Thursday, June 17, 2021

H C Verma solutions, ELECTRIC CURRENT IN CONDUCTORS, Exercises, Q71-Q80, Chapter-32, Concepts of Physics, Part-II

Electric Current in Conductors


Exercises, Q71 - Q80


   71.  How many time constants will elapse before the charge on a capacitor falls to 0.1% of its maximum value in a discharging RC circuit?  



ANSWER: Suppose the required time is t'.

Maximum charge on a capacitor  =Q 

At time t the charge on the capacitor is,

q =Qe-t/RC  

Here q =(0·1/100)Q =0·001Q 

Thus, 0·001Q =Qe-t'/RC 

→e-t'/RC =0·001

→-t'/RC = ln(0·001)

→t' = 6·9 RC. 

                          




 

   72.  How many time constants will elapse before the energy stored in a capacitor reaches half of its equilibrium value in a charging RC circuit?  



ANSWER: Let the equilibrium value of charge on the capacitor =Q. 

  The energy stored in the capacitor at equilibrium, U =Q²/2C, where C is the capacitance of the capacitor.  

    Suppose the charge on the capacitor at the time t when the energy stored is half of the equilibrium value is q. At this time, the energy stored U' =q²/2C.  

    But, U' = U/2

→q²/2C =Q²/4C.  

→q = Q/√2

   The number of time constants elapsed to reach this point, n =  t/RC.

For a charging circuit,  

→q = Q(1 -e-t/RC) =Q(1 -e-n)

→Q/√2 =Q(1 -1/en)  

→1/√2 = 1 -1/en

→1/en = (√2-1)/√2   

→en =√2/(√2-1) = 3·414

→n =ln 3·414 = 1·23.

                




 

   73.  How many time constants will elapse before the power delivered by the battery drops to half of its maximum in an RC circuit?   



ANSWER: In a charging circuit, the charge on the capacitor at time t is given as,      

q =ℰC(1 -e-t/RC)

The current at time t,

i = dq/dt =  (ℰ/R)e-t/RC

   It is clear that the current decreases as time increases. The maximum current is at t=0,

I = ℰ/R,

    The maximum power delivered,

 P =ℰI =ℰ²/R

    Suppose at time t', the power delivered is P/2 and the current in the circuit is i'. The number of time constants, n =t'/RC.  

So, i' = (ℰ/R)e-n 

Now power delivered by the battery,

=ℰi' 

So, P/2 = ℰi'    

→ℰ²/2R = (ℰ²/R)*e-n  

→en = 2

→n = ln 2 = 0·69. 

                                  





   74.  A capacitor of capacitance C is connected to a battery of emf ℰ at t =0 through a resistance R. Find the maximum rate at which the energy is stored in the capacitor. When does the rate have this maximum value?  



ANSWER: The energy stored in a capacitor, U = q²/2C.  

  At time t, q =ℰC(1 -e-t/RC) 

Hence U =½ℰ²C(1 -e-t/RC)²  

Rate of storage of energy, r =dU/dt 

=½*2ℰ²C(1 -e-t/RC)*{-e-t/RC*(-1/RC)}

=(ℰ²/R)e-t/RC*(1-e-t/RC) 

For r to be maximum or minimum,

 dr/dt =0 

→d²U/dt² =0, 

→(ℰ²/R){e-t/RC*e-t/RC/RC +(1-e-t/RC)e-t/RC*(-1/RC)}=0

→(1/RC){e-2t/RC -e-t/RC(1-e-t/RC) =0

→e-t/RC{e-t/RC -1 +e-t/RC} =0

→e-t/RC{2e-t/RC -1} =0, 

Either, e-t/RC =0, 

This gives t =infinity, it is the condition for minimum because the current then is negligible. 

Or, 2e-t/RC =1 

→e-t/RC = 1/2, 

→-t/RC = ln (1/2) = -0.69  

→t = 0.69RC 

Now the maximum rate of storage of energy,                                              

r =(ℰ²/R)e-0.69*(1-e-0.69) 

 =(ℰ²/R)*(0.5)²  

 =ℰ²/4R. 

As we have calculated that this maximum rate is at a time, 

t =0.69RC =RC.ln2 






 

   75.  A capacitor of capacitance 12.0 µF is connected to a battery of emf 6.00 V and internal resistance 1.00 Ω through resistance-less leads. 12.0 µs after the connections are made, what will be the (a) the current in the circuit, (b) the power delivered by the battery, (c) the power dissipated in heat, and (d) the rate at which the energy stored in the capacitor is increasing.       


ANSWER: ℰ =6.0 V, R =1 Ω, C =12 µF, t =12.0 µs. 

(a) The charge on the capacitor at time t is,   

q =ℰC(1 -e-t/RC), where Q is the charge at equilibrium.

Hence the current in the circuit,

i =dq/dt

→i =(ℰ/R)e-t/RC 

At t =12 µs,

t/RC =12/(1*12) =1, Thus the current at this time =(6/1)*e⁻¹ =6/e =2·21 A.


(b)  The power delivered by the battery at this instant, P = ℰi =6*2·21 W 

→P =13·26 W.   


 (c) As we have derived in Q-74, the rate of energy storage

r =(ℰ²/R)e-t/RC*(1-e-t/RC)  

 =(6²/1)*(1/e)*(1-1/e) 

=8·38 W ----------------- (i) 

So the power being stored =8·38 W 

The power dissipated in heat, 

=13·26 -8·38 =4·88 W.        

Alternatively, 

   The power dissipated in heat is through the resistance of 1 Ω, 

=i²R 

=(2·21)²*1 W

= 4·88 W.   


(d) As calculated in (c) (i) the rate is 8.38 W.   





