# H C VERMA Solutions for Class 12-science Physics Chapter 10 - Electric Current in Conductors

## Chapter 10 - Electric Current in Conductors Exercise 198

The amount of charge passed in time t through a cross-section of a wire is

(a) Write the dimensional formulae for A, B and C.

(b) If the numerical value of A, B and C are 5, 3 and 1 respectively in SI units, find the value of current at

(a) [Q]=[A][T]2=[B][T]=[C]

[I][T]=[A][T]2=[B][T]=[C]

[A]=[IT-1]

[B]=[I]

[C]=[IT]

(b)

I=2At+B

I=2(5)(5)+3

I=53A

An electron gun emits electrons per second. What electric current does this correspond to?

The electric current existing in a discharge tube is 2.0 μA. How much charge is transferred across a cross-section of the tube in 5 minutes?

The current through a wire depends on time as

Where and. Find the charge crossed through the section of the wire in 10 seconds.

_{ }

_{}

A current of 1.0 A exists in a
copper wire of cross-section 1.0mm^{2}. Assuming one free electron
per atom calculate the drift speed of the free electrons in the wire. The
density of copper is 9000kg/m^{3}.

Now,

A wire of length 1m and radius 0.1mm has a resistance of 100Ω. Find the resistivity of the material.

Ω.m

A uniform wire of resistance 100Ω is melted and recast in a wire of length double that of the original. What would be the resistance of the wire?

Ω

Consider a wire of length 4m and
cross-sectional area 1mm^{2} carrying a current of 2A. If each cubic
meter of the material contains free electrons, find the
average time taken by an electron to cross the length of the wire.

m/s

Now,

What length of a copper wire of
cross-sectional area 0.01mm^{2}
will be needed to prepare a resistance of 1 kΩ? Ω-m.

Figure (32-E1) shows a conductor of length l having a circular cross-section. The radius of cross-section varies linearly from a to b. the resistivity of the material is ρ. Assuming that b-a<<l, find the resistance of the conductor.

Resistance of the small strip

-(i)

Now,

Differentiate with respect to x

Now,

A copper wire of radius 0.1mm and resistance 1kΩ is connected across a power supply of 20 V. (a) How many electrons are transferred per second between the supply and the wire at one end? (b) Write down the current density in the wire.

(a)

(b)

Calculate the electric field in a
copper wire of cross-sectional area 2.0mm^{2} carrying a current of
1A. The resistivity of copper=1.7×10^{-8}Ω-m.

A wire has a length of 2.0m and a resistance of 5.0Ω. Find the electric field existing inside the wire if it carries a current of 10A.

The resistances of an iron wire and a copper
wire at 20°C are 3.9Ω and 4.1 Ω
respectively. At what temperature will the resistances be equal? Temperature
coefficient of resistivity for iron is 5.0×10^{-3} K^{-1}
and for copper it is 4.0×10^{-3}
K^{-1}. Neglect any thermal expansion.

Resistance at temperature T is given by

Let, at temperature T

The current in a conductor and the potential difference across its ends are measured by an ammeter and a voltmeter. The meters draw negligible currents. The ammeter is accurate but the voltmeter has a zero error (that is, it doesn't read zero when no potential difference is applied). Calculate the zero error if the readings for two different conditions are 1.75 A, 14.4 V and 2.75 A, 22.4 V

Voltmeter reading=zero error + potential due to current

14.4=V+(1.75)R (i)

22.4=V+(2.75)R (ii)

Subtract (ii) from (i)

8=R

Put value in (i)

14.4=V+(1.75)8

V=0.4 volt

Figure (32-E2) shows an arrangement to measure the emf ε and internal resistance r of battery. The voltmeter has a very high resistance and ammeter also has some resistance. The voltmeter reads 1.52 V when the switch S is open. When the switch is closed the voltmeter reading drops to 1.45 V and ammeter reads 1.0 A. find the emf and internal resistance of the battery.

