Sunday, May 23, 2021

H C Verma solutions, ELECTRIC CURRENT IN CONDUCTORS, Exercises, Q31-Q40, Chapter-32, Concepts of Physics, Part-II

Electric Current in Conductors


Exercises, Q31 - Q40


   31. A wire of resistance 15.0 Ω is bent to form a regular hexagon ABCDEFA. Find the equivalent resistance of the loop between the points (a) A and B, (b) A and C and (c) A and D.      
The diagram for Q-31


ANSWER: (a) Between A and B there are two parallel resistances. AB and BCDEFA. Resistance of the whole wire =15 Ω. Hence the resistance of each side =15/6 =2·5 Ω. Hence,

Resistance of AB =2·5 Ω 

Resistance of BCDEFA =2·5*5 =12·5 Ω

The equivalent resistance between A and B = R is given as,

1/R =1/2·5 +1/12·5

→1/R =(5+1)/12·5 =6/12·5 

→R =12·5/6 =2·08 Ω. 


(b) Between A and C, there are two parallel resistances, 

ABC =2·5*2 =5 Ω, and

CDEFA =2·5*4 =10 Ω. 

Hence equivalent resistance R is given as, 

1/R =1/5 +1/10 =3/10

→R =10/3 Ω =3·33 Ω.  


(c) Between points A and D, there two parallel resistances,

Resistance of ABCD =2·5*3 =7·5 Ω, and

Resistance of DEFA =2·5*3 =7·5 Ω. 

Equivalent resistance R is given as,

1/R =1/7·5 +1/7·5 =2/7·5

→R =7·5/2 =3·75 Ω.                    







   32. Consider the circuit shown in figure (32-E8). Find the current through 10 Ω resistor when the switch S is (a) open (b) closed.       
Diagram for Q-32



ANSWER: (a) When switch S is open, there are two resistors 10 Ω and 20 Ω connected in series. Equivalent resistance of these two resistances =10+20 =30 Ω. 

The potential difference across it = 3 V. 

Hence current in the circuit 

= 3/30 =0·1 A. 

Since 10 Ω is in series, hence the same current 0·1 A will flow through it. 


(b) When the switch S is closed, the current through 10 Ω will not enter to 20 Ω and only one resistor 10 Ω remains in the circuit.

Hence the current through 10 Ω resistor

=3/10 A =0·3 A.    





 

   33. Find the currents through the three resistors shown in figure (32-E9).
Diagram for Q-33

   ANSWER: The 4 Ω resistor is bypassed, hence the current through it will always be zero. Now only 4 Ω and 6 Ω resistors remain in the circuit and the same current i will flow through each of them. Equivalent resistance R =4+6 =10 Ω. Since the two cells connected in series are reversely connected, net emf =4-2 =2 V. 

The current in the circuit,

i =2/10 =0·2 A. 

Hence 0·2 A current will flow through each of 4 Ω and 6 Ω resistances.         






  

   34. Figure (32-E10) shows a part of an electric circuit. The potentials at points a, b and c are 30 V, 12 V and 2 V respectively. Find the currents through the three resistors. 
The figure for Q-34

      

ANSWER: Let the current through 10 Ω resistor =i. The potential drop across it =i*10 =10 i volts. Hence the potential at the junction of resistors =Vₐ-10i =30-10i volts.

Potential diffrence across 20 Ω 

=30-10i-12 =18-10i volts.

Current through it, i' =(18-10i)/20 A

Potential difference across 30 Ω resistor 

=30-10i-2 =28-10i volts. 

Current through it, i" =(28-10i)/30 A 

From the Kirchoff's junction law

i'+i" =i 

→(18-10i)/20 +(28-10i)/30 =i 

→{3(18-10i)+2(28-10i)}/60 =i 

→60i =54-30i+56-20i 

→60i =110-50i 

→110i =110

→i =110/110 =1 A.

