Sastry CBSE

Electrostatic Potential and Capacitance Important Questions for CBSE Class 12 Physics Capacitance

1.Conductors and Insulators Conductor contains a large number of free charge carriers to conduct electricity while insulator does not contain any free charge carriers to conduct electricity.
Examples of conductors are metals and graphite.
Examples of insulators are plastic rod and nylon.
NOTE Inside a conductor, the electrostatic field is zero.

2.Free Charges and Bound Charges inside a Conductor
(i)In a metal, the outer (valence) electrons are free to move. These electrons are free for moving within the metal but not free to leave the metal. These free electrons are free charges inside a conductor and are cause of conduct of electricity by conductors.
(ii)The bound charges are those positive ions which are made up of the nuclei and the bound electrons remain in their fixed positions.
Some important results regarding electrostatics of conductors are given as below:
(i) Inside a conductor, the electric field is zero.
(ii)The interior of a conductor can have no excess charge in static situation.
(iii)Electric field just outside a charged conductor is perpendicular to the surface of the conductorat every point.

3.Dielectric and Electric Polarisation
(i) In an External Electric Field When a conductor is placed in an external electric field, the free charge carriers adjust itself in such a way that the electric field due to induced charges and external field cancel each other and the net field inside the conductor is zero.
In case of dielectric however, the opposing field so induced does not exactly cancel the external field.

(ii)A net dipole moment is developed by an external field in either case, whether polar or non-polar dielectric. The dipole moment per unit volume is called polarisation and it is denoted by
4.Capacitor and Capacitance A capacitor is a device which is used to store electrostatic potential energy or charge. It comprises of two conductors separated by an insulating medium and capacitance of a conductor is the ability to store the electric charge.

5.Capacitance of a Conductor If charge q is given to an insulated conductor, it leads to increase its electric potential by V such that

where, C is known as capacitance of a conductor. The capacitance depends on the shape, size and geometry of conductor, nature of surrounding medium and presence of other conductor in the neighbourhood of it.Its SI unit is farad (F).

Farad is a very large unit of capacitance. So, pF is usually taken.

6.Principle of Capacitor The capacitance of an insulated conductor is increased significantly when an earthed uncharged conductor is placed near it. Such an arrangement is called capacitor.

9.Parallel Plate Capacitor The most common among all capacitors is parallel plate capacitor. It comprises of two metal plates of area A and separated by distance d filled with air or some other dielectric medium. The capacitance of air filled parallel plate capacitor is given by

When a dielectric of dielectric constant K is filled fully between the plates, then

10. Dielectric When a dielectric slab is introduced between the plates of charged capacitor or in the region of electric field, an electric field E_p induces inside the dielectric due to induced charge on dielectric in a direction opposite to the direction of applied external electric field. Hence, net electric field inside the dielectric gets reduced to (E₀ – E_p), where, E₀ is external electric field. The ratio of applied external electric field and reduced electric field is known as dielectric constant K of dielectric medium, i.e.

Previous Year Examination Questions

1 Mark Questions

1.The given graph shows the variation of charge q versus potential difference V for two capacitors C_l and C₂. Both the capacitors” have same plate separation but plate area of C₂ is greater than that C_x .Which line (A or B) corresponds to and why?[All India 2014 C]

Ans.Line B corresponds to C-, because slope (q/v) of B is less than slope of A.

2.A capacitor has been charged by a DC source. What are the magnitude of conduction and displacement current when it is fully charged?[Delhi 2013]
Ans.

3.Distinguish between a dielectric and a conductor.  [Delhi 2012]
Ans.Dielectrics are non-conductors and do not have free electrons at all. While conductor has free electrons in its any volume which makes it able to pass the electricity through it.

4.Define the dielectric constant of a medium. What is its unit? [Delhi 2011c]
Ans.Dielectric When a dielectric slab is introduced between the plates of charged capacitor or in the region of electric field, an electric field E_P induces inside the dielectric due to induced charge on dielectric in a direction opposite to the direction of applied external electric field. Hence, net electric field inside the dielectric gets reduced to £₀ – £_P, where £₀ is external electric field. The ratio of applied external electric field and reduced electric field is known as dielectric constant K of dielectric medium, i.e.

5.A metal plate is introduced between the plates of a charged parallel plate capacitor. What is its effect on the capacitance of the capacitor?[Foreign 2009]
Ans.If a metal plate is introduced between the plates of a charged parallel plate capacitor. The capacitance of parallel plate capacitor will becomes infinite.

6.In the figure given below X, Y represent parallel plate capacitors having the same area of plates and the same distance of separation between them. What is the relation between the energies stored in the capacitors?

