Sunday, June 13, 2021

H C Verma solutions, ELECTRIC CURRENT IN CONDUCTORS, Exercises, Q61-Q70, Chapter-32, Concepts of Physics, Part-II

Electric Current in Conductors


Exercises, Q61 - Q70


   61.  A capacitance C, a resistance R and an emf ℰ are connected in series at t = 0. What is the maximum value of (a) the potential difference across the resistor, (b) the current in the circuit, (c) the potential difference across the capacitor, (d) the energy stored in the capacitor, (e) the power delivered by the battery (f) the power converted into heat?    



ANSWER: (a) Before connections are made, the whole circuit is at the same potential, suppose these are at zero potential. Once the key K is switched on the positive charge rushes to end B of the resistor, (see diagram below). 
The diagram for Q-61

    So at that instant, the potential of end B is ℰ while the end A is still at zero potential. At this instant the potential difference across the resistor = ℰ. As time passes the potential of the whole resistor becomes =ℰ and the potential difference across the resistor becomes zero. Thus the maximum value of potential difference across the resistor is = ℰ.  


 (b) The maximum value of the current, 

i = maximum P.D/resistance = ℰ/R. 


 (c) Right-hand plate of the capacitor will attain the potential of the positive plate of the battery while the left-hand plate will attain the potential of the negative plate of the battery. Thus the maximum potential difference across the capacitor =ℰ. 


  (d) The maximum value of the energy stored in the capacitor will be at the maximum potential difference ℰ. Thus, 

Maximum E =½Cℰ².  


   (e) The maximum power delivered by the battery, P =ℰ*maximum current

→P = ℰ*ℰ/R = ℰ²/R.  


   (f) The maximum power converted into heat by the resistor = Maximum Current in the resistor*maximum potential difference across the resistor 

= (ℰ/R)*ℰ =ℰ²/R            





 

   62.  A parallel-plate capacitor with a plate area of 20 cm² and plate separation of 1.0 mm is connected to a battery. The resistance of the circuit is 10 kΩ. Find the time constant of the circuit.    


ANSWER: A = 20 cm² =0·002 m², d = 1 mm =0·001 m. Hence the capacitance,

C = ԑ₀A/d = 8.85x10⁻¹²*0.002/0.001 F

 =1·77x10⁻¹¹ F  

R =10 kΩ =1x10⁴ Ω

Hence the time constant of the circuit,

 𝜏 =RC =1x10⁴*1·77x10⁻¹¹ s

   =1·77x10⁻⁷ s

   ≈0·18x10⁻⁶ s

   =0·18 µs.  




 

   63.  A capacitor of capacitance 10 µF is connected to a battery of emf 2 V. It is found that it takes 50 ms for the charge on the capacitor to become 12.6 µC. Find the resistance of the circuit.     


ANSWER: The charge at time t on a capacitor is given as, 

q = ℰC(1 -e-t/RC) 

→12·6 =2*10*(1-e-t/RC) 

→e-t/RC =1-0·63 =0·37 

→-t/RC = ln 0·37 ≈ -1

→RC = t, {Given t =50 ms =0.05 s}

→R =0·05/10x10⁻⁶ Ω  

→R = 5000 Ω = 5 kΩ.             





 

   64.  A 20 µF capacitor is joined to a battery of emf 6·0 V through a resistance of 100 Ω. Find the charge on the capacitor 2·0 ms after the connections are made.    



ANSWER:   Given ℰ = 6·0 V, C = 20 µF, R =100 Ω, t = 2·0 ms =0·002 s and q = ?  

The value of t/RC =0·002/(100*20x10⁻⁶)

= 1 

Now,

q = ℰC(1 -e-t/RC) 

  =6*20(1-e⁻¹) µC

  =120*(1-0·37)  

  =120*0·63 µC

  ≈76 µC.                  





 

   65.  The plates of a capacitor of capacitance 10 µF, charged to 60 µC, are joined together by a wire of resistance 10 Ω at t = 0. Find the charge on the capacitor in the circuit at (a) t = 0, (b) t = 30 µs, (c) t = 120 µs and (d) t =1.0 ms.    


