Q: Three ammeters −1,2 and 3 have different internal resistances r1,r2 and r3 respectively. Internal resistance r1 is known but r2 and r3 are unknown. The angular deflection of pointer in each ammeter is proportional to the current. Initially, the three ammeters were connected in series to a voltage source (Fig. a) and deflections for the three ammeters were θ1,θ2 and θ3 respectively. The three ammeters were then connected in parallel to the same voltage source (Fig. b). This time the deflections were observed to be θ1′,θ2′ and θ3′ respectively. (a) Find r2 and r3. (b) If θ2=θ3 but θ3′>θ2′ then which one is larger r2 or r3 ?

(a) In series current i=k1θ1=k2θ2=k2θ3 In parallel V1k1θ1′⋅r1=V2=V3=k2θ2′r2=k3θ3′r3

Q: The capacitor A shown in Fig. has a capacitance C1=3μF. The dielectric filled in it has a breakdown voltage of 40 V and it has a resistance of 3MΩ. The capacitor B has a capacitance of C2=2μF and dielectric in it has a resistance of 2MΩ. Breakdown voltage for B is 50 V. The switch is closed at t=0. Will there be breakdown of any capacitor after the switch is closed? If yes, which will breakdown first and at what time?

The final voltage across A will be V1=C1+C2C2V=3+22×100=40 V and across B it will be V2=C1+C2C1⋅V=3+23×100=60 V Obviously, capacitor B will breakdown first as soon as voltage across it reaches 50 V (at that time voltage across A will be less than 40 V ] The effective capacitance of the circuit is C=C1+C2C1C2=3+23×2=56μF The effective resistance is R=3+2=5MΩ. Time constant of the circuit τ=RC=6sec Charge on capacitors as a function of time qq0q=q0[1−e−t/τ]=CV=56μF×100 V=120μC=(120μC)(1−e−t/τ) Breakdown of B will take place when Q=2μF×50 V=100μC 100μC=(120μC)(1−e−t/τ) ⇒65=1−e−t/τ ⇒e−t/τ=61 τt=ln6 t=τℓn6=6ln6sec.

Q: A parallel plate capacitor has its two plates connected to an ideal spring of force constant K. Relaxed length of spring is L and it is made of non conducting material. The area of each plate is A. The capacitor has a charge q0 on it. To discharge the capacitor through the resistance R, switch S is closed. If the time constant of the circuit is very large and discharge process is very slow, how much heat will be dissipated in the resistance? Assume that there is no friction and the plates always remains parallel to each other.

Force between capacitor plates is F=2ϵ0Aq02 The spring must be compressed by x such that kx=F ⇒x=2A∈0kq02 ∴ Separation between plates d=L−x=L−2A∈0kq02 Capacitance has initial value of C0=d∈0A Energy stored in the capacitor is U0=2C0q02=2ϵ0Aq02d=2ϵ0Aq02(L−2Aϵ0kq02)=2ϵ0Aq02L−4ϵ02A2kq04 Energy stored in the spring is Us=21kx2=8ϵ02A2kq04 The entire energy stored in the capacitor and the spring gets dissipated as heat in the resistor. Heat =U0+Us=2ϵ0Aq02L−8∈02A2kq04

Q: A charged capacitor (C1=3μF) is getting discharged in the circuit shown. When the current I was observed to be 2.5 A, switch ' S ' was opened. Determine the amount of heat that will be liberated in the circuit after ' S ' is opened.

Just before the switch is opened VR=IR=10Volt. ∴ p.d. across C1 at this instant V0=VR=10 Volt. Energy stored in C1 at this instant is U1=21C1V02=21×3×10−6×102=150μJ. Charge on C1 at this instant Q0=3×10−6×10=30μC Now this charge gets shared between C1 and C2 so that p.d across both of them becomes equal. Final common p.d V=C1+C2Q0=4μF30μC=7.5 Volt. ∴ Final energy stored in the capacitor system U2=21(C1+C2)V2=21×4×10−6×(7.5)2=28.13μJ ∴ Heat liberated = Energy lost =150−28.13=121.87μJ

Q: Consider two circuits given below. When switches S1 and S2 are closed at t=0, the charge on capacitors C1 and C2 change with time as shown in graph 1 and 2 respectively. The current i1 and i2, in the two circuits change as shown in graph 1′&2′ respectively. Write the variation of current in the second circuit as function of time after the switch is closed at t=0.e−1=0.37

q0=C1V1=200μC Maximum charge on C2 is (1−e−1)q0=0.63×200=126μC ∴ 2V2i0=126μC⇒C2=3μF=R1V1=0.2 A Maximum current in the second circuit is 200126×0.20.126AR2=R2V2=R242=31000Ω Time constant of the second circuit is τ2=R1C2=31000×3μs ii=100μs=10−3 s=0.126e−t/τ2=126e−1000t [ t is in second ]

Question 1

An ohm-meter is a device that measures an unknown resistance. A simple ohm-meter can be constructed using a galvanometer as shown in the figure. The cell used in the circuit has emf E=20 volt. The full scale deflection current and resistance of the galvanometer are 2 mA and 20Ω respectively. R0​ is a fixed resistance and R is the unknown resistance whose value is directly given by the galvanometer scale. The galvanometer scale is shown in figure. When an unknown resistance R is placed in the circuit, the pointer deflects by θ=90∘. Find R.