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7.The following graph shows the variation of charge Q with voltage V for two capacitors K and L In which capacitor is more electrostatic energy stored?

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8.Define dielectric strength of a dielectric. [Delhi 2008 C]
Ans.The dielectric strength of a dielectric is the maximum value of applied electric field required to just breakdown of the dielectric material.

2 Marks Questions

9.A parallel plate capacitor of capacitance C is charged to a potential V. It is then connected to another uncharged capacitor having the same capacitance. Find out the ratio of the energy stored in the combined system to that stored initially in the single capacitor.[All India 2014]
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10.Two parallel plate capacitors of capacitances C_x and C₂ such that C_x =2C₂ are connected across a battery of V volt as shown in the figure Initially, the key (k) is kept closed to fully charge the capacitors.The key is now thrown open and a dielectric slab of dielectric constant K is inserted in the two capacitors to completely fill the gap between the plates. Find the ratio of (i) the net capacitance and (ii) the energies stored in the combination before and after the introduction of the dielectric slab.[Delhi 2014 C]

Ans.

11.Two parallel plate capacitors of capacitances Qand C₂ such that q =C₂ /2 are connected across a battery of V volts as shown in the figure. Initially, the key (k) is kept closed to fully charge the capacitors.The key is now thrown open and a dielectric slab of dielectric constant K is inserted in the two capacitors to completely fill the gap between the plates. Find the ratio of (i) the net capacitance and (ii) the energies stored in the combination before and after the introduction of the dielectric slab.[Delhi 2014 C]

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12.Find the charge on the capacitor as Shown in the circuit. [Foreign 2014]

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13.A slab of material of dielectric constant K has the same area as that of the plates of a parallel plate capacitor, but has the thickness d/2, where d is the separation between the plates. Find out the expression for its capacitance when the slab is inserted between the plates of the capacitor.  [Delhi 2013]
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14.Determine the potential difference across the plates of the capacitor C_x of the network shown in the figure, (assume, E₂ >E1 ) [All India 2013]

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17.Two identical parallel plate (air) capacitors C_x and C₂ have capacitance C each. The space between their plates is now filled with dielectrics as shown in the figure. If the two capacitors still have equal capacitance, they obtain the relation between dielectric constants K, K_x andK₂.[Foreign 2011]

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18.You are given an air filled parallel plate capacitor C₁. The space between its plates is now filled with slabs of dielectric constants K_x and K₂ as shown in figure. Find the capacitance of the capacitor C₂ if area of the plates is A and distance between the plates is d.

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19.Figure shows a sheet of aluminium foil of negligible thickness placed between the plates of a capacitor. How will its capacitance be affected if
(i)the foil is electrically insulated?
(ii)the foil is connected to the upper plate with a conducting wire?[Foreign 2011]

Ans.

20.(i)Obtain the expression for the energy stored per unit volume in a charged parallel plate capacitor.
(ii) The electric field inside a parallel plate capacitor is E. Find the amount of work done in moving a charge q over a closed rectangular loop abcda. [Delhi 2014]

Ans.(i) The energy of a charged capacitor is measured by the total work done in charging the capacitor to a given potential.

3 Marks Questions

21.(i) Derive the expression for the capacitance of a parallel plate capacitor having plate area A and plate separation d.
(ii) Two charged spherical conductors of radii and 1^ when connected by a conducting plate respectively. Find the ratio of their surface charge densities in terms of their radii. [Delhi 2014]
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24.A capacitor of 200 pF is charged by a 300 V battery. The battery is then disconnected and the charged capacitor is connected to another uncharged capacitor of 100 pF. Calculate the difference between the final energy stored in the combined system and the initial energy stored in the single capacitor.[Foreign 2012]
Ans.

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26.A parallel plate capacitor is charged by a battery. After sometime, the battery is disconnected and a dielectric slab with its thickness equal to the plate separation is inserted between the plates. How will
(i)the capacitances of the capacitor,
(ii)potential difference between the plates and
(iii)the energy stored in the capacitors be affected? Justify your answer in each case.[Delhi 2010]
Ans.

27.A parallel plate capacitor, each with plate area A and separation d is charged to a potential difference V. The battery used to charge it remains connected. A dielectric slab of thickness d and dielectric constant K is now placed between the plates. What change if any will take place in
(i)charge on plates?
(ii)electric field intensity between the plates?
(iii)capacitance of the capacitor? Justify your answer in each case.[Delhi 2010]
Ans.

