Wednesday, May 19, 2021

H C Verma solutions, ELECTRIC CURRENT IN CONDUCTORS, Exercises, Q21-Q30, Chapter-32, Concepts of Physics, Part-II

Electric Current in Conductors


Exercises, Q21 - Q30


   21. Consider N =n₁n₂ identical cells, each of emf Ɛ and internal r. Suppose n₁ cells are joined in series to form a line and n₂ such lines are connected in parallel. The combination drives a current in an external resistance R. (a) Find the current in the external resistance. (b) Assuming that n₁ and n₂ can be continuously varied, find the relation between n₁, n₂, R and r for which the current in R is maximum. 



ANSWER: In a row, the cells are connected in series, so the emf of a row will be the sum of the emf of all cells in a row. Here emf of a row = n₁Ɛ.

  Since the rows of the cells are joined in parallel, the emf of the combination will be the same as the emf of a row =n₁Ɛ.

  The internal resistance of a cell =r. The internal resistance of one row =n₁r. Now n₂ such resistances are joined in parallel. Hence the total internal resistance of the combination R' is given as,

1/R' =n₂(1/n₁r)

→R' = n₁r/n₂

The total resistance of the circuit =R+R'

=R +n₁r/n₂


(a) Hence the current in the circuit

i =(n₁Ɛ)/(R+n₁r/n₂)


(b) i =(n₁n₂Ɛ)/(n₁r+n₂R)

i will be maximum when n₁r+n₂R is minimum. Now it can be written as,

n₁r+n₂R = {√(n₁r)-√(n₂R)}²+2√(n₁n₂rR)

Hence n₁r+n₂R will be minimum if

√(n₁r)-√(n₂R) is minimum i.e. equal to zero. Hence the required condition is 

√(n₁r)-√(n₂R) =0

→√(n₁r) = √(n₂R) 

→n₁r = n₂R.          




 

   22. A battery of emf 100 V and a resistor of resistance 10 kΩ are joined in series. This system is used as a source to supply current to an external resistance R. If R is not greater than 100 Ω, the current through it is constant up to two significant digits. Find its value. This is the basic principle of a constant-current source.  



ANSWER: If R =1 Ω, Current i =100/(10001) = 0.009999 A. If R = 2 Ω then current =100/10002 =0.009998 A.

For R=10 Ω, current=100/10010 =0.00999 A. Similarly, for R= 99 Ω

the corrent =100/10099 =0.00990 A. When R = 100 Ω,

The current =100/10100 =0.00990 A 

 So we see that for R = 1 Ω to 100 Ω, the value of current is constant up to two significant digits. (When the digits before the decimal are zero, the digits starting after the right of the zeros after the decimal are significant digits. Here are 9 and 9 at third and fourth place after the decimal.) 

  So the value of the constant current for R up to 100 Ω 

=0.0099 A ≈0.01 A =10 mA.     





 

  23. If the reading of ammeter A₁ in figure (32-E4) is 2.4 A, what will the ammeters A₂ and A₃ read? Neglect the resistance of ammeters.    
Figure for Q-23



ANSWER: Since A₁ and A₂ are in parallel, the potential difference across each of them will be the same. Hence for the reading in ammeter A₂ = i, 

(2.4 A)*(20 Ω) = i*(30 Ω)

→i =2.4*2/3 A = 1.6 A. 


From Kirkhoff's junction law, algebraic sum of current at a junction point is zero, i.e. the total current towards the point is equal to the total current coming out of the point. Applying this law at the junction of 10 Ω, 20 Ω and A₂, 

i' = 2.4 + 1.6 A

   =4.0 A.

Hence the current in resistor 10 Ω and reading of A₃ = 4.0 A.            





 

  24. The resistance of the rheostat shown in figure (32-E5) is 30 Ω. Neglecting the meter resistance, find the minimum and maximum currents through the ammeter as the rheostat is varied.    
The figure for Q-24



ANSWER: If the resistance of the rheostat =R, then the equivalent resistance of the combination R' is given as,

R' = (20*10)/(20+10) +R

    =R + 20/3

Now the current through the ammeter is equal to the current in the circuit,

i = V/R'

  =5·5/(R +20/3)

Since R is in the denominator, the minimum value of i will be for a maximum value of R, i.e. R =30 Ω. So

iₘᵢₙ = 5·5/(30+20/3) A

    =5·5*3/110 A

    =3/20 A

    =0·15 A  


And i will be maximum when R is minimum i.e. R =0. Hence

iₘₐₓ = 5·5/(20/3) A

     =3*5·5/20 A

     =16·5/20 A

     =0·825 A ≈ 0·83 A.

 



 

  25. Three bulbs each having a resistance of 180 Ω, are connected in parallel to an ideal battery of emf 60 V. Find the current delivered by the battery when (a) all the bulbs are switched on, (b) two of the bulbs are switched on and (c) only one bulb is switched on.    