Initially, no current flows in circuit when switch is open.

So, emf of cell=voltmeter reading

ε=1.52V

Now, when switch is closed, cell is discharging

So, TPD=ε-ir

1.45=1.52-(1)(r)

r=0.07Ω

The potential difference between the terminals of a battery of emf 6.0 V and internal resistance 1 Ω drops to 5.8 V when connected across an external resistor. Find the resistance of an external resistor.

For battery

TPD=ε-ir

5.8=6-i(1)

i=0.2A

for external resistor

V=IR

5.8=(0.2)(R)

R=29Ω

## Chapter 10 - Electric Current in Conductors Exercise 199

The potential difference between the terminals of a 6.0 V battery is 7.2 V when it is being charged by a current of 2.0 A. What is the internal resistance of the battery?

TPD=ε+ir

7.2=6+(2)(r)

r=0.6Ω

The internal resistance of an accumulator battery of emf 6V is 10Ω when it is fully discharged. As the battery gets charged up, its internal resistance decreases to 1Ω. The battery in its completely discharged state is connected to a charger which maintains a constant potential difference of 9V. find the current through the battery. (a)Just after the connections are made and (b) After a long time when it is completely charged.

(a) Net emf across battery=9-6=3V,

R_{eq}=10Ω

current=3/10=0.3A

(b) Net emf across battery =9-6=3V

Resistance = 1Ω

Current=3/1=3A

Find the value of i_{1}/i_{2}
in figure (32-E3) if (a) R=0.1Ω (b)R=1Ω (c)
R=10Ω.Note from your answers that in
order to get more current from a combination of two batteries they should be
joined in parallel if the external resistance is small and in series if the
external resistance is large as compared to the internal resistances.

Cells are in series combination

So,

cells are in parallel combination

Divide (i) and (ii)

(a)

(b)

(c)

Consider identical cells, each of emf ε
and internal resistance r. Suppose n_{1 }cells are joined in series
to form a line and n_{2} such lines are connected in parallel. The
combination drives a current in an external resistance R. (a) Find the
current in external resistance. (b) Assuming that n_{1} and n_{2 }can
be continuously varied, find the relation between n_{1,} n_{2},
R and r for which the current in R is maximum.

For I to be maximum, will be minimum

To minimize,

Differentiate wrt

The battery of emf 100 V and a resistor of resistances 10kΩ are joined in series. This system is used as a source to supply current to an external resistance R. If R is not greater than 100Ω, the current through it is constant up to two significant digits. Find its value. This is the basic principle of a constant-current source

Emf=100V

R_{eq}= 10000+R

Current

For R=1Ω

For R=100Ω

So up to R=100Ω the current does not change up to 2 significant figures.

If the reading of ammeter A_{1}
in figure (32-E4) is 2.4 A, what will the ammeters A_{2} and A_{3}
read? Neglect the resistances of the ammeter.

Voltage drop access A_{1}=voltage
drop access A_{2}

I_{1}R_{1}=I_{2}R_{2}

(2.4)(20)=I_{2}(30)

I_{2}=1.6A

Reading in A_{3}=reading in
A_{1}+reading in A_{2}

=(2.4)+(1.6)

=4A

The resistance of the rheostat shown in figure (32-E5) is 30 Ω. Neglecting the meter resistance; find the minimum and maximum currents through the ammeter as the rheostat is varied.

Emf=5.5V

for minimum current

Now, for maximum current

Three bulbs, each having a resistance of 180Ω,are connected in parallel to an ideal battery of emf 60 V. find the current delivered by the battery when (a) all the bulbs are switched on, (b) two of the bulbs are switched on and (c) Only one bulb is switched on.

For each bulb

Voltage applied=60V

Resistance=180Ω

Current I=

(a) All bulbs switched on

I_{N}=I_{1}+I_{2}+I_{3}

I_{N}=

(b) Two bulbs switched on

I_{N}=I_{1}+I_{2}

=

=0.67A

(c) Only one bulb switched on

I_{N}==0.33A

Suppose you have three resistors of 20Ω, 50Ω and 100Ω. What minimum and maximum resistances can you obtain from these resistors?