ANSWER:  For the discharging capacitor, the charge on it at time t is given as, q =Qe-t/RC, hence 

(a) at t = 0, q =Q*1 =Q =60 µC, 


(b) at t = 30 µs,

-t/RC =-(30 µs)/(10 Ω*10 µF) 

  = -0·3

Hence, q =Qe-0·3 

      =(60 µC)*0·74

      =44 µC.  


(c) At t = 120 µs, 

-t/RC = -(120 µs)/(10 Ω*10 µF) 

     =-1·2  

Hence, q =Qe-1·2 

→q =(60 µC)*0·30 = 18 µC. 


(d) At t =1.0 ms =0.001 s

-t/RC =-0·001/(10*10x10⁻⁶) 

         = -10 

Hence q =Qe-10 

       =(60 µC)*4·54x10⁻⁵ 

       ≈0·003 µC.        




 

   66.  A capacitor of capacitance 8·0 µF is connected to a battery of emf 6·0 V through a resistance of 24 Ω. Find the current in the circuit, (a) just after the connections are made and (b) one time constant after the connections are made.    


ANSWER: (a) Just after the connections are made, the potential difference across the resistor is maximum and equal to the emf. So V' =6 V, R =24 Ω, hence the current in the circuit, 

i =V'/R =6/24 A =0·25 A. 


(b) Given C =8·0 µF 

Time constant 𝜏 = RC  

After one time constant i.e t =RC, the charge on the capacitor,

q = ℰC(1 -e-t/RC) 

→q =ℰC(1 -e⁻¹) 

→q =6*8*0·63 µC = 30·24 µC 

The potential difference across the capacitor, V' =q/C =30·24/8·0 =3·78 V.

Let the current in the circuit = i. 

From the Kirchhoff's loop law, 

iR +V' - ℰ = 0 

→i*24 +3·78 -6 =0  

→i = 2·22/24 A =0·09 A.           






 

   67.  A parallel plate capacitor of plate area 40 cm² and separation between the plates 0.10 mm is connected to a battery of emf 2.0 V through a 16 Ω resistor. Find the electric field in the capacitor 10 ns after the connections are made.    



ANSWER: Given A =40 cm² =0·004 m², Separation d =0·10 mm =0·0001 m, the emf ℰ =2·0 V, resistance R =16 Ω. 

The capacitance C = ԑ₀.A/d 

→C = 8·85x10⁻¹²*0.004/0.0001 

   =3.54x10⁻¹⁰ F 

t =10 ns =10x10⁻⁹ s 

t/RC =10x10⁻⁹/(16*3·54x10⁻¹⁰) 

   =1·77 

Charge on the capacitor after 10 ns, 

q = ℰC(1-e-t/RC)

   =2*3·54x10⁻¹⁰(1-e-1·77) 

   =5·88x10⁻¹⁰ C 

The potential difference across the capacitor, V' =q/C 

→V' = 5·88x10⁻¹⁰/3.54x10⁻¹⁰ 

    =1.66 V 

Hence the electric field in the capacitor,

E =V'/d  

  =1.66/0.0001 V/m

  ≈1.7x10⁴ V/m                           





 

   68.  A parallel plate capacitor has a plate area of 20 cm², plate separation 1.0 mm and a dielectric slab of dielectric constant 5·0 filling up space between the plates. This capacitor is joined to a battery of emf 6·0 V through a 100 kΩ resistor. Find the energy of capacitor 8.9 µs after the connections are made.    


ANSWER: Given A =20 cm² =2x10⁻³ m², d =1·0 mm =1x10⁻³ m, K =5·0. Hence the capacitance, C =Kԑ₀A/d

→C =5*8.85x10⁻¹²*2x10⁻³/(1x10⁻³)

→C =88.5x10⁻¹² F

→C ≈8.9x10⁻¹¹ F

R =100 kΩ =1x10⁵ Ω

ℰ = 6·0 V

t =8·9 µs =8.9x10⁻⁶ s

t/RC =8·9x10⁻⁶/(1x10⁵*8·9x10⁻¹¹)

   =1

The charge on the capacitor after 8.9 µs,

q =ℰC(1 -e⁻¹)

  =6*8.9x10⁻¹¹*0.63 C

  =3.36x10⁻¹⁰ C 

Hence the energy of the capacitor at this instant,

U =q²/2C

=(3.36x10⁻¹⁰)²/(2*8.9x10⁻¹¹) J

=0.63x10⁻⁹ J

=6.3x10⁻¹⁰ J

        





 

   69.  A 100 µF capacitor is joined to a 24 V battery through a 1.0 MΩ resistor. Plot qualitative graphs (a) between current and time for the first 10 minutes and (b) between charge and time for the same period.      


