ² 

=½*1.0*(0.10)² J

=0.005 J 

=5 mJ.          




 

     16.  An inductance of 2.0 H. a capacitance of 18 µF and a resistance of 10 kΩ are connected to an AC source of 20 V with adjustable frequency. (a) What frequency should be chosen to maximize the current in the circuit? (b) What is the value of this maximum current? 


ANSWER: (a) The impedance of the circuit, 

Z =√{R²+(Xc -Xi)²},

Since the current in the circuit, 

i =E/Z, 

To maximize the current we need to keep Z at a minimum. Minimum Z will be when reactance is zero i.e. Xc =Xi. 

→1/ωC =ωL

→ω² =1/LC

→(2πf)² =1/(2.0*18x10⁻⁶)

→2πf =√(2.78x10⁴)

→f =167/2π

   ≈27 Hz.


(b) Since the maximum current reactance of the circuit is zero and it has only resistance in effect. Hence the maximum current,

Iₘ =E/R

  =20/(1.0x10⁴) A

  =2.0x10⁻³ A

  =2.0 mA.

        




 

     17.  An inductor coil, a capacitor, and an AC source of rms voltage 24 V are connected in series. When the frequency of the source is varied, a maximum rms current of 6.0 A is observed. If the inductor coil is connected to a battery of emf 12 V and internal resistance of 4.0 Ω, what will be current?   


ANSWER: Suppose that the inductor coil has resistance R. When the frequency of the source is varied to get maximum current, the reactance of the circuit is zero. Only resistance R is effective. 

R =Eᵣₘₛ/Iᵣₘₛ

   =24/6 =4 Ω 

When this coil is connected to a battery of V =12 volts and internal resistance r =4 Ω, the total resistance of the circuit, 

=R+r =4 +4 =8 Ω. 

Hence current =12/8 A =1.5 A.   

     




 

     18.  Figure (39-E1) shows a typical circuit for a low-pass filter. An AC input Vi =10 mV is applied at the left end and the output Vₒ is received at the right end. Find the output voltage for 𝝂 =10 kHz, 100 kHz, 1.0 MHz, and 10.0 MHz. Note that as the frequency is increased the output decreases and hence the name low-pass filter. 
Figure for Q-18

 

ANSWER: R =1000 Ω. C =10x10⁻⁹ F. 

Reactance of the circuit X =1/ωC. 

→X =1.0x10⁸/ω. 

Impedance of the circuit, 

Z =√(R²+X²) 

   =√(1000²+10¹⁶/ω²) 

   =(10³/ω)√(ω²+10¹⁰) 

Hence current in the circuit, 

i =Vᵢ/Z 

   =10x10⁻³/{(10³/ω)√(ω²+10¹⁰)} 

   =1.0x10⁻⁵ω/√(ω²+10¹⁰) 

Potential difference across the capacitor,

Vₒ =i*X

={1.0x10⁻⁵ω/√(ω²+10¹⁰)}*1x10⁸/⍵ 

=1000/√(ω²+10¹⁰)

=1000/√(4π²𝝂²+10¹⁰)


For 𝝂 =10 kHz =10x10³ Hz

Vₒ =1000/√(39.48*10⁸+10¹⁰)

 =0.0085 volts

 =8.5 mV.


For 𝝂 =100 kHz =1.0x10⁵ Hz

Vₒ =1000/√(39.48x10¹⁰+10¹⁰)

  =0.0016 volts

  =1.6 mV.


For 𝝂 =1.0 MHz =1.0x10⁶ Hz

Vₒ =1000/√(39.48x10¹²+10¹⁰) volts

  =0.00016 V

  =0.16 mV.


For 𝝂 =10.0 MHz =1.0x10⁷ Hz

Vₒ =1000/√(39.48x10¹⁴+10¹⁰) volts

  =1.6x10⁻⁵ V 

 =16 µV.      




 

     19.  A transformer has 50 turns in the primary and 100 in the secondary. If the primary is connected to a 220 V DC supply, what will be the voltage across the secondary?  


ANSWER: A transformer can not work when connected to a DC supply, because the DC supply has a constant emf and a constant steady-state current in the primary coil. There will be no change of flux in the primary coil that is necessary to induce an emf in the secondary coil. Thus the voltage across the secondary coil will be zero.  

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Links to the Chapters






CHAPTER- 34- Magnetic Field

CHAPTER- 29- Electric Field and Potential











CHAPTER- 28- Heat Transfer

OBJECTIVE -I







EXERCISES - Q51 to Q55


CHAPTER- 27-Specific Heat Capacities of Gases

CHAPTER- 26-Laws of Thermodynamics


CHAPTER- 25-CALORIMETRY

Questions for Short Answer

OBJECTIVE-I

OBJECTIVE-II


EXERCISES - Q-11 to Q-18


CHAPTER- 24-Kinetic Theory of Gases







CHAPTER- 23 - Heat and Temperature






CHAPTER- 17 - Light Waves




CHAPTER- 14 - Fluid Mechanics



CHAPTER- 13 - Fluid Mechanics


CHAPTER- 12 - Simple Harmonic Motion








CHAPTER- 11 - Gravitation




CHAPTER- 10 - Rotational Mechanics




CHAPTER- 9 - Center of Mass, Linear Momentum, Collision


CHAPTER- 8 - Work and Energy

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Click here for → Exercises (1-10)

Click here for → Exercises (11-20)

Click here for → Exercises (21-30)

Click here for → Exercises (31-42)

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CHAPTER- 7 - Circular Motion

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CHAPTER- 6 - Friction

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For more practice on problems on friction solve these- "New Questions on Friction".

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CHAPTER- 5 - Newton's Laws of Motion


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Click here for→Newton's Laws of Motion,Exercises(Q.No. 13 to 27)

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CHAPTER- 4 - The Forces

The Forces-

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CHAPTER- 3 - Kinematics - Rest and Motion

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CHAPTER- 2 - "Physics and Mathematics"

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