Thursday, June 3, 2021

H C Verma solutions, ELECTRIC CURRENT IN CONDUCTORS, Exercises, Q51-Q60, Chapter-32, Concepts of Physics, Part-II

Electric Current in Conductors


Exercises, Q51 - Q60


   51.  An ammeter is to be constructed which can read currents up to 2·0 A. If the coil has a resistance of 25 Ω and takes 1 mA for full-scale deflection, what should be the resistance of the shunt used?       


ANSWER: At the full-scale deflection the current in the ammeter i' = 1 mA = 0·001 A. Resistance of the coil, r =25 Ω. Hence the potential difference across the coil =i'*r = 0·001*25 =0·025 V. 

     Maximum current to be measured, i =2·0 A. From this current i' =0·001 A is diverted to ammeter coil and the rest i" =i -i' is diverted to the shunt. 

     Let the resistance of the shunt =R, then, R =0·025/(i -i')

→R =0·025/(2·0-0·001) 

      = 0·0125 Ω

      =1·25x10⁻² Ω.          




 

   52.  A voltmeter coil has a resistance of 50.0 Ω and a resistor of 1.15 kΩ is connected in series. It can read potential differences up to 12 volts. If this same coil is used to construct an ammeter which can measure currents up to 2.0 A, what should be the resistance of the shunt used?     


ANSWER: The equivalent resistance of the coil and the resistor R 

=50 +1·15x10³ Ω =1200 Ω. 

     It can read a potential difference of 12 V. At this potential difference, the current through the coil and resistor,

i' = 12/1200 =0·01 A. 

     So the coil can take a maximum current of 0.01 A. At the maximum current (to be measured) the potential difference across the coil,

E' =i'*50 =0.01*50 =0·5 V.

   This potential difference will also be for the shunt of the ammeter to be constructed. Since the ammeter has to measure a current of 2·0 A, out of this the coil takes a current of 0.01 A, so the shunt has to carry a current of,

i" =i -i' =2·0 -0·01 A 

→i" =1·99 A. 

      Hence the resistance of the shunt should be, R =E'/i"

→R =0·5/1·99 = 0·251 Ω.               





 

   53.  The potentiometer wire AB shown in figure (32-E26) is 40 cm long. Where should the free end of the galvanometer be connected on AB so that the galvanometer may show zero deflection?      
The figure for Q-53


ANSWER: Suppose the point where the free end of the galvanometer touches AB for zero deflection is C. Now the resistances 8 Ω, 12 Ω, AC and CB make a balanced Wheatstone bridge. The wire AB is uniform, hence its resistance is proportional to the length. Suppose the resistance of the wire is k per unit length, then the resistance of length AC =k·AC and that of CB =k·CB. 

   For the above balanced Wheatstone bridge,

k·AC/k·CB =8/12

→AC =CB*2/3

{But CB =AB-AC =40 -AC cm}

→AC =(40 -AC)*2/3

→3AC =80 -2AC

→5AC =80

→AC =80/5 =16 cm.

So for the zero deflection in the galvanometer, the free end of it should be connected at a point 16 cm from end A on AB.             





 

   54.  The potentiometer wire AB shown in figure (32-E27) is 50 cm long. When AD = 30cm, no deflection occurs in the galvanometer. Find R.       
The figure for Q-54



ANSWER: AB =50 cm, AD =30 cm, hence DB =50 -30 =20 cm. 

   The wire AB has a uniform section and its resistance is proportional to its length. The arrangement given in the figure is a balanced Wheatstone bridge hence the four resistances are proportional. Thus,

 AD/DB = 6/R 

→30/20 =6/R 

→R =6*2/3 =4 Ω.               





 

   55.  A 6-volt battery of negligible internal resistance is connected across a uniform wire AB of length 100 cm. The positive terminal of another battery of emf 4 V and internal resistance 1 Ω is joined to the point A as shown in figure (32-E28). Take the potential at B to be zero. (a) what are the potentials at points A and C? (b) At which point D of the wire AB, the potential is equal to the potential at C? (c) If the points C and D are connected by a wire, what will be the current through it? (d) If the 4 V battery is replaced by a 7·5 V battery, what would be the answers of parts (a) and (b)?       
The figure for Q-55


ANSWER: (a) The potential at B is the same as the potential of the negative terminal of the 6 V battery, here it is given zero. The potential of the positive terminal of the 6 V battery is the same as that of point A. Since the potential of the positive terminal of the 6 V battery is 6 V higher than the negative terminal (and point B), hence the potential at A is 6 V. 

      Now the potential of the positive terminal of the 4 V battery is also the same as A, i.e. 6 V. Since point C is free, there is no current in this branch. The negative terminal and C are at a potential 4 V lower than the positive terminal. Hence the potential at C =6 V -4 V =2 V. 


