Saturday, May 28, 2022

H C Verma solutions, ALTERNATING CURRENT, Chapter-39, EXERCISES, Q1 to Q10, Concepts of Physics, Part-II

Alternating Current


EXERCISES, Q1 to Q10


     1.  Find the time required for a 50 Hz alternating current to change its value from zero to the rms value.  


ANSWER: Frequency of the alternating current, f =50 Hz. For an alternating current, the instantaneous current is given as, 

i = iₒ sin ωt 

And rms current, 

iᵣₘₛ = iₒ/√2 

At t = 0, i =0. 

If at t =t', i =iᵣₘₛ, then

iᵣₘₛ =iₒ sin ωt'

→iₒ/√2 =iₒ sin ωt'

→sin ωt' =1/√2 =sin π/4

→ωt' = π/4

→2πft' =π/4

→t' =1/(8f)

      =1/(8*50) s

      =0.0025 s

      =2.5 ms. 





 

     2.  The household supply of electricity is at 220 V (rms value) and 50 Hz. Find the peak voltage and the least possible time in which the voltage can change from the rms value to zero.  


ANSWER: Vᵣₘₛ =220 V, f =50 Hz.

Since Vᵣₘₛ =Vₒ/√2

→Peak voltage, Vₒ =√2iᵣₘₛ

→Vₒ =√2*220 volts

      =311 V.

In a cycle when E begins with zero, it reaches rms value and then to peak value. Now E begins to decrease, reaches rms value, and then to zero. So the least time to reach from Eᵣₘₛ to zero is in this decreasing phase. This time is the same when E changes from zero to Eᵣₘₛ, so we calculate this time. 

Eᵣₘₛ =Eₒ sin ωt

→Eₒ/√2 =Eₒ sin ωt

→sin ωt =1/√2 =sin π/4

→ωt =π/4

→2πft =π/4

→t =1/(8f)

     =1/400 s

     =0.0025 s

     =2.5 ms.

 




 

     3.  A bulb rated 60 W at 220 V is connected across a household supply of alternating voltage of 220 V. Calculate the maximum instantaneous current through the filament.  


ANSWER: From the bulb's rating, it is clear that the bulb consumes a power of 60 W at rms current of 220 V. Let the resistance of the bulb =R. We know,  

Power, P = V²/R, hence

60 =(Eᵣₘₛ)²/R =220²/R

→R =220²/60 Ω

     =807 Ω

Maximum instantaneous current will be at a time when the voltage is maximum, i.e. at peak voltage. So,

iₒ =Eₒ/R

   =√2*Eᵣₘₛ/R

   =√2*220/807 A

   =0.39 A.

       




 

     4.  An electric bulb is designed to operate at 12 volts of DC. If this bulb is connected to an AC source and gives normal brightness, what would be the peak voltage of the source?  


ANSWER: Let a constant current i pass through the bulb when connected to the DC circuit. If it gives the same brightness when connected to an AC circuit, it means that the rms value of this current is equal to the constant current i because both produce the same amount of Joule heating in a given time period. So for the AC circuit, 

iᵣₘₛ = i

→iₒ/√2 =i 

→iₒR/√2 =iR,

{where R =bulb's resistance,iₒ =peak current and Eₒ =peak voltage of the source}

→Eₒ/√2 =E, 

→Eₒ =√2E 

      =√2*12 V 

      =17 V.     

  




 

     5.  The peak power consumed by a resistive coil when connected to an AC source is 80 W. Find the energy consumed by the coil in 100 seconds which is many times larger than the time period of the source.  


ANSWER: Let the resistance of the coil =R. Hence the peak power consumed by the coil,

Pₒ =iₒ²R

→iₒ² =Pₒ/R 

→(√2iᵣₘₛ )² =80/R

→iᵣₘₛ² =40/R

Energy consumed by the resistive coil in a time period t is given as,

=iᵣₘₛ²Rt

=(40/R)*R*100 J

=4000 J

=4.0 kJ.

     




 

     6.  The dielectric strength of air is 3.0x10⁶ V/m. A parallel plate air capacitor has an area of 20 cm² and plate separation of 0.10 mm. Find the maximum rms voltage of an AC source that can be safely connected to this capacitor.   


ANSWER:  Given for the capacitor, 

Plate area, A =20 cm² =0.002 m², 

Separation, d =0.10 mm

                   =1.0x10⁻⁴ m. 

Dielectric strength of air,

 E' =3.0x10⁶ V/m.

It means that if the electric field inside the capacitor is equal to the dielectric strength of the air, it is at the limit of safe working. Let at this electric field the potential difference across the plate is V, then

E' =V/d

→V =E'd

  =3.0x10⁶*1.0x10⁻⁴ volts

  =300 V.

So this should be the maximum potential difference across the capacitor i.e. equal to the peak voltage Eₒ of the AC source. Since we need to know the rms voltage corresponding to this peak voltage, we have,

Eᵣₘₛ =Eₒ/√2. 

     =300/√2 volts

     ≈210 V.

      




 

     7.  The current in a discharging LR circuit is given by i =iₒe^-t/𝝉 where 𝝉 is the time constant of the circuit. Calculate the rms current for the period t =0 to t =𝝉.  


ANSWER: rms current is the square root of the mean square current. So mean square current =iᵣₘₛ².