For maximum resistance all will be in series combination

R_{eq}=20+50+100=170Ω

For minimum resistance all will be in parallel combination

A bulb is made using two filaments. A switch selects whether the filaments are used individually or in parallel. When used with a 15 V battery, the bulb can be operated at 5 W, 10 W or 15 W. What should be the resistances of the filaments?

When bulbs are used in parallel, the equivalent resistance will be less than their individual resistance

∴ two resistances are 45Ω and 22.5Ω

Figure (32-E6) shows a part of a circuit. If a current of 12 mA exists in the 5kΩ resistor, find the currents in the other three resistors. What is the potential difference between the points A and B?

Let I be the current in 20kΩ

So(12-I) will be current in 10kΩ

Now, V_{20k}_{Ω}=V_{10k}_{Ω}(parallel combination)

I(20)=(12-I)(10)

I=4mA

So,

I_{20k}_{Ω}=4mA

I_{10k}_{Ω}=8mA

I_{100k}_{Ω}=12mA

Equivalent resistance between AB

V_{AB}=I

=(12×10^{-3})(111.67×10^{3})

V_{AB}=1340V

An ideal battery sends a current of 5A in a resistor. When another resistor of value 10Ω is connected in parallel, the current through the battery is increased to 6A. Find the resistance of the first resistor.

Initially,

V=IR

V=5R -(i)

Now, when 10Ω is connected in parallel to R

V=IR_{eq}

From (i) and (ii)

R=2Ω

Find the equivalent resistance of the network shown in figure (32-E7) between the points a and b.

A wire of resistance 15.0Ω 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.

Resistance of each side of hexagon=15/6=2.5Ω

(a)

R_{AB}=

(b)

R_{AC}=

(c)

R_{AD}=

Consider the circuit shown in the figure (32-E8). Find the current through the 10Ω resistor when the switch S is (a) Open and (b) Closed.

(a) R_{eq}=10+20=30Ω

V=3 volt

V=IR

3=I(30)

I=0.1A

(b) R_{eq}=10Ω

V=IR

3=I(10)

I=0.3A

## Chapter 10 - Electric Current in Conductors Exercise 200

Find the currents through the three resistors shown in figure (32-E9).

Emf_{eq}=4-2=2V

R_{eq}=4+6=10Ω

Current

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

Potential at X

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

(a) For resistor a,

For resistor b,

(b)

For resistor a,

For resistor b,

Find the potential difference (V_{a}-V_{b}) in the circuits shown in
figure (32-E12).

(a) Both cells are in parallel combination

Internal resistance of cell will be

Equivalent Resistance of circuit

So, current

Voltage difference across ab

(b) The circuit can be redrawn as shown in figure

It is similar to above circuit in (a).

Hence same answer.

In the circuit shown in figure (32-E13), ε_{1}
= 3V, ε_{2} = 2V, ε_{3}
= 1V and r_{1}=r_{2}=r_{3}=1Ω. Find the potential difference between the points A and B and the
current through each branch.

KVL for loop 1

3-I_{1}(1)+I_{2}(1)-2=0

I_{1}-I_{2}=1 -(i)

KVL for loop 2

2-I_{2}(1)-( I_{1}+I_{2})(1)-1=0

I_{1}-2I_{2}=1 -(ii)

Solving (i) and (ii)

I_{1}=1A and I_{2}=0A

Now,

V_{AB}=TPD for cell of 2V

= ε-ir

= 2-(0)(1)

V_{AB }= 2V

Find the current through the 10Ω resistor shown in figure (32-E14)

Potential at point of X volt

Solving, X=3V

Current in 10Ω resistance=

Find the current in the three resistors shown in figure (32-E15).