(b) The resistance of the uniform wire AB is proportional to its length. Hence potential from A (6 V) to B (zero) decreases uniformly. The potential of D is equal to the potential of C =2 V. Thus, 

BD/AB = 2/6 

→BD =AB/3 =100/3 =33·33 cm. 

Hence AD =100 -33·33 =66·7 cm. 

So the required point is 66·7 cm from A.   


(c) As given in (b) the potentials of D and C are the same. Hence there is no potential difference between them. It means no current through the connected wire CD. Zero. 


(d) When the 4 V battery is replaced by a 7.5 V battery.

There will be no change in the potential of A which is at 6 V. The positive terminal of the new battery and A are at the same potential i.e. at 6 V but the negative terminal and C are now at a potential 7·5 V less than A and the positive terminal of the new battery. Hence the potential of C =6 V -7·5 V = -1·5 V.

   Since the potential on AB varies from zero to 6 V, from B to A, there is no point D on AB that has a potential of -1·5 V (equal to C).              





  

   56.  Consider the potentiometer circuit arranged as in figure (32-E29). The potentiometer wire is 600 cm long. (a) At what distance from point A should the jockey touch the wire to get zero deflection in the galvanometer? (b) If the jockey touches the wire at a distance of 560 cm from A, what will be the current in the galvanometer?       
The figure for Q-56


ANSWER: (a) The total resistance of the circuit =r +15r = 16r.  

    The current in the circuit, i =Ɛ/16r.

 If we assume the potential at B to be zero, then the potential of the negative plate of the battery = 0-ir =-ir,

  =-Ɛr/16r = -Ɛ/16.

   And the potential of the positive terminal of the battery (as well as that of A) =Ɛ +(-Ɛ/16) =15Ɛ/16.

   So the potential from B to A uniformly varies at the rate of α =15Ɛ/(16*600) volts per centimeter.

   There is no current in the lower branch and the potential of the positive terminal of the battery having emf Ɛ/2 is the same as that of A =15Ɛ/16. Thus the potential from the negative terminal of this battery to the jockey 

=15Ɛ/16 -Ɛ/2

=(15Ɛ -8Ɛ)/16 

=7Ɛ/16.

   To get a zero deflection in the galvanometer, the point where the jockey will touch the wire should have the same potential as the jockey, i.e. =7Ɛ/16.

  The rate of variation of the potential on the wire is α = 15ℰ/(16*600). Hence the point on AB having a potential of 7ℰ/16 from B = (7ℰ/16)÷(15ℰ/(16*600) cm

=7ℰ*16*600/(15ℰ*16) cm

=7*600/15 cm

=7*40 cm

=280 cm.

        Hence the distance of this point from A =600-280 cm =320 cm.


(b) When the jockey touches the wire at 560 cm from A, there will be a current in the lower loop. Also, the current in the upper loop will change. Let us draw a diagram as follows:-  
The diagram for Q-56b

  Point D divides the wire into two resistances. Resistance of the part AD =(560/600)*15r 

=14r. 

Resistance of the other part DB =15r-14r =r. 

Let the current in the galvanometer =i'.

The current in AD =i" and the current in DB =i as shown in the diagram. 

  With Kirchhoff's junction law at A,

i =i' +i".

Let us apply loop law in the upper loop anticlockwise,  

ir +ir -ℰ +i"*14r =0 

→14i"r +2ir =ℰ 

→14i"r +2i'r +2i"r =ℰ

→16i"r =ℰ-2i'r

→i" =(ℰ-2i'r)/16r

In the lower loop,

ℰ/2 +i'r -i"*14r =0 

→i'r -14i"r +ℰ/2 =0

→i'r -(14rℰ -28i'r²)/16r +ℰ/2 =0

→16i'r² -14rℰ +28i'r² +8rℰ = 0 

→44i'r² -6rℰ =0

→r(22i'r -3ℰ) =0

Since r is not zero, hence,

221i'r -3ℰ =0

→i' =3ℰ/22r.

So the current in the galvanometer will be =3ℰ/22r.                  





 

   57.  Find the charge on the capacitor shown in the figure (32-E30).     
The figure for Q-57


ANSWER: The potential difference across the capacitor will be the same as the potential difference across the 10 Ω resistor. The 10 Ω resistor and the 20 Ω resistor are connected in series. The equivalent resistance R =10 +20 =30 Ω.  ℰ =2 V. 

Hence the current in the circuit, 

i =2/30 =0·067 A. 

    So the current through the 10 Ω resistor is 0·067 A. Hence the potential difference across 10 Ω resistor = i*10

=0·067*10 =0·67 V.

   It is also the potential difference across the capacitor. Capacitance, C = 6 µF. 

Hence the charge on the capacitor, 

Q =CV =(6 µF)*(0·67 V) 

→Q =4.02 µC ≈4 µC.    