28.A parallel plate capacitor is charged to a potential difference V by a DC source. The capacitor is then disconnected from the source. If the distance between the plates is doubled, state with reason, how the following will change?
(i)Electric field between the plates
(ii)Capacitance
(iii)Energy stored in the capacitor [Delhi 2010]
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29.Find the ratio of the potential differences that must be applied across the parallel and the series combination of two identical capacitors so that the energy stored in the two cases becomes the same.
[Foreign 2010]
Ans.

30.(i) How is the electric field due to a charged parallel plate capacitor affected when a dielectric slab is inserted between the plates fully occupying the intervening region?
(ii) A slab of material of dielectric constant K has the same area as the plates of a parallel plate capacitor but has thickness 1/2 d,
where d is the separation between the plates. Find the expression for the capacitance when the slab is inserted between the plates. [Foreign 2010]
Ans.

31.(i) Plot a graph comparing the variation of potential V and electric field E due to a point charge 0 as a function of distance R from the point charge.
(ii) Find the ratio of the potential differences that must be applied across the parallel and the series
combination of two capacitors, C_l and C₂ with their capacitances in the ratio 1 : 2, so that the energy stored in the two cases becomes the same[Foreign 2010]
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33.A parallel plate* capacitor is charged by a battery. After sometime, the battery is disconnected and a dielectric slab of dielectric constant K is inserted between the plates. How would
(i)the electric field between the plates
(ii)the energy stored in the capacitor be affected? Justify your answer. [All India 2009]
Ans.(i) The total charge on the capacitor remains conserved on introduction of dielectric slab. Also, the capacitance of capacitor increases to K times of original values.

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37.A system of capacitors connected as shown in the figure has a total energy of 160 mJ stored in it. Obtain the value of the equivalent capacitance of this system and the value of Z. [All India 2009 c]

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38.The two plates of a parallel plate capacitor are 4 mm apart. A slab of dielectric constant 3 and thickness 3 mm is introduced between the plates with its faces parallel to them. The distance between the plates is so adjusted that the capacitance of the capacitor becomes (2/3 )rd of its original value. What is the new distance between the plates?[All India 2008 C]
Ans.

5 Marks Questions

39.(i) A parallel plate capacitor is charged by a battery to a potential. The battery is disconnected and a dielectric slab is inserted to completely fill the space between the plates.
How will
(a)its capacitance
(b)electric field between the plates and
(c)energy stored in the capacitor be affected? Justify your answer giving necessary mathematical expressions for each case.
(ii) (a) Draw the electric field lines due to a conducting sphere.
(b) Draw the electric field lines due to a dipole.
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41.(i)Show that in a parallel plate capacitor if the medium between the plates of a capacitor is filled with an insulating substance of dielectric constant K, its capacitance increases.
(ii) Deduce the expression for the energy stored in a capacitor of capacitance C with charge Q.[Delhi 2009 C]
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42.A small sphere of radius a carrying a positive charge q is placed concentrically inside a larger hollow conducting shell of radius b(b> a). This outer shell has charge Q on it. Show that if these spheres are connected by a conducting wire, charge will always flow from the inner sphere to the outer sphere irrespective of the magnitude of the two charges.
Ans.

Ans.(i) An expression for energy stored in a charged capacitor. (ii) Connected battery ensures same potential difference (V) across capacitor even after reducing the spacing between plates halved and introducing the dielectric slab.

44.Define the terms (i)capacitance of a capacitor (ii)dielectric strength of a dielectric
(iii)When a dielectric is inserted between the plates of a charged parallel plate capacitor fully occupying the intervening region, how does the polarisation of the dielectric medium affect the net electric field? For a linear dielectric,show that the introduction of the dielectric increases its capacitance by a factor K which is a characteristic of the dielectric.   [Delhi 2008 C]
Ans.

Important Questions for Class 12 PhysicsClass 12 PhysicsNCERT Solutions Home Page

Electrostatic Potential and Capacitance Important Questions for CBSE Class 12 Physics Electrostatic Potential

1.Electrostatic Potential The electrostatic potential at any point in an electric field is equal to the amount of work done per unit positive test charge or in bringing the unit positive test charge from infinite to that point, against the electrostatic force without acceleration.

NOTE Electrostatic potential is a state dependent function as electrostatic forces are conservative forces

2.Electrostatic Potential Difference The electrostatic potential difference between two points in an electric field is defined as the amount of work done in moving a unit positive test charge from one point to the other point against of electrostatic force without any acceleration (i.e. the difference of electrostatic potentials of the two points in the electric field).

where, is work done in taking charge q₀ from A to B against of electrostatic force.
Also, the line integral of electric field from initial position A to final position B along any path is termed as potential difference between two points in an electric field, i.e.