Potential difference across each resistance is 0

So, current =0

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

The circuit is in wheat stone symmetry

So, for all values of R, current=0.

Find the equivalent resistance of the circuits shown in figure (32-E17) between the points a and b. Each resistor has a resistance r.

(a)

(b)

Find the current measured by the ammeter in the circuit shown in the figure (32-E18).

The 50R resistors are in wheat stone symmetry, so no current flows in them

## Chapter 10 - Electric Current in Conductors Exercise 201

Consider the circuits shown in (32-E19a). Find

(a) The current in the circuit,

(b) The potential drop across the 5Ω resistor,

(c) The potential drop across the 10Ω resistor.

(d) Answer the parts (a), (b) and (c) with reference to figure (32-E19b).

(a) (Emf)_{eq} = 12+6 = 18V

R_{eq} = 10+5 = 15Ω

Current I =

(b) Now,

(c)

(d) Circuit in figure (b) is same as in figure (a)

Twelve wires, each having equal resistance r, are joined to form a cube as shown in figure (32-E20). Find the equivalent resistance between the diagonally opposite points a and f.

Apply KVL for abcfa

Find the equivalent resistances of the networks shown in figure (32-E21) between the points a and b.

(a)

(b)

(c)

(d)

(e)

An infinite ladder is constructed with 1Ω and 2Ω resistors as shown in figure (32-E22).

(a) Find the effective resistance between the points A and B.

(b) Find the current that passes through the 2Ω resistor nearest to the battery.

(a)

Let equivalent resistance between AB be x

(b)

I=V/R=6/2=3A

Current in 2Ω

I_{2}_{Ω}=3/2=1.5A

The emf ε and the internal resistance r of the battery shown in figure (32-E23) are 4.3V and 1.0Ω respectively. The external resistance R is 50Ω. The resistances of the ammeter and voltmeter are 2.0Ω and 200Ω respectively.

(a) Find the readings of the two meters.

(b) The switch is thrown to the other side. What will be the readings of the two meters now?

(a)

Current, (Ammeter reading)

Now,50Ω and 200Ω are in parallel combination

So, current is divided in inverse Ratio

current in voltmeter=

= =0.02A

Voltmeter reading =(0.02)(200)=4V

(b)

Current, (Ammeter reading)

TPD across cell=voltmeter reading

Voltmeter reading= ε-Ir=4.3-0.08(1)=4.2V

A voltmeter of resistance 400Ω is used to measure the potential difference across the 100Ω resistor in the circuit shown in figure (32-E24).

(a) What will be the reading of voltmeter?

(b) What was the potential difference across 100Ω before the voltmeter was connected?

(a)

Since, 100Ω and 400Ω are in parallel combinationso, current is divided in inverse Ratio of resistances.

current in voltmeter I_{v}=

Voltmeter reading

=

=24V

(b)

So,

V_{100Ω}=IR=

The voltmeter shown in figure (32-E25) reads 18V across the 50Ω resistor. Find the resistance of the voltmeter.

I_{24Ω}=

I_{50Ω}=

I_{24}_{Ω}= I_{50Ω}+I_{v} [by KCL]

0.5=0.36+I_{v}

I_{v}=0.14A

For voltmeter,

V_{v}=I_{v}R_{v}

18=0.14R

R_{v}≃130Ω

A voltmeter consists of a 25Ω coil connected in series with a 575Ω resistor. The coil takes 10 mA for full scale deflection. What maximum potential difference can be measured on this voltmeter?

R_{eq}=25+575=600Ω;I=10mA

V=IR

V=

V=6 volt

An ammeter is to be constructed which can read currents up to 2.0A. If the coil has a resistance of 25Ω and takes 1mA for full-scale deflection, what should be the resistance of the shunt used?

I_{g}R_{g}=(I-I_{g})R_{s}

(10^{-3})(25)=(2-10^{-3})R_{s}

R_{s}=1.25×10^{-2}R

A voltmeter coil has resistance 50.0Ω and resistor of 1.15kΩ 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 2.0A, what should be the resistance of the shunt used?