       




 

   58.  (a) Find the current in the 20 Ω resistor shown in figure (32-E31). (b) If a capacitor of capacitance 4 µF is joined between points A and B, what would be the electrostatic energy stored in it in a steady-state?       
The figure for Q-58


ANSWER: (a) From the symmetry, the current in each of the 10 Ω resistor = i (say) in the direction as shown in the diagram. From the junction law at A, the current in 20 Ω resistor, i' =2i, i.e. i =i'/2.  
The diagram for Q-58a

Applying Kirchhoff's loop law in the left loop, clockwise:- 

 i*10 +i'*20 -5 =0

 →10*i'/2 +20i' =5  

→25i' =5

→i' =5/25 =0·2 A.

So the current in the 20 Ω resistor =0·2 A. 


(b) When a capacitor is joined between points A and B the potential difference across it is the same as the potential difference across the 20 Ω resistor. This potential difference, V =i'*20 

→V =0.2*20 =4 volts.  
The diagram for Q-58b

The capacitance of the capacitor (given), C =4 µF.  

Hence the electrostatic energy stored, 

E =½CV²  

→E =½*4*4² µJ  

→E =32 µJ.

                   


 


 

   59.  Find the charges on the four capacitors of capacitances 1 µF, 2 µF, 3 µF, and 4 µF shown in figure (32-E32).  
The figure for Q-59



ANSWER: We know that in a steady-state no current flows across the plates of the capacitors. Hence in the given arrangement, the current supplied by the battery will only go through the resistors. Let us first name the junctions and capacitors for our convenience as follows:-
Ch-32, Q-59

  In the upper loop of resistors, a current will flow in the loop AKLBA. The potential difference across KL = 6 V. The equivalent resistance of resistors connected in series between KL = 1+2 =3 Ω. Hence the current between KL,  

i =(6 V)/(3 Ω) =2 A. 

The potential of G and K is the same and also the potential of M and N are the same. Hence the potential difference across the capacitor ① is the same as the resistor KN which will be equal to,  

V' =i*(1 Ω) =(2 A)*(1 Ω) =2 volts.

Thus the charge on the capacitor ①,

=(1 µF)*(2 V) =2 µC. 

  Similarly the potential difference across the capacitor ② = i*(2 Ω) =(2 A)*(2 Ω) =4 V. Hence the charge on the capacitor ② =(2 µF)*(4 V) =8 µC.      


  Now consider the lower loop AEFBA. The potential difference across CD =6 V. 

 Equivalent resistance across CD =3 Ω+ 3 Ω =6 Ω. Hence the current in section CD, i' =(6 V)/(6 Ω) = 1 A. 

  The potential difference across the resistor CP = the potential difference across the capacitor ③ =i'*(3 Ω)

=(1 A)*(3 Ω) = 3 V. 

Thus the charge on the capacitor ③,

=(3 µF)*(3 V) =9 µC. 

        

Similarly, the potential difference across the capacitor ④ =i'*(3 Ω) =3 V.

The capacitance of ④ =4 µF, 

The charge on the capacitor ④ =4*3 µC

=12 µC.    





 

   60.  Find the potential difference between points A and B and between B and C of the figure (32-E33) in a steady-state.     
The figure for Q-60


ANSWER: In the given figure all the elements above points A and C are capacitors. In a steady-state, there is no current flow through a capacitor. Hence the potential of a continuous wire will be the same.

     Two 3 µF capacitors are connected in parallel between A and B, Their equivalent capacitance =3 +3 =6 µF. Similarly, two 1 µF capacitors are in parallel between B and C. Their equivalent capacitance =1 +1 = 2 µF.            Now the arrangement simplifies as below:-
The diagram for Q-60

  Capacitors AB and BC are in series. Their equivalent capacitance C' is given by,

1/C' = 1/6 +1/2 =4/6 =2/3

→C' =3/2 µF.

C' and 1 µF are in parallel hence the total equivalent capacitance,

C =1 +3/2 =5/2 =2·5 µF. 

    The potential across this equivalent capacitor =100 V. Thus the total charge on this capacitor,

Q =CV =2·5*100 µC =250 µC. 

The potential difference across both capacitors C' and 1 µF is 100 V due to the parallel connection. Thus the charge stored on 1 µF capacitor,

q =1*100 =100 µC. 

Rest of the charge is on C' i.e.,

q' =Q -q =250 -100 =150 µC. 

C' comprises of two capacitors in series whose equivalent capacitances are 6 µF and 2 µF. Since in series connection the charge on each of the capacitor is the same, hence the charge on the 6 µF capacitor between A and B is also q' =150 µC. Now the potential difference across 6 µF capacitor = The potential between A and B = V' =q'/6 =150/6 V =25 V. 

             Similarly the potential difference across 2 µF capacitor = The potential difference between B and C =q'/2 =150/2 V =75 V.            


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Links to the Chapters



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

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

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CHAPTER- 5 - Newton's Laws of Motion


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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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CHAPTER- 3 - Kinematics - Rest and Motion

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CHAPTER- 2 - "Physics and Mathematics"

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