NOTE As, work done on a test charge by the electrostatic field due to any given charge configuration is independent of the path, hence potential difference is also same for any path.
For the diagram given as below, potential difference between points A and B will be same for any path.

3.Electrostatic potential due to a point charge q at any point P lying at a distance r from it is given by
4.The potential at a point due to a positive charge is positive while due to negative charge, it is negative.

5.When a positive charge is placed in an electric field, it experiences a force which drives it from points of higher potential to the points of lower potential. On the other hand, a negative charge experiences a force driving it from lower potential to higher.

10.Graphical representation of variation of electric potential due to a charged shell at a distance r from centre of shell is given as below

11.Equipotential Surface A surface which have same electrostatic potential at every point on it, is known as equipotential surface.
The shape of equipotential surface due to
(i) line charge is cylindrical.
(ii) point charge is spherical as shown along side:

• Equipotential surfaces do not intersect each other as it gives two directions of electric field E at intersecting point which is not possible.
• Equipotential surfaces are closely spaced in the region of strong electric field and vice-versa.
• Electric field is always normal to equipotential surface at every point of it and directed from one equipotential surface at higher potential to the equipotential surface at lower potential.
• Work done in moving a test charge from one point of equipotential surface to other is zero.

where, negative sign indicates that the direction of electric field is from higher potential to lower potential, i.e. in the direction of decreasing potential.
NOTE (i) Electric field is in the direction of which the potential decreases steepest.
(ii) Its magnitude is given by the change in the magnitude of potential per unit displacement normal to the equipotential surface at the point.

13. Electrostatic Potential Energy The work done against electrostatic force gets stored as potential energy. This is called electrostatic potential energy

14.The work done in moving a unit positive test charge over a closed path in an electric field is zero. Thus, electrostatic forces are conservative in nature.

15.Electrostatic potential energy of a system of two point charges is given by

Putting the values of charge with their signs.

16.Electrostatic potential of a system of n point charges is given by

17.Potential Energy in an External Field
(i)Potential Energy of a single charge in external field Potential energy of a single charge q at a point with position vector r, in an external field is qV(r),where V(r) is the potential at the point due to external electric field E.
(ii)Potential Energy of a system of two charges in an external field

where, q_x and q₂ – two point charges at position vectors r_: and r₂, respectively
V (r_x) = potential at r_x due to the external field
V(r₂) = potential at r₂ due to the external field
18.Potential energy of a dipole in a uniform electric field E is given by Potential energy = -p .E

19.Electrostatic Shielding The process which involves the making of a region free from any electric field is known as electrostatic shielding

It happens due to the fact that no electric field exist inside a charged hollow conductor. Potential inside a shell is constant. In this way we can also conclude that the field inside the shell (hollow conductor) will be zero

Previous Year Examination Questions

1 Mark Questions

1.The figure shows the field lines of a positive charge. Is the work done by the field is moving a small positive charge from Q to P positive or negative? [Foreign 2014]

Ans.

2.For any charge configuration,equipotential surface through a point is a normal to the electric field. Justify.    [Delhi 2014]
Ans.

Ans.

4.Why electrostatic potential is constant throughout the volume of the conductor and has the same value as on its surface? [Delhi 2012]
Ans. Since, electric field intensity inside the conductor is zero. So, electrostatic potential is a constant.

5.Why is the potential inside a hollow spherical charged conductor is constant and has the same value as on its surface?[Foreign 2012]
Ans.Electric field inside the hollow spherical charged conductor is zero. So, no work is done in moving a charge inside the shell.
This implies that potential is a constant and therefore, equal to its value at the surface, i.e.

6.Why there is no work done in moving a charge from one point to another on an equipotential surface?      [Foreign 2012]
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7.A hollow metal sphere of radius 5 cm is charged such that potential on its surface is 10 V. What is the potential at the centre of the sphere?      [All India 2011]
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8.Can two equipotential surface intersect each other? Justify your answer. [Delhi 2011 c]
Ans.(i) No, two equipotential surfaces cannot intersect each other because two normals can be drawn at intersecting point on two surfaces which give two directions of E at the same point which is impossible.
(ii) Also, two values ot potential at the same point is not possible.

9.A point charge Q is placed at point O as shown in the figure. Is the potential  difference  (V_A       – V_B)
positive, negative or zero if Q is (i)positive (ii) negative

Ans.

10.Draw equipotential surfaces due to a single point charge. [All India 2011 c]
Ans.Equipotential surfaces due to a single point charge are concentric sphere having charge at the centre.

