Monday, May 3, 2021

H C Verma solutions, ELECTRIC CURRENT IN CONDUCTORS, Exercises, Q11-Q20, Chapter-32, Concepts of Physics, Part-II

Electric Current in Conductors


Exercises, Q11 - Q20


   11. A copper wire of radius 0.1 mm and resistance 1 kΩ is connected across a power supply of 20 V. (a) How many electrons are transferred per second between the supply and the wire at one end? (b) Write down the current density in the wire.    



ANSWER: (a) Current in the wire,

i = V/R =20/1000 A

→i =0.02 A =0.02 C/s

The number of electrons transferred per second between the supply and the wire at one end =0.02/(1.602x10⁻¹⁹)

  = 1.25x10¹⁷. 


(b) Cross-section of the wire A =πr²

=π*(1x10⁻⁴)² m²

=3.14x10⁻⁸ m² 

The current density in the wire,

j =i/A =0·02/(3.14x10⁻⁸) A/m²

  =6.37x10⁵ A/m².       





 

   12. Calculate the electric field in a copper wire of cross-sectional area 2.0 mm² carrying a current of 1 A. The resistivity of copper = 1.7x10⁻⁸ Ω-m.  



ANSWER: Area of cross-section, A =2.0 mm² =2.0x10⁻⁶ m².

Resistance of the wire R =ρl/A

Current in the wire i = 1.0 A

The potential difference between the ends of the wire = V = iR

The electric field in the wire E =V/l

→E =iR/l

     =iρl/Al

     =iρ/A

     =1.0*1.7x10⁻⁸/(2.0x10⁻⁶) V/m

    =0.85x10⁻² V/m

    =8.5x10⁻³ V/m

    =8.5 mV/m. 




 

   13. A wire has a length of 2·0 m and a resistance of 5·0 Ω. Find the electric field existing inside the wire if it carries a current of 10 A.    



ANSWER:  The current, i = 10 A.

Resistance R =5·0 Ω. Length l =2·0 m.

From Ohm's law, potential difference,

V =iR =10*5·0 volts =50 V,

The electric field in the wire, E =V/l

→E = 50/2 V/m

      = 25 V/m. 

     




 

   14. The resistance of an iron wire and a copper wire at 20°C are 3·9 Ω and 4·1 Ω respectively. At what temperature will the resistances be equal? The temperature coefficient of resistivity for iron 5·0x10⁻³ K⁻¹ and for copper it is 4·0x10⁻3 K⁻¹. Neglect any thermal expansion.  



ANSWER: At 20°C resistance of iron, R =3·9 Ω. 

Resistance of copper wire R' =4·1 Ω.    

Let after a temperature difference of ΔT, the resistances of both wires become equal. 

   At that temperature the resistance of the iron wire, Ř =R(1+α.ΔT)

and the resistance of the copper wire, Ř' =R'(1+α'.ΔT)

Since Ř =Ř', so

R(1+α.ΔT) =R'(1+α'.ΔT)

→3·9(1+0·005*ΔT) =4·1(1+0·004*ΔT)

→3·9 +0.0195*ΔT =4·1 +0·0164*ΔT

→(0·0195-0·0164)ΔT =4·1-3·9

→ΔT =0·2/0·0031 °C

        =64·5°C

      Hence the required temperature 

=20°C +ΔT

=20°C +64·5°C

=84·5°C.


 



 

   15. The current in a conductor and the potential difference across its ends are measured by an ammeter and a voltmeter. The meters draw negligible currents. The ammeter is accurate but the voltmeter has a zero error (that is, it does not read zero when no potential difference is applied). Calculate the zero error if the readings for two different conditions are 1·75 A, 14·4 V and 2·75 A, 22·4 V.  



ANSWER: Suppose the reading in the voltmeter is x when no potential difference is applied across the ends of the conductor. Suppose the resistance of the conductor = R.

    In the first case, R = P.D./Current

→R =(14·4-x)/1·75 Ω

{with correction for zero error}

Since the resistance of the wire is constant, in the second case,

R =(22·4-x)/2·75 Ω

Equating both expressions, 

(14·4-x)/1·75 =(22·4-x)/2·75 

→11(14·4-x) =7(22·4-x)

→158·4 -11x =156·8 -7x

→4x =1·6

→x =1·6/4 =0·4 V. 






