Wednesday, May 18, 2022

H C Verma solutions, ALTERNATING CURRENT, Chapter-39, Questions for Short Answer, Concepts of Physics, Part-II

Alternating Current


Questions for Short Answer


     1.  What is the reactance of a capacitor connected to a DC source?  


ANSWER: The reactance of a capacitor is given as, 

Xₖ = 1/⍵C

For a DC source, ⍵ = 0. 

Hence, Xₖ = 1/0 =∞.

So the reactance of a capacitor connected to a DC source is infinite.


 





     2.  The voltage and current in a series AC circuit are given by

V =Vₒcos ⍵t and i = iₒsin ⍵t

What is the power dissipated in the circuit? 


ANSWER: The average power delivered by AC source is,  

P = iᵣₘₛ Vᵣₘₛ cos φ

where iᵣₘₛ =root mean square value of the current,

 Vᵣₘₛ = root mean square value of the voltage,

and φ = phase difference between the current and the voltage. 

  In the given AC circuit, 

i = iₒ sin ωt 

V = Vₒ cos ωt =Vₒ sin(ωt +π/2) 

Hence the phase difference between the current and the voltage is φ =π/2. 

 Thus the power, 

P =iᵣₘₛ Vᵣₘₛ cos (π/2)

   =iᵣₘₛ Vᵣₘₛ *0

   =zero. 

So no power is dissipated in the circuit.  


     




     3.  Two alternating currents are given by

i₁ = iₒsin ⍵t and i₂ = iₒsin {⍵t+π/3}.

Will the rms values of the currents be equal or different?    


ANSWER: The rms value of a current is, 

iᵣₘₛ = iₒ/√2

So the rms value of a current only depends upon the peak value of the current iₒ. The given currents have the same peak value with a phase difference of π/3. Hence the rms values of the currents will be equal.


    




     4.  Can the peak voltage across the inductor be greater than the peak voltage of the source in an LCR circuit?  


ANSWER: The magnitude of the voltage across an inductor is, 

V =L*di/dt 

   =L*d{iₒ sin(ωt+φ)}/dt 

   =Liₒω cos(ωt+φ)

   =Vₒ cos(ωt+φ) 

So the peak voltage Vₒ =Lωiₒ

In an LCR circuit, impedance 

Z =√{R²+(1/ωC -ωL)²} 

At the resonant frequency, 1/ωC =ωL, and Z=R. So iₒ =Eₒ/R.

So the peak voltage across the inductor

Vₒ =Lωiₒ

   =LωEₒ/R

   =(Lω/R)Eₒ

Thus in a case where Lω >R in an LCR circuit, the factor Lω/R >1. In such a situation, Vₒ > Eₒ. So the peak voltage across an inductor may be greater than the peak voltage of the source in an LCR circuit.

  

         




     5.  In a circuit containing a capacitor and an AC source, the current is zero at the instant the source voltage is maximum. Is it consistent with Ohm's law?   


ANSWER: No, it is not consistent with Ohm's law because Ohm's law is valid only for a resistive circuit. It is not valid if there is a reactive element (reactance, capacitive, or inductive) in the circuit.    




     6.  An AC source is connected to a capacitor. Will the rms current increase, decrease, or remain constant if a dielectric slab is inserted into the capacitor?  


ANSWER: Reactance of a capacitor, 

X = 1/ωCₒ. 

If a dielectric slab is inserted into the capacitor, the capacitance is given by, 

C =KCₒ

where K =dielectric constant of the inserted slab,

 So reactance of the capacitor is now,

X =1/ωKCₒ, where Cₒ is the capacitance without a dielectric slab. Since the dielectric constant K >1, reactance X will decrease with the insertion of the dielectric slab.

Now the rms current,

iᵣₘₛ =iₒ/√2

    =Eₒ/(X√2)

With the decrease in reactance X, iᵣₘₛ will increase.

 




     7.  When the frequency of the AC source in an LCR circuit equals the resonant frequency, the reactance of the circuit is zero. Does it mean that there is no current through the inductor or the capacitor?  


ANSWER: At the resonant frequency in an LCR circuit, the reactance of the circuit is zero but there is resistance in the circuit.     

So the current in the circuit is 

i =(Eₒ/R) sin ωt. 

Thus there is a current through the inductor or capacitor. The only thing is that the potential difference across the combined inductor and capacitor will be zero because the current leads the emf by π/2 in the capacitor but it lags by π/2 in the inductor.