11.Name the physical quantity whose SI  unit is J/C. Is it a scalar or a vector quantity? [All India 2010]
Ans. J/C is the SI unit of electric potential. It is a scalar quantity.

12.What is the work done in moving a test charge q through a distance of 1 cm along the equatorial axis of an electric dipole?  [All India 2009]
Ans.

13.Define the term potential energy for charge q at a distance r in an external field.  [All India 2009]
Ans.The electric potential energy at any point lying at a distance r from the source charge q is equal to the amount of work done in moving unit positive test charge from infinity to that point without any acceleration against of electrostatic force.

14.The potential due to a dipole at any point on its axial line is zero. [All India 2009 C]
Ans.Wrong, as potential due to an electric dipole is zero on equatorial line in spite of axial line.
The potential due to a dipole at any point on equatorial line is zero, not an axial line.

15.What is the electric potential due to an electric dipole at an equatorial point?[All India 2009]
Ans.Zero, as potential on equatorial point due to charges of electric dipole, is equal to magnitude but opposite in nature and hence, their resultant is zero

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2 Marks Questions

17.Two point charges q₁ and q₂ are located at q and r₂, respectively in an external electric field E. Obtain the expression for the total work done in assembling this configuration. [Delhi 2014 C]
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19.A test charge q is moved without acceleration from A to C along the path from A to B and then from B to C in electric field E as shown in the figure.
(i)Calculate the potential difference between A and C.
(ii)At which point (of the two) is the electric potential more and why?

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20.Graph the electric potential (V) with distance r due to a point charge Q.[Delhi 2012]
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22.A dipole with its charge -q and +q located at the points (0, – b,0) and (0,+ 5,0) is present in a uniform electric field E. The equipotential surfaces of this field are planes parallel to the Y Z-plane.
(i)What is the direction of the electric field E?
(ii)How much torque would the dipole experience in this field?[Delhi 2010 C]
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24.Find out the expression for the potential energy of a system of three charges q₁, q₂ and g₃ located at q , r₂ and r₃ with respect to the common origin O. [Delhi 2010 c]
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25.Draw three equipotential surfaces corresponding to a field that uniformly increases in magnitude but remains constant  along Z-direction. How are  these surfaces different from that of a constant electric field along Z-direction? [Foreign 2008; All imiia 2009]
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In case of constant electric field along Z-direction,the perpendicular distance between equipotential surfaces remains same whereas for field of increasing magnitude, equipotential surfaces get closer as we go away from the origin.In both cases, surfaces be planes parallel to XY- plane.

26.(i)Can two equipotential surfaces intersect each other? Give reasons,
(ii)Two charges – qand + qare located at points A (0, 0, – a) and B{0, 0, +a) respectively. How much work is done in moving a test charge from point P{7, 0, 0) to Q(- 3, 0, 0)?  [Delhi 2009]
Ans.(i) No, two equipotential surfaces cannot intersect each other because two normals can be drawn at intersecting point on two surfaces which give two directions of E at the same point which is impossible.

27.(i) Can two equipotential surfaces intersect each other? Give reasons, (ii) Two charges + q and – q are located at points A(0, 0, – 2) and B (0, 0, 2) respectively. How much work will be done in moving a test charge from point P{7,0,0) to Q (- 3, 0, 0)?  [Delhi 2009]
Ans.(i) No, two equipotential surfaces cannot intersect each other because two normals can be drawn at intersecting point on two surfaces which give two directions of E at the same point which is impossible.

28.(i)Write two characteristics of equipotential surfaces.
(ii) Draw the equipotential surfaces due to an electric dipole. [All India 2009 C]
Ans.(i) Two equipotential surfaces do not intersect each other as normals at intersecting points on two surfaces will give two directions of electric field which is impossible.
(ii) Closely spaced equipotential surfaces represent strong electric field and vice-versa.

29.Two point charges, q_t =10xl0^-8 C and q₂ =-2×10 ~⁸ C are separated by a distance of 60 cm in air.
(i)What a distance from the 1st charge q_x would the electric potential be zero?
(ii)Also, calculate the electrostatic potential energy of the system.[All India 2007,2008]
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30.Two point charges 40, O are separated by lm in air. At what point on the line joining the charges, is the electric field intensity zero? Also calculate the electrostatic potential energy of the system of charges taking the value of charge, O = 2 X 10~⁷ C. [All India 2008]
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31.Define the dipole moment of an electric dipole. How does the electric potential due to a dipole vary on the dipole axis as a function of r distance of the point from the mid-point of the dipole at large distances? [All India 2008 C]
Ans.Electric field inside the hollow spherical charged conductor is zero. So, no work is done in moving a charge inside the shell.
This implies that potential is a constant and therefore, equal to its value at the surface, i.e.

variation of electric potential due to dipole on axial lines when point is at large distance from mid-point of the dipole, occurs as per relation below.