Hence i' =(18-10*1)/20 =8/20 =0·4 A. 

andi" =(28-10*1)/30 =18/30 =0·6 A.   

So the current through 10 Ω =1 A, through 20 Ω =0·4 A and through 30 Ω =0·6 A.                   




  


   35. Each of the resistors shown in figure (32-E11) has a resistance of 10 Ω and each of the batteries has an emf of 10 V. Find the currents through the resistors a and b in the two circuits.    
The figure for Q-35


ANSWER: In figure (a), resistor a has a potential difference of 10 V across it.

 Resistance of a = 10 Ω. Hence current through it, 

i =(10 V)/(10 Ω) =1 A. 

The potential difference across b 

=10 V-10V

=0 

Hence current through b =zero. 


In figure (b), the potential difference across a =10 V. 

Resistance of a = 10 Ω. 

Hence current through a 

=(10 V)/(10 Ω)

=1 A. 

Potential difference across b =10-10 =0 V.

Hence current through b = zero.     





 

   36. Find the potential differences Vₐ-Vᵦ in the circuits shown in figure (32-E12).
The figure for Q-36


ANSWER: Consider figure (a) on the left. Let current i₁ in R₁ towards left and current i₂ in R₂ towards left. Then current in R₃ is i₁+i₂ towards the right (considering the junction point b).

   Applying Kirchoff's voltage law in a loop. Going anticlockwise in the upper loop.

 (i₁+i₂)R₃+i₁.R₁-Ԑ₁ =0 ---- (i)

Similarly clockwise in lower loop,

(i₁+i₂)R₃+i₂.R₂ -Ԑ₂ =0 ----- (ii) 

Subtracting (ii) from (i) we get,

i₁R₁-i₂R₂-Ԑ₁+Ԑ₂ =0 

→i₁R₁-i₂R₂ =Ԑ₁-Ԑ₂ ----- (iii)

Adding (i) and (ii) we get,

2(i₁+i₂)R₃+i₁.R₁-Ԑ₁+i₂R₂-Ɛ₂=0

→(R₁+2R₃)i₁+(R₂+2R₃)i₂=Ɛ₁+Ɛ₂ ---(iv)

We solve it for i₁ and i₂.

Multiply (iii) by (R₂+2R₃) and multiply (iv) by R₂, then add. We get,

i₁R₁(R₂+2R₃)+i₁R₂(R₁+2R₃)=(Ɛ₁-Ɛ₂)(R₂+2R₃)+(Ɛ₁+Ɛ₂)R₂ 

→i₁(2R₁R₂+2R₁R₃+2R₂R₃)=R₂(Ɛ₁-Ɛ₂+Ɛ₁+Ɛ₂)+2R₃(Ɛ₁-Ɛ₂)

→2i₁(R₁R₂+R₂R₃+R₃R₁)=2R₂Ɛ₁+2R₃Ɛ₁-2R₃Ɛ₂

→2i₁(R₁R₂+R₂R₃+R₃R₁)=2(R₂+R₃)Ɛ₁-2R₃Ɛ₂

→i₁={2(R₂+R₃)Ɛ₁-2R₃Ɛ₂}/{2(R₁R₂+R₂R₃+R₁R₃)}

Similarly we can find out

i₂={2(R₁+R₃)Ɛ₂-2R₃Ɛ₁}/{2(R₁R₂+R₂R₃+R₃R₁)}

Hence,

i₁+i₂ =(R₁Ɛ₂+R₂Ɛ₁)/(R₁R₂+R₂R₃+R₃R₁)

Now, Vₐ-Vᵦ=(i₁+i₂)R₃

=(R₁Ɛ₂+R₂Ɛ₁)R₃/(R₁R₂+R₂R₃+R₃R₁)

Dividing numerator and denominator by R₁R₂R₃, we get

Vₐ-Vᵦ=(Ɛ₁/R₁+Ɛ₂/R₂)/(1/R₁+1/R₂+1/R₃)


The circuit in figure (b) is the same as in (a) except that between a and b resistor R₁ is left to Ɛ₁ but we shall get the same set of equations. Hence the answer is the same.     

