Monday, May 23, 2022

H C Verma solutions, ALTERNATING CURRENT, Chapter-39, OBJECTIVE-I, Concepts of Physics, Part-II

Alternating Current


Objective-I


     1.  A capacitor acts as an infinite resistance for 

(a) DC   

(b) AC 

(c) DC as well as AC 

(d) neither AC nor DC    


ANSWER: (a).  

 

EXPLANATION: For a DC, angular frequency ⍵ =0. Hence, the reactance of the capacitor,

X =1/⍵C

   =∞.

Hence the capacitor acts as an infinite resistance for DC. 





     2.  An AC source producing emf 

Ɛ =Ɛₒ[cos(100π s⁻¹)t +cos(500π s⁻¹)t]

is connected in series with a capacitor and a resistor. The steady-state current in the circuit is found to be 

i =i₁ cos[(100π s⁻¹)t+φ₁]+i₂cos[(500π s⁻¹)t+φ₂]

(a) i₁ > i₂

(b) i₁ = i₂

(c) i₁ < i₂

(d) the information is insufficient to find the relation between i₁ and i₂. 


ANSWER: (c).   


EXPLANATION: In a steady state, the charge on a capacitor is, 

Q =CƐ

  =CƐₒ[cos(100π s⁻¹)t+cos(500π s⁻¹)t]

Hence the current,

i =dQ/dt

   =-100πCƐₒsin(100π s⁻¹)t-500πCƐₒsin(500π s⁻¹)t]

  =100πCƐₒcos{(100π s⁻¹)t+π/2}+500πCƐₒcos{(500π s⁻¹)t+π/2}

Comparing with the given current,

i₁ =100πCƐₒ

i₂ =500πCƐₒ

Hence i₁ < i₂. 

So option (c) is correct.

 




     3.  The peak voltage in a 220 V AC source is

(a) 220 V

(b) about 160 V 

(c) about 310 V 

(d) 440 V.     


ANSWER: (c).  

 

EXPLANATION: vᵣₘₛ =220 V.

We know that rms voltage vᵣₘₛ =vₒ/√2. 

Hence peak voltage, 

vₒ =√2*vᵣₘₛ   

   =√2*220 ≈310 V. 

Hence option (c) is correct. 




     4.  An AC source is rated 220 V, 50 Hz. The average voltage is calculated in a time interval of 0.01 s. It 

(a) must be zero 

(b) may be zero 

(c) is never zero

(d) is (220√2) V  


ANSWER: (b).  

 

EXPLANATION: The frequency is given,

f = 50 Hz. 

Time period, T =1/f =1/50 s =0.02 s.

So to complete one cycle time taken is 0.02 s.    
Diagram for Q-4
We have been given a time interval of 0.01 s which is half-cycle time. If this time interval starts from O, then we have time from O to C (above diagram). Clearly, in this interval, the source emf is positive everywhere. Thus the average voltage in this interval will sure be positive (non-zero). So option (a) is not correct. 

   Now consider that the time starts a quarter cycle after O, at A. Then the time interval ends at B. The source emf in this interval decreases from Vₒ at A to zero at C in a time interval of 0.005 s. Now the direction of emf reverses and increases from zero at C to Vₒ at B in the next 0.005 s. If we calculate the average voltage in 0.01 s from A to B, it will be zero. So option (c) is also not correct. 

   Since the average voltage in 0.01 s depends upon the start of the time, it has no fixed value. Option (d) is also not correct. 

   The average voltage may be zero is the only correct option, i.e. (b).    




     5.  The magnetic field energy in an inductor changes from maximum value to minimum value in 5.0 ms when connected to an AC source. The frequency of the source is 

(a) 20 Hz 

(b) 50 Hz

(c) 200 Hz

(d) 500 Hz.   


ANSWER: (b).  

 

EXPLANATION: The magnetic field energy in an inductor is, 

E =½Li². 

So it will be maximum when 'i' has a peak value and minimum when i =0. We can see in the diagram of Problem-4 (assuming it drawn for current) that it takes a quarter cycle time for the current to change from maximum to minimum i.e. from A to C. 

  Thus in this case T/4 =5.0 ms 

→T = 20 ms =20/1000 s 

→T =1/50 s. 

Thus the frequency of the source, 

f = 1/T 

  =1/(1/50) s⁻¹ 

  =50 Hz. 

Option (b) is correct.         