 

   16. Figure (32-E2) shows an arrangement to measure the emf Ԑ and internal resistance r of a battery. The voltmeter has a very high resistance and the ammeter also has some resistance. The voltmeter reads 1·52 V when the switch S is open. When the switch is closed the voltmeter reading drops to 1·45 V and the ammeter reads 1·0 A. Find the emf and internal resistance of the battery.     
Figure for Q-16



ANSWER: (a) Since the voltmeter has a very high resistance, nearly zero current passes through the voltmeter when the switch S is open. The circuit through the ammeter has no current. So the voltage drop due to the internal resistance of the battery is zero. Thus the voltmeter reads the emf of the battery. Ԑ =1·52 V.      

(b) When S is closed, current flows through the lower loop with ammeter. i =1·0 A. P.D. across the battery with internal resistance =1·45 V. Hence,

Ԑ -ir =1·45

→1.52 -1.0*r = 1.45

→r =1.52-1.45 =0.07 Ω   



 

   17. The potential difference between the terminals of a battery of emf 6·0 V and internal resistance 1 Ω drops to 5·8 V when connected across an external resistor. Find the resistance of the external resistor.   



ANSWER: Given that, E =6·0 V. Internal resistance, r = 1 Ω. Let external resistor =R. With connection the P.D. =5·8 V.     

The current in the circuit,

i =E/(R+r) =6/(R+1)

Now, E -ir =V 

→6 -6/(R+1) =5·8

→0.2 =6/(R+1)

→R+1 =30

→R = 29 Ω.      




 

   18. The potential difference between the terminals of a 6·0 V battery is 7·2 V when it is being charged by a current of 2·0 A. What is the internal resistance of the battery?   



ANSWER:  The battery is being charged. emf of the battery, E =6·0 V. The current through it, i = 2·0 A.

  The potential difference across the battery will be equal to the emf of the battery plus voltage drop across the internal resistance of the battery.

   V =E +ir

→7·2 =6 +2r

→2r =1·2

→r =0.6 Ω.       




 

   19. The internal resistance of an accumulator battery of emf 6 V is 10 Ω when it is fully discharged. As the battery gets charged up, its internal resistance decreases to 1 Ω. The battery in its completely discharged state is connected to a charger which maintains a constant potential difference of 9 V. Find the current through the battery (a) just after the connections are made and (b) after a long time when it is completely charged.  



ANSWER: (a) The applied potential difference across the battery = 9 V.

emf of the battery = 6 V.

Net emf across the internal resistance of the battery = 9 - 6 = 3 V.

Just at the beginning, internal resistance, r = 10 Ω.

So the current through the battery,

i =3/10 =0·3 A.        


(b) When the battery is fully charged, r = 1 Ω.

The current through the battery,

i =3/1 = 3 Ω.  





 

   20. Find the value of i₁/i₂ in figure (32-E3) if (a) R =0·1 Ω. (b) R =1 Ω (c) R =10 Ω. Note from your answers that in order to get more current from a combination of two batteries they should be joined in parallel if the external resistance is small and in series if the external resistance is large as compared to the internal resistances.  
The figure for Q-20



ANSWER:   Applying Kirchhoff's loop law in the first circuit,

i₁R +i₁*1-6 +i₁*1 -6 =0

→i₁(R+2) =12

→i₁ =12/(R+2)

The diagram for Q-20


In the second circuit, loop ABCD

i₂R+i'*1-6 =0

→i₂R+i' =6 ----- (i)

Loop CDEF

i'*1-6+6-(i₂-i')*1 =0

→i'-i₂+i' =0

→2i' =i₂

→i' =i₂/2

Putting in (i)

i₂R+i₂/2 =6  

→i₂(R+½)=6

→i₂ =6/(R+½)

Hence

i₁/i₂={12/(R+2)}/{6/(R+½)}

=2(R+½)/(R+2)

=(2R+1)/(R+2) 


(a) R = 0·1 Ω

i₁/i₂=(0·2+1)/(0·1+2)

      =1·2/2·1

      =0·57.


(b) R =1 Ω

i₁/i₂ =(2*1+1)/(1+2)

       =3/3

       = 1.


(c) R = 10 Ω

i₁/i₂ =(2*10+1)/(10+2)

       =21/12

       =1·75.  


From the results, it is clear that when R is small in comparison to r, as in the case (a), more current is drawn in the second circuit where batteries are in parallel combination.

But when R is large in comparison to r, as in (c), more current is drawn in the first circuit where the batteries are in series combination.  


 

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Links to the Chapters



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


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CHAPTER- 24-Kinetic Theory of Gases







CHAPTER- 23 - Heat and Temperature






CHAPTER- 17 - Light Waves




CHAPTER- 14 - Fluid Mechanics



CHAPTER- 13 - Fluid Mechanics


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CHAPTER- 11 - Gravitation




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