ANSWER: Given C =100 µF =1x10⁻⁴ F, ℰ =24 V, R =1.0 MΩ =1x10⁶ Ω. 

RC =1x10⁶*1x10⁻⁴ s

    =100 s.

   At time t the charge on the capacitor,  

q = ℰC(1 -e-t/RC) 

→q =24*1x10⁻⁴(1- e-t/100) --- (i)

→q =24x10⁻⁴ -24x10⁻⁴e-t/100  


(a) Final time =10 min =600 s,

The current i =dq/dt 

→i = (-24x10⁻⁴e-t/100)*(-1/100) 

 =24x10⁻⁶*e-t/100 

 =24x10⁻⁶/et/100   

 At t =0, i =24x10⁻⁶ A 

At t =100 s, i =24x10⁻⁶*0.37 A

At t =200 s, i =24x10⁻⁶*0.14 A
At t =300 s, i =24x10⁻⁶*0.05 A
At t =400 s, i =24x10⁻⁶*0.02 A 
At t =500 s, i =24x10⁻⁶*0.006 A 
At t =600 s, i =24x10⁻⁶*0.002 A.
The time-current graph will be as below,
The graph for Q-69(a)

      

(b ) The on the capacitor at time t, 

q =24x10⁻⁴(1 -e-t/100) --(i)

At t =0, q = 0,

At t =100 s, q =24x10⁻⁴*0.63 C
At t =200 s, q =24x10⁻⁴*0.86 C
At t =300 s, q =24x10⁻⁴*0.95 C
At t =400 s, q =24x10⁻⁴*0.98 C
At t =500 s, q =24x10⁻⁴*0.994 C
At t =600 s, q =24x10⁻⁴*0.998 C
So at t =600 s the charge is very very close to the capacitance of the capacitor.
   The graph can be plotted as below:-
The graph for Q-32-69(b)

    

  



   70.  How many constants will elapse before the current in a charging RC circuit drops to half of its initial value? Answer the same question for a discharging RC circuit.     


ANSWER: For a charging circuit, the charge on the capacitor at the time t 

q =Q(1-e-t/RC) 

The current,   

i =dq/dt  

→i =-Qe-t/RC*(-1/RC)  

→i =Q/(RCet/RC) 

Current at t = 0,

i' =Q/RC          ------------- (i)

Let the current at time t' is i'/2, then

i'/2 =Q/(RCet'/RC) -------- (ii)  

Dividing (i) by (ii), 

2 =et'/RC   

→et'/RC = 2    

→t'/RC = ln 2 

→t' = 0.69*RC 

RC = time constant. So 0.69 time constants will elapse before the current in the circuit drops to half of its initial value. 


For the discharging circuit:- 

The charge on the capacitor at time t is, 

q = Qe-t/RC 

The current at time t, 

i =dq/dt =-Q/(RCet/RC)

The negative sign is for the direction compared to the charging circuit.

At time t = 0, i' =-Q/RC  -----(i) 

Let at time t', i =i'/2, then 

i'/2 =-Q/(RCet'/RC)   ---- (ii)  

Dividing (i) by (ii), 

2 = et'/RC 

→t'/RC = ln2 

→t' = (ln 2)*RC     

   =0.69*Time-constant  

So the 0.69 time constant will elapse before the current in the discharging circuit drops to half of its initial value.           

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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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CHAPTER- 5 - Newton's Laws of Motion


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CHAPTER- 4 - The Forces

The Forces-

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CHAPTER- 3 - Kinematics - Rest and Motion

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