     6.  Which of the following plots may represent the reactance of a series LC combination? 
Figure for Q-6


ANSWER: (d).  

 

EXPLANATION: The reactance of an LC circuit is, 

X =⍵L -1/⍵C

→X =2πfL -1/2πfC  ----(i)

If 2πfL =1/2πfC, then X =0

i.e. for f =1/{2π√(LC)}, X =0.

Also from (i) it is clear that the plot between X and f is not a straight line. Thus the given plots 'a' and 'b' are not correct. The non-straight line plot 'c' has never a zero value, thus it is also not correct. Only plot 'd' fulfills the conditions, hence option (d) is correct.

  




     7.  A series AC circuit has a resistance of 4 Ω and a reactance of 3 Ω. The impedance of the circuit is 

(a) 5 Ω 

(b) 7 Ω 

(c) 12/7 Ω 

(d) 7/12 Ω.      


ANSWER: (a).  

 

EXPLANATION: Given, R =4 Ω, X =3 Ω

Hence the impedance of the circuit 

Z =√(R²+X²)  

   =√(4²+3²) Ω 

   =√25 Ω 

   =5 Ω. 

Option (a) is correct.  


     



     8.  Transformers are used  

(a) in DC circuits only 

(b) in AC circuits only 

(c) in both DC and AC circuits

(d) neither in DC nor in AC circuits.     


ANSWER: (b).  

 

EXPLANATION: The transformer works on the principle that due to a changing emf in the primary coil, a changing magnetic flux is generated. This changing magnetic flux induces emf in the secondary coil. The emf in the secondary coil depends upon the ratio of the number of turns of the secondary coil to the primary coil. 

   Since the emf changes only in AC circuits, the transformers can be used only in AC circuits. 

  Option (b) is correct.   


  



     9.  An alternating current is given by 

i =i₁ cos ⍵t +i₂ sin ⍵t.

The rms current is given by 

(a) (i₁+i₂)/√2

(b) |i₁+i₂|/√2

(c) √{(i₁²+i₂²)/2}

(d) √{(i₁²+i₂²)/√2}

 


ANSWER: (c).  

 

EXPLANATION: The rms value of current is given as, 

iᵣₘₛ =√(∫i²dt/∫dt)

The limit of integration will be from 0 to t=T=2π/ω where T is the time period.

Now i² =(i₁cos ωt +i₂sin ωt)²

 =i₁²cos²ωt+i₂²sin²ωt+2i₁i₂cosωt.sinωt

 =½i₁²(cos2ωt+1)+½i₂²(1-cos2ωt)+i₁i₂sin2ωt

 =(i₁²+i₂²)/2+½(i₁²-i₂²)cos2ωt+i₁i₂sin2ωt

So,

∫i²dt=(i₁²+i₂²)t/2+¼(i₁²-i₂²)sin2ωt/ω-i₁i₂cos2ωt/2ω

Putting the limits for t, we get,

∫i²dt =(i₁²+i₂²)π/ω+0-i₁i₂/2ω-0-0+i₁i₂/2ω

   =(i₁²+i₂²)π/ω

And, ∫dt =t

Putting limits,

∫dt =T =2π/ω

Hence iᵣₘₛ =√[{(i₁²+i₂²)π/ω}/(2π/ω)]

→iᵣₘₛ =√{(i₁²+i₂²)/2}

Option (c) is correct.   




      10.  An alternating current having a peak value of 14 A is used to heat a metal wire. To produce the same heating effect, a constant current 'i' can be used where i is  

(a) 14 A 

(b) about 20 A

(c) 7 A

(d) about 10 A.       


ANSWER: (d).  

 

EXPLANATION: The constant current that produces the same heating effect that of an AC is called the rms current. 

iᵣₘₛ =iₒ/√2

   =14/√2 A

   ≈10 A.

Option (d) is correct. 




     11.  A constant current of 2.8 A exists in a resistor. The rms current is 

(a) 2.8 A

(b) about 2 A 

(c) 1.4 A

(d) undefined for a direct current.   


ANSWER: (a).  

 

EXPLANATION: Since rms current is given as 

iᵣₘₛ =√(∫i²dt/∫dt) 

Since i is constant here,   

iᵣₘₛ =√(i²∫dt/∫dt) 

    =√i² 

    =i =2.8 A. 

Hence option (a) is correct.   


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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

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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

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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

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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"

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