The variation of electric potential with distance r is shown as below:

32.Derive an expression for the electric potential at any point along the axial line of an electric dipole. [Delhi 2008]
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33.Two charges of 5 nC and – 2 nC are placed at points (5 cm, 0, 0) and (23 cm, 0, 0) in the region of space, where there is no other external field. Calculate the electrostatic potential energy of this charge system.[Delhi 2008 C]
Ans.

3 Marks Questions

34.A wire AB is carrying a steady current of 12 A and is lying on the table. Another wire CD carrying 5 A is held directly above AB at a height of 1mm. Find the mass per unit length of the wire CD, so that it remains suspended at its position when left free. Give the direction of the current flowing in CD with respect to that in [Take the value of g =10 ms^-2][All India 2013]
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35.(i) Depict the equipotential surfaces for a system of two identical positive point charges placed a distance d apart.
(ii) Deduce the expression for the potential energy of a system of two point charges q₁ and q₂ brought from infinity to the points with positions and r₂ respectively, in presence of external electric field E. [Delhi 2010]
Ans.

By definition, electric potential energy of any charge q placed in the region of electric field is equal to the work done in bringing charge q from infinity to that point and given by U =qV
where, V is the electric potential (as potential at infinity is assumed to be zero) where, the charge q is placed.Now, considering the electric potentials at positions r1 and r₂ as V1, and V₂, respectively. Therefore, total potential energy of the system of two charges q•, and q₂ placed at points with position vectors r, and r₂ in the region of E is given by U= work done in bringing charge q from infinite to that position in E is equal to work done for charge q₂ from infinite to that position in E +work done to that of charge q₂ at these positions in presence of q1

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37.(i) Two points charges q_x and q₂ initially at infinity, are brought one by one to points P_x and P₂ specified by position vectors r_t and r₂ relative to same origin. What is the potential energy of this charge configuration? Define an equipotential surface.
(ii) Draw schematically the equipotential surface corresponding to a field that uniformly increases in magnitude but remains constant in direction.  [Delhi 2008]
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Important Questions for Class 12 PhysicsClass 12 PhysicsNCERT Solutions Home Page

Alternating Current Important Questions for CBSE Class 12 Physics AC Devices

1.Transformer It is a device which converts high voltage AC into low voltage AC and vice-versa. It is based upon the principle of mutual induction.

Working When an alternating current is passed through the primary coil, the magnetic flux through the iron core changes, which does two things, produces emf in the primary coil and an induced emf is set up in the secondary coil. If we assume that the resistance of primary coil is negligible, then the back emf will be equal to the voltage applied to the primary coil.

2.Energy losses in a transformer

(i) Eddy Current Loss Eddy current in iron core of transformer facilitate the loss of energy in the form of heat.

(ii) Flux Leakage Total fluxes linked with primary do not completely pass through the secondary which denotes the loss in the flux.

(iii) Copper Loss Due to heating, energy loss takes place in copper wires of primary and secondary coils.

(iv) Hysterisis Loss The energy loss takes place in magnetizing and demagnetizing the iron core oven every cycle.

(v) Humming Loss The magnetostriction effect leads to set core in vibration which in turn produced the sound. This loss is referred as humming loss.

3.Types of Transformer There are two types of transformers which are given as below:

(i) Step-up transformer (N _s >N_P) It converts low alternating voltage into high alternating voltage.

(ii) Step-down transformer (N_s > N_p) It converts high alternating voltage into low alternating voltage.

Previous Year Examination Questions

1 Mark Questions

1. Mention the two characteristic properties of the material suitable for making core of a transformer.             [All India 2012]

Ans. (i) Low retentivity or coercivity.

(ii) Low hysterisis loss or high permeability and susceptibility.

2. What is the function of a step-up transformer? [All India 2011C]

Ans. Step-up transformer converts low alternating voltage into high alternating voltage and high alternating current into low alternating current. The secondary coil of step-up transformer has a greater number of turns that the primary (N_s > N_P)

3. Write any two factors responsible for energy losses in actual transformers.      [Delhi 2009c]

Ans. Two factors for energy losses are

(i) eddy currents

(ii) hysterisis loss

2 Marks Questions

4.State the underlying principle of a transformer. How is the large scale transmission of electric energy over long distances done with the use of transformers? [All India 2012]

Ans. Principle of transformer A transformer is based on the principle of mutual induction, i.e. whenever the amount of magnetic flux linked with a coil changes, an emf is induced in the neighbouring coil.