Accepted Answer

The ohm meter reads R=0Ω when full scale deflection current passes through it.
∴∴​2 mAR0​​=R0​+20Ω20 volt ​=9980Ω​
When R=∞, current in the galvanometer is zero and there is no deflection.
For θ=90∘, current through the galvanometer is
∴⇒⇒​R​i9980+20+R20 volt ​13333=3333Ω​=120∘2 mA​×90∘=1.5 mA=1.5×10−13=10,000+R​

Question 2

Three ammeters −1,2 and 3 have different internal resistances r1​,r2​ and r3​ respectively. Internal resistance r1​ is known but r2​ and r3​ are unknown. The angular deflection of pointer in each ammeter is proportional to the current. Initially, the three ammeters were connected in series to a voltage source (Fig. a) and deflections for the three ammeters were θ1​,θ2​ and θ3​ respectively. The three ammeters were then connected in parallel to the same voltage source (Fig. b). This time the deflections were observed to be θ1′​,θ2′​ and θ3′​ respectively.
(a) Find r2​ and r3​.
(b) If θ2​=θ3​ but θ3′​>θ2′​ then which one is larger r2​ or r3​ ?

Accepted Answer

(a) In series
current
i=k1​θ1​=k2​θ2​=k2​θ3​
In parallel
V1​k1​θ1′​⋅r1​​=V2​=V3​=k2​θ2′​r2​=k3​θ3′​r3​​

Question 3

The capacitor A shown in Fig. has a capacitance C1​=3μF. The dielectric filled in it has a breakdown voltage of 40 V and it has a resistance of 3MΩ. The capacitor B has a capacitance of C2​=2μF and dielectric in it has a resistance of 2MΩ. Breakdown voltage for B is 50 V. The switch is closed at t=0. Will there be breakdown of any capacitor after the switch is closed? If yes, which will breakdown first and at what time?

Accepted Answer

The final voltage across A will be
V1​=C1​+C2​C2​​V=3+22​×100=40 V
and across B it will be
V2​=C1​+C2​C1​​⋅V=3+23​×100=60 V
Obviously, capacitor B will breakdown first as soon as voltage across it reaches 50 V (at that time voltage across A will be less than 40 V ]
The effective capacitance of the circuit is
C=C1​+C2​C1​C2​​=3+23×2​=56​μF
The effective resistance is
R=3+2=5MΩ. 
Time constant of the circuit τ=RC=6sec
Charge on capacitors as a function of time
qq0​q​=q0​[1−e−t/τ]=CV=56​μF×100 V=120μC=(120μC)(1−e−t/τ)​
Breakdown of B will take place when
Q=2μF×50 V=100μC
100μC=(120μC)(1−e−t/τ)
⇒65​=1−e−t/τ
⇒e−t/τ=61​
τt​=ln6
t=τℓn6=6ln6sec.

Question 4

In the circuit shown in fig. the switch is kept closed in position 1 for a long time. At time t=0 the switch is moved to position 2. Write the dependence of voltage (VC​) across the capacitor as a function of time (t). Take VC​ to be positive when plate a is positive.
Given: R1​=20Ω,R2​=60Ω,R3​=400kΩ,V1​=40 V, V2​=90 V and C=0.5μF.

Accepted Answer

When switch is in position 1 , potential difference across R2​ is
V=R1​+R2​R2​​V1​=20+6060​×40=30 volt. 
This is also the potential difference across the capacitor with plate b positive. Charge on capacitor is
q0​=CV=(0.5μF)(30 V)=15μC. 
As per sign convention given in the problem, initial charge (q0​) on the capacitor is negative Let charge on it be q at time t. Then
R3​dtdq​+Cq​R3​dtdq​∫q0​q​V2​−Cq​dq​​=V2​=V2​−Cq​=R3​1​∫0t​dt​
⇒[ln(V2​−Cq​)]q0​q​=−R3​Ct​
⇒ln(V2​−Cq​)−ln(V2​−Cq0​​)=−R3​Ct​
⇒V2​−q0​/CV2​−q/C​=e−t/R3​C
⇒V2​−Cq​=(V2​−Cq0​​)e−R3​Ct​
⇒q=C[V2​−(V2​−Cq0​​)e−t/R3​c]
∴q=CV2​−(CV2​−q0​)e−t/R3​C
CV2​=0.5μF×90 V=45μC
q0​=−15μC
R3​C=(400kΩ)(0.5μF)=0.2 s
R3​C1​=0.21​=5
q=[45−60e−5t]μC.
∴ Potential difference across C is
V=Cq​=(90−120e−5t) volt.