 

 


 

 


   37. In the circuit shown in figure (32-E13), Ԑ₁ =3 V, Ԑ₂ =2 V, Ԑ₃ =1 V and r₁ =r₂ =r₃ =1 Ω. Find the potential difference between points A and B and the current through each branch. 
The figure for Q-37


ANSWER: Let the current through r₁ and r₃ be i₁ and i₃ towards left respectively. The current through r₂ is i₂ towards the right.

  Applying Kirchhoff's loop law anticlockwise in the upper loop, we get 

i₁r₁+i₂r₂+Ɛ₂-Ɛ₁ =0

→i₁r₁+i₂r₂ =Ɛ₁-Ɛ₂ 

→i₁+i₂ =1, -----(i)

{after putting the values} 

   In the lower loop, clockwise

i₂r₂+Ɛ₂-Ɛ₃+i₃r₃ =0

→i₂r₂+i₃r₃ =Ɛ₃-Ɛ₂

→i₂+i₃ = -1 

→-i₃ =1+i₂

At junction A, 

i₁+i₃ =i₂ 

→i₁-i₂ =-i₃ = 1+i₂

→i₁-2i₂ = 1 ------ (ii) 

Subtract (ii) from (i) 

3i₂ = 0

→i₂ = 0.

Considering the middle resistor and cell, the potential difference between A and B, 

=i₂r₂ +Ɛ₂ = 0+2 =2 V. 

From (ii),

i₁ = 1 A.

Since i₂+i₃= -1,

i₃ = -1 A.            





 

   38. Find the current through the 10 Ω resistor shown in figure (32-E14). 
The figure for Q-38


ANSWER: Suppose the current in all the resistors is from left to the right. We denote current in 3 Ω resistor =i₁, in 6 Ω resistor =i₂ and in 10 Ω resistor =i₃. Applying Kirchhoff's junction law at the junction of three resistors, we get

i₁ =i₂ +i₃

Considering the voltage drop in the lower loop clockwise,

i₁*3+i₂*6 -4.5=0

→i₁+2i₂ =1.5  

→i₂+i₃+2i₂ =1.5

→3i₂ +i₃ = 1.5 

→3i₂ =1.5-i₃ ---------- (i)


Considering the voltage drop in the upper loop clockwise,

i₃*10+3-i₂*6 =0

→6i₂-10i₃ =3  

→3 -2i₃-10i₃ =3

→12i₃ = 0

→i₃ = 0.

    Hence the current through 10 Ω resistor is zero.       





 

   39. Find the current in three resistors shown in figure (32-E15). 
The figure for Q-39


ANSWER: Assume that the current in three resistors from left to right is i₁, i₂ and i₃ all downwards. Consider the left loop for voltage drop clockwise (Kirchhoff's Voltage law).

2 +i₁*1-2 = 0

→i₁ =0.

Similarly in the middle loop clockwise,

2+i₂*1-2-i₁*1 =

→i₂ =i₁ =0.

Going clockwise in the right loop,

2+i₃*1-2-i₂*1 =0

→i₃ =i₂ = 0.


Hence the current in all three resistors is zero.     





 

   40. What should be the value of R in figure (32-E16) for which the current in it is zero? 
The figure for Q-40


ANSWER: The connections of the four resistors are like a Wheatstone bridge, only there is resistor R in place of a galvanometer. For the current in R to be zero the bridge should be balanced, i.e. the ratio of the upper two resistors is equal to the ratio of the lower two resistors. Here we see that ratio of the upper two resistors =10 Ω/5 Ω =2:1 and the ratio of the lower two resistors =10 Ω/5 Ω =2:1.

       Both ratios are the same, hence the bridge is balanced. Therefore there will be no current through R whatever be the value of resistance of R.    

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CHAPTER- 28- Heat Transfer

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