V=I_{g}(R_{g}+R)

12=I_{g}(50+1150)

I_{g}=0.01A

Now, I_{g}R_{g}=(I-I_{g})R_{s}

(0.01)(50)=(2-0.01)R_{s}

R_{s}=0.25R

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

At null deflection, it will be in wheat-stone symmetry.

Let l be the balancing length from end A

## Chapter 10 - Electric Current in Conductors Exercise 202

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

A 6 volt battery of negligible internal resistance is connected across a uniform wire AB of length 100cm. The positive terminal of another battery of emf 4V and internal resistance 1Ω is joined to the point A as shown in the figure (32-E28). Take the potential at B to be zero.

(a) What are the potentials at the 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 4V battery is replaced by 7.5V battery, what would be the answers of parts (a) and (b)?

(a)

V_{A}=6V

V_{C}=2V

(b) Potential across AD=Potential across AC=4V

⇒ Potential across DB=2V

(c) No current will flow as V_{C}=V_{D}

(d) Potential difference across AC=potential at A-potential at C

7.5=6-V_{c}

V_{C}=-1.5V

No points exist on wire AB as potential difference. Wire isalways greater than or equal to zero.

Consider the potentiometer circuit arranged as in figure (32-E29). The potentiometer wire is 600cm long.

(a) At what distance from the 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 560cm from A, what will be the current in the galvanometer?

(a) R_{eq}_{}for primary circuit=r+15r=16r

Current=

V_{AB}=IR

V_{AB}=

Potential gradient

Let balancing length be l from end A.

Emf=xl

(b)

Resistance of 560cm wire =

Resistance of 40cm wire=15r-14r=r

Nodal analysis at X

Potential difference across resistance r

Find the charge on the capacitor shown in figure (32-E30).

In steady state, no current flows through capacitor

R_{eq}=10+20=30Ω

V=IR

2=I(30)

I=

Voltage across capacitor=voltage across 10Ω

=IR

=

=

Q=CV

Q=6

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

Nodal analysis at X

For 20Ω resistor

Find the chargers on the four capacitors of capacitances 1 μF, 2 μF, 3 μF and 4 μF shown in figure (32-E32).

1Ω and 2Ω resistors are in series combination

So, potential across 1=potential for 1Ω

=1×2=2volt

So, charge=CV=(1)(2)=2μC

Potential across 2μF=potential diff for 2Ω

=2×2=4V

So, charge=CV=(2)(4)=8μC

Now, for 3Ω-3Ω resistors,

Potential drop across each resistance is 3V

Charge on 3μF capacitor=CV

=(3μF)(3)=9μC

Charge on 4μF capacitor=CV

=(4μF)(3)=12μC

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

No current flows in steady state

Nodal analysis at X

(X-100)3+(X-100)3+(X-0)1+(X-0)1=0

8X=600

X=75

V_{A}-V_{B}=100-75

=25V

V_{B}-V_{C}=75-0

=75V

A capacitance C, a resistance R and an emf ε 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 and

(f) The power converted into heat.

(a) Maximum potential difference across resistor=ε

When charge on capacitor is 0

(b) At t=0, capacitor acts as short circuit

V_{R}=ε=IR

(c) At t=∞, capacitor acts as open circuit

V_{max}_{-cap.}=ε

(d) (at t=∞,V_{cap}=ε)

(e) Power=εI

P_{max}=εI_{max}

P_{max}=

(f) Power dissipated=

A parallel-plate capacitor with plate area 20cm^{2} and plate separation 12.0mm is connected to a battery. The
resistance of the circuit is 10 kΩ. Find the
time constant of the circuit.

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

At 20 μF capacitor is joined to a battery of emf 6.0V through a resistance of 100Ω. Find the charge on the capacitor 2.0 ms after the connections are made.