→iᵣₘₛ² =∫i²dt/∫dt

  =∫(iₒe^-t/𝝉)²dt/∫dt

Keeping the limit for t from 0 to 𝝉, we get

  =(iₒ²/𝝉)[-𝝉e^-2t/𝝉/2]

 =(iₒ²/2)[-e^-2t/𝝉] 

 =½iₒ²[-e⁻²+1], putting limits.

 =½iₒ²(e²-1)/e²

Hence iᵣₘₛ =√{½iₒ²(e²-1)/e²}

      =(iₒ/e)√{(e²-1)/2}.

      




 

     8.  A capacitor of capacitance 10 µF is connected to an oscillator giving an output voltage Ɛ =(10 V)sin ⍵t. Find the peak currents in the circuit for ⍵ =10 s⁻¹, 100 s⁻¹, 500 s⁻¹, 1000 s⁻¹.  


ANSWER: Reactance of a capacitor connected to an AC source of emf, 

Ɛ =Ɛₒ sin ωt, is

X =1/ωC 

  Hence the peak current,

 iₒ = Ɛₒ/X

   =Ɛₒ/(1/ωC)

   =CƐₒω.

From the given data, Ɛₒ =10 V.

Hence iₒ =10Cω.

    =10*10x10⁻⁶ω,

   =1x10⁻⁴ω.

For ω =10 s⁻¹,

Peak current =1.0x10⁻⁴*10 A

        =1.0x10⁻³ A.

For ω =100 s⁻¹,

Peak current =1.0x10⁻⁴*100 A

        =0.01 A.

For ω =500 s⁻¹,

Peak current =1.0x10⁻⁴*500 A

        =0.05 A.

For ω =1000 s⁻¹,

Peak current =1.0x10⁻⁴*1000 A

        =0.1 A.


   




 

     9.  A coil of inductance 5.0 mH and negligible resistance is connected to the oscillator of the previous problem. Find the peak currents in the circuit for 

⍵ = 100 s⁻¹, 500 s⁻¹, 1000 s⁻¹.   


ANSWER: Inductance of the coil, 

L =5.0 mH =5.0x10⁻³ H. 

Reactance of the inductive coil, 

Xi =ωL

Hence peak current iₒ =Ɛₒ/ωL

→iₒ =10/(5.0x10⁻³ω)

     =2.0x10³/ω


For ω =100 s⁻¹,

iₒ =2.0x10³/100 A

  =20 A.


For ω =500 s⁻¹,

iₒ =2.0x10³/500 A

  =4.0 A.


For ω =1000 s⁻¹,

iₒ =2.0x10³/1000 A

   =2.0 A.

 




 

     10.  A coil has a resistance of 10 Ω and an inductance of 0.4 henries. It is connected to an AC source of 6.5 V, 30/π Hz. Find the average power consumed in the circuit.  


ANSWER: R =10 Ω, L =0.4 H, f =30/π Hz. 

Reactance of the coil X =ωL

→X =2πfL

   =2π*(30/π)*0.4 Ω

   =24 Ω.

Impedance of the coil, 

Z =√(R²+X²)

   =√(10²+24²)

   =√676 Ω

Average power is given as,

P =EᵣₘₛIᵣₘₛcos φ

Now, Eᵣₘₛ =6.5 V, Iᵣₘₛ =Eᵣₘₛ/Z, and 

cos φ =R/Z. Thus,

P =6.5*(6.5/√676)*10/√676 W 

   =(422.5/676) W 

   =4225/6760 W 

   =325/520 W,  {division by 13} 

   =25/40 W,  {division by 13} 

   =5/8 W.     

 

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Links to the Chapters




CHAPTER- 39- Alternating Current


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

Click here for → Question for Short Answers

Click here for → OBJECTIVE-I

Click here for → OBJECTIVE-II

Click here for → Exercises (1-10)

Click here for → Exercises (11-20)

Click here for → Exercises (21-30)

Click here for → Exercises (31-42)

Click here for → Exercise(43-54)

CHAPTER- 7 - Circular Motion

Click here for → Questions for Short Answer 

Click here for → OBJECTIVE-I

Click here for → OBJECTIVE-II

Click here for → EXERCISES (1-10)

Click here for → EXERCISES (11-20)

Click here for → EXERCISES (21-30)

CHAPTER- 6 - Friction

Click here for → Questions for Short Answer

Click here for → OBJECTIVE-I

Click here for → Friction - OBJECTIVE-II

Click here for → EXERCISES (1-10)

Click here for → Exercises (11-20)

Click here for → EXERCISES (21-31)

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

---------------------------------------------------------------------------------

CHAPTER- 5 - Newton's Laws of Motion


Click here for → QUESTIONS FOR SHORT ANSWER

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

-------------------------------------------------------------------------------

CHAPTER- 4 - The Forces

The Forces-

"Questions for short Answers"    


Click here for "The Forces" - OBJECTIVE-I


Click here for "The Forces" - OBJECTIVE-II


Click here for "The Forces" - Exercises


--------------------------------------------------------------------------------------------------------------

CHAPTER- 3 - Kinematics - Rest and Motion

Click here for "Questions for short Answers"


Click here for "OBJECTIVE-I"


Click here for EXERCISES (Question number 1 to 10)


Click here for EXERCISES (Question number 11 to 20)


Click here for EXERCISES (Question number 21 to 30)


Click here for EXERCISES (Question number 31 to 40)


Click here for EXERCISES (Question number 41 to 52)


CHAPTER- 2 - "Physics and Mathematics"

Click here for "Questions for Short Answers"


Click here for "OBJECTIVE-II"

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