ANSWER: (a) When all the bulbs are switched on, the resistances of each of them are connected in parallel. Effective resistance R is given as,

1/R =1/180 +1/180 +1/180

→1/R =3/180 =1/60

→R = 60 Ω.

Emf of battery = 60 V 

Current delivered by the battery,  

i =(60 V)/(60 Ω) = 1 A. 


(b) When only two of the bulbs are switched on, the effective resistance of the bulbs,

R =180/2 =90 Ω

Now the current delivered by the ideal battery, i =(60 V)/(90 Ω) 

  =0·67 A. 


(c) When only one bulb is switched on, the resistance R =180 Ω. Hence the current delivered in this case,

i =(60 V)/(180 Ω) =0·33 A.       

 

          



 

  26. Suppose you have three resistors of 20 Ω, 50 Ω, and 100 Ω. What minimum and maximum resistance can you obtain from three resistors?    



ANSWER: (a) The minimum resistance can be obtained by connecting them in parallel. Let R be the equivalent resistance in this case. R can be given as,

1/R =1/20 +1/50 +1/100

      =(5+2+1)/100

      =8/100 

→R =100/8 Ω =25/2 Ω =12·5 Ω. 


(b) The maximum resistance can be obtained by connecting them in series. Equivalent resistance in this case, 

R' =20 Ω +50 Ω +100 Ω

    =170 Ω.                   





 

  27. A bulb is made using two filaments. A switch selects whether the filaments are used individually or in parallel. When used with a 15 V battery, the bulb can be operated at 5 W, 10 W, or 15 W. What should be the resistance of the filaments?   



ANSWER: The power P is given as

P = V²/R 

Hence R = V²/P

Given V =15 volts,

For power 5 W,

R = 15²/5 Ω =45 Ω

For Power 10 W,

R' =15²/10 Ω =22·5 Ω 

For power 15 W, 

R" =15²/15 Ω =15 Ω. 

Assume the resistances of the filaments = r and r'. These are connected in parallel. Suppose r > r', then maximum resistance R =45 Ω when the only r is switched on. Hence r =45 Ω.

     Minimum equivalent resistance R"= 15 Ω will be when r and r' are both switched on because they are in parallel. R" will be less than the smaller resistance. Hence the smaller resistance filament, 

r' =22·5 Ω.






 

  28. Figure (32-E6) shows a part of a circuit. If a current 12 mA exists in the 5 kΩ resistor, find the currents in the other three resistors. What is the potential difference between points A and B?     
The figure for Q-28



ANSWER:   Current in the 5 kΩ resistor =12 mA. Let the current in the 10 kΩ resistor = i mA. Then the current in the 20 kΩ resistor = 12-i mA. The 10 kΩ and the 20 kΩ resistors are in parallel hence the potential difference across them is the same. Hence,  

i*10 =(12-i)*20, 

→10i+20i =240,  

→30i =240, 

→i =240/30 =8 mA. 

Thus the current in 10 kΩ resistor =8 mA. So the current in 20 kΩ resistor =12-i =12-8 = 4 mA. 


Since the 100 kΩ resistor is connected in series with 5 kΩ resistor, hence the same current 12 mA flow through it.

The potential difference between A and B =12*5+20*4+12*100 V

=60+80+1200 V  

=1340 V.


Note: In the above potential difference the units mA*kΩ is the same as A*Ω =V.                        






 

  29. An ideal battery sends a current of 5 A in a resistor. When another resistor of value 10 Ω is connected in parallel, the current through the battery is increased to 6 A. Find the resistance of the first resistor.    



ANSWER:   Let the resistance of the first resistor =R. Current through it is given as equal to 5 A. Hence the potential difference across R =5R volts.

    Since the battery is ideal, the potential difference 5R remains the same in the second case when 10 Ω resistor is connected in parallel. The equivalent resistance in this case R' is given as,   

1/R' =1/R +1/10 =(R+10)/10R

→R' =10R/(R+10) 

Now the current delivered by the battery 

i = 5R/{10R/(R+10) 

  =(R+10)/2

But given that i =6 A, hence 

(R+10)/2 =6

→R =6*2-10 =2 Ω.              

        




 

  30. Find the equivalent resistance of the network shown in figure (32-E7) between points a and b.    
The figure for Q-30


ANSWER: The arrangement in the given picture can be simplified as below without changing its property.
The diagram for Q-30

We see that the three resistors each having resistance r are connected in parallel between a and b. Hence the equivalent resistance between a and b is given as

1/R =1/r+1/r+1/r =3/r

→R = r/3.  

     

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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


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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CHAPTER- 5 - Newton's Laws of Motion


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CHAPTER- 4 - The Forces

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CHAPTER- 3 - Kinematics - Rest and Motion

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