Q = Q_{0}(1-)

= (20

= 12

= 76μC

The plates of a capacitor of capacitance 10μF, charged to 60μC, are joined together by a wire of resistance 10Ω 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

Q = Q_{0}

Q = 60

Q = 6

(a) t = 0

Q = 6

(b) t = 30μs

Q = 6

Q = 44μC

(c) t = 120μs

Q = (6

= 18μC

(d) t = 1ms

Q = (6

= 0.003μC

A capacitor of capacitance 8.0 μF is connected to a battery of emf 6.0V through a resistance of 24Ω. Find the current in the circuit

(a) Just after the connections are made and

(b) One time constant after the connections are made

(a)

(b)

## Chapter 10 - Electric Current in Conductors Exercise 203

A parallel-plate capacitor has plate area 40cm^{2} and separation between the plates 0.10mm is connected to a
battery of emf 2.0V through a 16Ω resistor. Find the electric field in the capacitor 10ns after the
connections are made.

Capacitance

Time constant

Electric field in capacitor =

)

A parallel-plate capacitor has plate area 20cm^{2}, plate separation 1.0mm and a dielectric slab of dielectric
constant 5.0 filling up the space between the plates. This capacitor is
joined to a battery of emf 6.0v through a 100kΩ resistor. Find the energy of the capacitor 8.9μs after the connections are made.

Capacitance

charge

time constant =RC

Energy stored=

A 100μF capacitor is joined to a 24V battery through a 1.0MΩ resistor. Plot qualitative graphs

(a) Between current and time for the first 10 minutes and

(b) Between charge and time for the same period

(a)

at t=0,

At t=10×60=600sec.

(b)

at

at

How many times 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.

For discharging, so, answer will be same.

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

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

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

Power=

VI=

I=

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

Energy stored,

Rate of energy stored=

For maximum rate of energy stored

Put in (i)

A capacitor of capacitance 12.0μF is connected to a battery of emf 6.00V and internal resistance1.00Ω through resistance less leads. 12.0μs after the connections are made, what will be

(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.

(a)

(b) P=VI

=(6)(2.21)

P=13.25W

(c) Power dissipated=I^{2}R

=(2.21)^{2}(1)

=4.87W

(d) Energy stored

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 and also by finding the decrease in the energy stored in capacitor.

Heat dissipated

By energy method:-

Heat dissipated= energy stored at t=0-energy stored at time t

=

By evaluating , 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.

Current in circuit at time t,

Heat dissipated

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

Capacitance

resistance of dielectric material,

time constant

It is independent of geometric parameters

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

So, charge on each capacitor=18.4/2=9.2μC

The Switch S shown in figure (32-E35) is kept closed for a long time and is then opened at. Find the current in the middle 10Ω resistor at

Initially, the capacitor was charged completely

Potential across capacitor=Potential across 10Ω

charge

=

Now, capacitor is discharged

A capacitor of capacitance 100μF is connected across a battery of emf 6.0 V through a resistance of 20kΩ for 4.0s. The battery is then replaced by a thick wire. What will be the charge on the capacitor 4.0s after the battery is disconnected?

During charging, charge developed in 4sec

Now discharging

Consider the situation shown in
figure (32-E36). The switch is closed at when the capacitors are
uncharged. Find the charge on the capacitor C_{1} as a function of
time t.

Equivalent capacitance

A capacitor of capacitance C is given a charge Q. At, it is connected to an uncharged capacitor of equal capacitance through a resistance R. Find the charge on second capacitor as a function of time.

Let charge on capacitor B at time t be q

Applying KVL,

A capacitor of capacitance C is given a charge Q. At, it is connected to an ideal battery of emf ε through a resistance R. Find the charge on capacitor at time t.

Initial charge given to the capacitor=Q

When the capacitor is connected by a battery, it will charge through the battery. The initial charge will also delay.

Growth of charge by battery

Delay of charge through capacitor

Net charge on capacitor at time

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