Power transmission Electric power is transmitted over long distances at high voltage.So, step-up transformers are used at power stations to increase the voltage of power whereas a series of step-down transformers are used to decrease the voltage up to 220 V.

5.State the principle  of    working of a transformer. Can a transformer be used to step-up or step-down a DC voltage? Justify your answer.            [All India 2011]

Ans. (i) For principle of transformer A transformer is based on the principle of mutual induction, i.e. whenever the amount of magnetic flux linked with a coil changes, an emf is induced in the neighbouring coil.

(ii) No, transformer cannot be used to change DC voltage. This happens because DC voltage cannot change flux linked with primary or secondary coils.

6. Mention various energy   losses in a transformer.     [All India 2011]

Ans. Energy losses in transformer:

(i) Eddy current loss Eddy current in iron core of transformer facilitates the loss of energy in the form of heat.

(ii) Flux leakage Total fluxes linked with primary do not completely pass through the secondary which denotes the loss in the flux.

(iii) Copper loss Due to heating, energy loss takes place in copper wires of primary and secondary Coils.

(iv) Hysterisis loss The energy loss takes place in magnetizing and demagnetising the iron core over every cycle.

(v) Humming loss The magnetostriction effect leads to set core in vibration which in turn produced the sound. This loss is referred as humming loss.

5 Marks Questions

7. (i) Draw a schematic arrangement for winding of primary and secondary coils in a transformer when the two coils are wound on top of each other.

(ii) State the underlying principle of a transformer and obtain the expression for the ratio of secondary to primary voltage in terms of the

• number of secondary and primary windings and
• primary and secondary currents.

(iii) Write the main assumption involved in deriving the  above relations.

(iv) Write any two reasons due to which energy losses may occur in actual transformers. [HOTS]

Ans.  (i) The schematic arrangement of a transformer is shown as below:

(iii) Main assumptions

• The primary resistance and current are small.
• The flux linked with primary and secondary coil is some where is no leakage of flux from the core.
• secondary current is small

(iv) Energy losses in transformer:

(i) Eddy current loss Eddy current in iron core of transformer facilitates the loss of energy in the form of heat.

(ii) Flux leakage Total fluxes linked with primary do not completely pass through the secondary which denotes the loss in the flux.

(iii) Copper loss Due to heating, energy loss takes place in copper wires of primary and secondary Coils.

(iv) Hysterisis loss The energy loss takes place in magnetizing and demagnetising the iron core over every cycle.

(v) Humming loss The magnetostriction effect leads to set core in vibration which in turn produced the sound. This loss is referred as humming loss.

8.  (i) State the principle of a step-up transformer. Explain with the help of a labelled diagram, its working.

(ii) Describe briefly and two energy losses giving the reasons for their occurrence in actual transformer. [Foreign 2012]

Ans. Principle of transformer A transformer is based on the principle of mutual induction, i.e. whenever the amount of magnetic flux linked with a coil changes, an emf is induced in the neighbouring coil.

 

9. With the help of a labelled diagram, describe briefly the underlying principle and working of a step-up transformer.

(ii) Write any two sources of energy loss in a transformer, (iii) A step-up transformer converts a low input voltage into a high output voltage. Does it violate law of conservation of energy? Explain.         [Delhi 2011]

Ans.(i) Principle of transformer A transformer is based on the principle of mutual induction, i.e. whenever the amount of magnetic flux linked with a coil changes, an emf is induced in the neighbouring coil.

(ii) Two factors for energy losses are

•  eddy currents
•  hysterisis loss

(iii) No, it does not violate the law of conservation of energy because voltage increase is accompanied by decrease in it. In alternative, currrent in such a way for an ideal transformer input power equals to the output power.

10. (i) Draw a schematic diagram of a step-up transformer. Explain its working principle. Assuming the transformer to be 100% efficient, obtain the relation for

• the current in the secondary in terms of the current in the primary and
• the number of turns in the primary’ and secondary windings.

(ii) Mention two important energy losses in actual transformers and state how these can be minimized? [All India 2009C; Foreign 2009]

Ans. Principle of transformer A transformer is based on the principle of mutual induction, i.e. whenever the amount of magnetic flux linked with a coil changes, an emf is induced in the neighbouring coil.

(ii) Two energy loses in the actual transformers

• Hysterisis loss Loss of energy in magnetising and demagnetising the core of transformer in every cycle. It can be minimised by taking core of soft iron having low coercivity, retentivity and hysterisis loop.