Question 5

A parallel plate capacitor has its two plates connected to an ideal spring of force constant K. Relaxed length of spring is L and it is made of non conducting material. The area of each plate is A. The capacitor has a charge q0​ on it. To discharge the capacitor through the resistance R, switch S is closed.
If the time constant of the circuit is very large and discharge process is very slow, how much heat will be dissipated in the resistance? Assume that there is no friction and the plates always remains parallel to each other.

Accepted Answer

Force between capacitor plates is
F=2ϵ0​Aq02​​
The spring must be compressed by x such that kx=F
⇒x=2A∈0​kq02​​
∴ Separation between plates d=L−x=L−2A∈0​kq02​​
Capacitance has initial value of C0​=d∈0​A​
Energy stored in the capacitor is
U0​​=2C0​q02​​=2ϵ0​Aq02​d​=2ϵ0​Aq02​​(L−2Aϵ0​kq02​​)=2ϵ0​Aq02​L​−4ϵ02​A2kq04​​​
Energy stored in the spring is
Us​=21​kx2=8ϵ02​A2kq04​​
The entire energy stored in the capacitor and the spring gets dissipated as heat in the resistor.
 Heat =U0​+Us​=2ϵ0​Aq02​L​−8∈02​A2kq04​​

Question 6

In the last problem calculate the amount of heat dissipated in the resistance assuming that the time constant of the circuit is very small and the discharge process is almost instantaneous.

Accepted Answer

The spring will get no time to expand during the discharge and the heat dissipated will be =U0​
=2ϵ0​Aq02​L​−4ϵ02​A2kq04​​

Question 7

A charged capacitor (C1​=3μF) is getting discharged in the circuit shown. When the current I was observed to be 2.5 A, switch ' S ' was opened. Determine the amount of heat that will be liberated in the circuit after ' S ' is opened.

Accepted Answer

Just before the switch is opened
VR​=IR=10Volt.
∴ p.d. across C1​ at this instant
V0​=VR​=10 Volt. 
Energy stored in C1​ at this instant is
U1​=21​C1​V02​=21​×3×10−6×102=150μJ. 
Charge on C1​ at this instant
Q0​​=3×10−6×10=30μC​
Now this charge gets shared between C1​ and C2​ so that p.d across both of them becomes equal.
Final common p.d
V=C1​+C2​Q0​​=4μF30μC​=7.5 Volt. 
∴ Final energy stored in the capacitor system
U2​​=21​(C1​+C2​)V2=21​×4×10−6×(7.5)2=28.13μJ​
∴ Heat liberated = Energy lost
=150−28.13=121.87μJ

Question 8

Consider two circuits given below. When switches S1​ and S2​ are closed at t=0, the charge on capacitors C1​ and C2​ change with time as shown in graph 1 and 2 respectively. The current i1​ and i2​, in the two circuits change as shown in graph 1′&2′ respectively. Write the variation of current in the second circuit as function of time after the switch is closed at t=0.e−1=0.37

Accepted Answer

q0​=C1​V1​=200μC
Maximum charge on C2​ is (1−e−1)q0​=0.63×200=126μC
∴
2​V2​i0​​=126μC⇒C2​=3μF=R1​V1​​=0.2 A​
Maximum current in the second circuit is
200126​×0.20.126AR2​​=R2​V2​​=R2​42​=31000​Ω​
Time constant of the second circuit is τ2​=R1​C2​=31000​×3μs
ii​=100μs=10−3 s=0.126e−t/τ2​=126e−1000t​
[ t is in second ]

Question 9

In the circuit shown in Fig. the capacitor is initially uncharged. Two way switch (s) is placed in position 1 for a very short interval of time (Δt) and then is moved to position 2 . The switch is held in position 2 for equally short interval Δt and is then moved back to position 1 . The process is repeated a large number of times until the charge on the capacitor stops changing. Find this final a value of charge on the capacitor. [Assume that in each contact the charge on the capacitor changes by a very small amount]

Accepted Answer

Initially when ' S ' is in position 1 or 2 the capacitor gets charged. This continues till p.d across capacitor becomes 3.5 volt.
Now, when in position 1 , the current is
i=1Ω0.5 V​=0.5 A. 
This current discharges the capacitor. Its charge gets reduced by Δq=iΔt=0.5Δt
When in position 2 , the current is once again i=1Ω0.5 V​=0.5 A but this time it charges the capacitor.
Charge gained in time Δt is
Δq=0.5Δt
Therefore, once the capacitor gets charged to 3.5 V, charge on it will become almost constant.

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Solutions for Problems in Physics II for IIT JEE by Shashi Bhushan Tiwari

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