  



     8.  When an AC source is connected to a capacitor there is a steady-state current in the circuit. Does it mean that the charges jump from one plate to the other to complete the circuit?  


ANSWER: When an AC source is connected to a capacitor there is a potential difference across the plates. The charge stored on the plates is proportional to the potential difference. Since the potential difference across the plates varies due to the AC source, the accumulated charge on the plates also varies. So in the steady state of current, charges do not jump from one plate to another but either accumulate to one plate or discharge from the other plate.  


   




     9.  A current i₁ =iₒsin ⍵t passes through a resistor of resistance R. How much thermal energy is produced in one time period? A current i₂ = -iₒsin⍵t passes through the resistor. How much thermal energy is produced in one time period? If i₁ and i₂ both pass through the resistor simultaneously, how much thermal energy is produced? Is the principle of superposition obeyed in this case?  


ANSWER: The thermal energy produced in one time period is,  

H =iᵣₘₛ²RT

   =(iₒ/√2)²R(2π/⍵)

   =πiₒ²R/⍵

For the current i₂ = -iₒsin ⍵t, thermal energy produced in one time period,

H' =iᵣₘₛ²RT

    =(-iₒ/√2)²R*(2π/⍵)

    =πiₒ²R/⍵

So H = H'. It means, that in both conditions, the thermal energy produced is the same.

When i₁ and i₂ both pass through the resistor simultaneously, the net current in the resistor is zero. Thus the thermal energy produced in the resistor is zero. So the principle of superposition is obeyed in determining the net current in the resistor.



 




     10.  Is energy produced when a transformer steps up the voltage?  


ANSWER: No. When a transformer steps up the voltage, the resulting current decreases. Thus following the principle of conservation of energy, no energy is produced when stepping up the voltage.  


  




     11.  A transformer is designed to convert an AC voltage of 220 V to an AC voltage of 12 V. If the input terminals are connected to a DC voltage of 220 V, the transformer usually burns. Explain.  


ANSWER: The coils of the transformers have inductance only and their resistance is very low. When it is connected to a DC source, initially the induced emf resists the current from increasing but it tries to reach a steady state. The steady-state current is given as,

i = E/R =220/R

Since R is very low here, the value of i will be a large one. Since the heat produced is proportional to the square of i, it is sufficient to burn the coil of the transformer. 



  




     12.  Can you have an AC series circuit in which there is a phase difference of 180° between the emf and the current? 120°?  


ANSWER: In a resistive circuit the current is in phase with the emf. For a capacitive circuit, the current leads the emf by π/2, and for an inductive circuit, the current lags the emf by π/2. For an LCR circuit the phase difference φ is in a range π/2 >φ > -π/2 because

tan φ =(1/⍵C -⍵L)/R. 

  So φ will never be 180° or 120°. 



      




     13.  A resistance is connected to an AC source. If a capacitor is included in the series circuit, will the average power absorbed by the resistance increase or decrease? If an inductor of small inductance is also included in the series circuit, will the average power absorbed increase or decrease further?  


ANSWER:  When a capacitor is included in the series circuit, the impedance of the circuit, 

Z =√(R²+Xc²). Where Xc is the reactance of the capacitor. 

   Since the impedance increases, rms current in the circuit will decrease. The average power absorbed by a resistor is given by,

P =iᵣₘₛ²RT.

  Hence the power absorbed by the resistor will decrease.

    Now if an inductor of small inductance is also included in the series circuit, the impedance now is,

Z =√{R²+(Xc -XL)²}

    Thus with the further inclusion of the inductor in the circuit, impedance reduces a little. So the rms current increases. The average power absorbed by the resistor will increase further.



 



     14.  Can a hot-wire ammeter be used to measure a direct current having a constant value? Do we have to change the graduations?  


ANSWER: A hot-wire ammeter measures the rms value of an AC current. Since rms current of AC circuit is equivalent to direct current i where both produce the same amount of joule heating in the same time interval. So the hot wire ammeter can be used to measure a direct current having a constant value.

  We need not to change the graduations. 

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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

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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CHAPTER- 6 - Friction

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Click here for → Friction - OBJECTIVE-II

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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)

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CHAPTER- 4 - The Forces

The Forces-

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Click here for "The Forces" - OBJECTIVE-I


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Click here for "The Forces" - Exercises


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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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