 

   76.  A capacitance C charged to a potential difference V is discharged by connecting its plates through a resistance R. Find the heat dissipated in one time constant after the connections are made. Do this by calculating ∫i²Rdt and also by finding the decrease in the energy stored in the capacitor.  



ANSWER: The energy stored in the capacitor initially U =½CV². 

Charge on the capacitor after time t, 

q =CVe-t/RC. 

Now the energy remaining, 

U' =q²/2C   

  =(C²V²e-2t/RC)/2C 

  =½CV²e-2t/R 

After one time constant, t/RC =1 

→U' =½CV²e⁻²

Hence the heat dissipated after one time constant =U -U'    

=½CV² -½CV²e⁻²

=½(1 -e⁻²)CV². 


Alternative method:-  

Current in the circuit at time t,

 i =dq/dt 

  =d(CVe-t/RC)/dt 

  = (-1/RC)CVe-t/RC 

  =-(V/R)e-t/RC 

Hence the heat dissipated in circuit, 

H =∫i²Rdt 

   =(V/R)²∫(e-t/RC)²R dt 

  ={(V/R)²*R/(-2/RC)}[e-2t/RC]

  =-½CV²[e-2t/RC] 

The limit of integration is from t =0 to t =RC. Hence, 

H = -½CV²[e⁻² -1] 

 =½(1 -e⁻²)CV².                                     




 

   77.  By evaluating ∫i²Rdt, show that when a capacitor is charged by connecting it to a battery through a resistor, the energy dissipated as heat equals the energy stored in the capacitor.  


ANSWER: When a capacitor is fully charged, the energy stored in it is,

U =½Cℰ²  ----------------------(i).

  For a charging capacitor at any time t, the charge on the capacitor 

q = ℰC(1 -e-t/RC)

Current in the circuit, i =dq/dt 

→i =(ℰ/R)e-t/RC 

The energy dissipated as heat through the resistor, U(H) =∫i²Rdt

=(ℰ²/R)∫e-2t/RCdt 

=-(RC/2)(ℰ²/R)[e-2t/RC] 

=-½Cℰ²[e-2t/C] 

Putting the limits of integration from t =0 to t = ∞ (because theoriticaaly the capacitor will be fully charged at time infinity though practically after a long time), 

U(H) =-½Cℰ²[e-2∞/RC -1]

        =-½Cℰ²(0 -1)

        =½Cℰ²      

It is the same as energy stored in the capacitor as in (i) = U,

Hence proved.                                   





 

   78.  A parallel-plate capacitor is filled with a dielectric material having a resistivity ⍴ and dielectric constant K. The capacitor is charged and disconnected from the charging source. The capacitor is already discharged through the dielectric. Show that the time constant of the discharge is independent of all geometrical parameters like the plate area or separation between the plates. Find this time constant.


ANSWER: Let the plate area of the capacitor = A and the distance between the plates =d. The capacitance of this capacitor,

C =KԑₒA/d 

The dielectric is the resistor through which the capacitor is discharged. It's resistance,  

R =⍴(d/A) 

The time constant of the discharge, 

𝛕 = RC 

  = ⍴(d/A)*KԑₒA/d 

  = ԑₒ⍴K.  

Clearly the time constant is independent of all geometrical parameters.                         




 

   79.  Find the charge on each of the capacitors 0.20 ms after the switch S is closed in figure (32-E34).  
The figure for Q-79


ANSWER: Two equal capacitors (each C =2.0 µF) are connected in parallel, hence the equivalent capacitance, 

C' = C +C =2C =4.0 µF

ℰ =6.0 V, R =25 Ω.

The charge on the equivalent capacitor after time t =0.20 ms,

Q' =ℰC(1 -e-t/RC) 

   =6*(4 µF)*(1 -e-{2.0x10⁻⁴/(25x4x10⁻⁶)} µC

  =24*(1 -e⁻²) µC

  =24(1 -0.14) µC 

  =20.64 µC  

 

If the charge on each capacitor = q, then,

q =CV, where V is the potential difference across the capacitors. 

→C =q/V

And, Q' =C'V, 

→C' =Q'/V, 

But, C' = 2C

→Q'/V =2q/V

→2q =Q'

→q =Q'/2

     =20.64/2 µC

     =10.32 µC.                      





 

   80.  The switch S shown in figure (32-E35) is kept closed for a long time and is then opened at t = 0. Find the current in the middle 10 Ω resistor at t =1·0 ms.   
The figure for Q-80


ANSWER: After a long time the capacitor is fully charged. When the switch is closed further, the current still flows through both of the resistors.   

   Suppose the potential difference across the capacitor and the middle 10 Ω resistor =V.

  The current in the middle resistor,

i =V/10. 

This will also be the current in the lower loop because no current passes through the capacitor. The resistance of the lower loop =10 Ω +10 Ω =20 Ω. (Series connection)

Hence the current in this circuit =12/20 =0·6 A 

So V/10 =0.6 

→V =6 Volts      

 The charge on the capacitor, 

Q = (25 µF)*(6 V) =150 µC. 

     When the switch is opened, the battery and lower 10 Ω resistor are out of the circuit. The charged capacitor is now being discharged through the middle 10 Ω resistor. For a discharging capacitor, charge remaining on the capacitor, 

q =Qe-t/RC

The current in the resistor at time t, 

i =dq/dt

  =-(Q/RC)e-t/RC 

At t =1.0 ms =0.001 s

t/RC =0.001/(10*25x10⁻⁶) =4 

i =-{150 µC/(10*25 µF)}e-4

  =-0.60*0.018 A   

  =-0.011 A  

  =-11 mA. 

The negative sign shows the current opposite of the charging direction.          

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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 → OBJECTIVE-II

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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


Click here for → QUESTIONS FOR SHORT ANSWER

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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