• Eddy current Energy loss in the form of heat due to eddy current. It can be minimised by taking laminated core consisting of insulated rectangular sheets, piled-up one over another.

11. Draw a schematic diagram of a step-up transformer. Explain its working principle. Deduce the expression for the secondary to primary voltage in terms of the number of turns in the two coils.

In an ideal transformer, how is this relation related to the currents in the two coils.

How is the transformer used in large scale transmission and distribution of electrical energy over the long distances?        [All India 2010]

Ans. 

Principle of transformer A transformer is based on the principle of mutual induction, i.e. whenever the amount of magnetic flux linked with a coil changes, an emf is induced in the neighbouring coil.

(ii) Two energy loses in the actual transformers

• Hysterisis loss Loss of energy in magnetising and demagnetising the core of transformer in every cycle. It can be minimised by taking core of soft iron having low coercivity, retentivity and hysterisis loop.

• Eddy current Energy loss in the form of heat due to eddy current. It can be minimised by taking laminated core consisting of insulated rectangular sheets, piled-up one over another.

Step-up transformers are used at generating stations so as to transmit the power at high voltage to minimise the loss in the form of heat, whereas series of step-up transformers are used at receiving ends

12 .A step-up down transformer operated on a 2.5 kV line. It supplies a load with 20 A. The ratio of the primary winding to the secondary is 10 :1. If the transformer is 90% efficient, calculate

• the power output
• the voltage and
• the current in the secondary coil. [Foreign 2010]

Important Questions for Class 12 PhysicsClass 12 PhysicsNCERT Solutions Home Page

Electric Charges and Fields Important Questions for CBSE Class 12 Physics Gauss’s Law

1.Area Vector The vector associated with every area element of a closed surface is taken to be in the direction of the outward normal.
Consider the diagram given as below.

Here, AS is the area vector in the direction of the unit vector n normal to the surface area AS

2.Electric Flux Electric flux linked with any surface is proportional to the total number of electric field lines that normally pass through that surface. It is a scalar quantity.
SI unit of electric flux is N – m² C^-1 or JmC^-1 or Vm.
CGS unit of electric flux is dyne – cm²/ stat -C.
Now, consider the electric flux linked with a surface when

Previous Year Examination Questions

1 Mark Questions

1.Consider two hollow concentric spheres S₁ and S₂ enclosing charges 20 and 40 respectively, as shown in the figure, (i) Find out the ratio of the electric flux through them, (ii) How will the electric flux through the spheres S₁ change if a medium of dielectric constant e_r is introduced in the space inside S₁ in place of air? Deduce the necessary expression. [All India 2014]

Ans.

2.Two charges of magnitudes –20 and + O are located at points (a, 0) and (4a, 0), respectively. What is the electric flux due to these charges through a sphere of radius 3a with its centre at the origin? [All India 2013]
Ans.

3.A charge q is placed at the centre of a cube of side L. What is the electric flux passing through each face of the cube? [All India 2010; Foreign 2010]
Ans.

4.Figure shows three point charges,+ 2q, – q and + 3q. Two charges + 2q and – q are enclosed within a surface S. What is the electric flux due to this configuration through the surface S?   [Delhi 2010]

Ans.

5.If the radius of the Gaussian surface enclosing a charge is halved, how does the electric flux through the Gaussian surface change?[All India 2009,2008]
Ans.Total charge enclosed by the Gaussian surface remains same even when radius is halved. Therefore, total electric flux remains constant as per Gauss’ theorem. There will not be any change in electric flux through the Gaussian surface.

2 Marks Questions

6.Given a   uniform  electric  held E =5 x 10³ i N/C, find the flux of this held through a square of 10 cm on a side whose plane is parallel to the YZ-plane. What would be the flux through the same square if the plane makes an angle of 30° with the X-axis?       [Delhi 2014]
Ans.

7.Given a   uniform    electric     held E =2 x 10³ i N/C, find the flux of this held through a square of side 20 cm, whose plane is parahel to the YZ-plane. What would be the flux through the same square if the plane makes an angle of 30° with the X-axis?      [Delhi 2014, HOTS]
Ans.Refer to ans. 6. (Ans 40 Nm²/C)

8.Given a   uniform    electric  held E = 4 x 10³ i N/C. Find the flux of this field through a square of 5 cm on a side whose plane is parallel to the YZ-plane. What would be the flux through the same square if the plane makes an angle of 30° with the X-axis?       [Delhi 2014, HOTS]
Ans.Refer to ans 6. (Ans. 5 Nm²/C)

9.A sphere S_t of radius q enclosed a net charge Q. If there is another concentric sphere S₂ of radius r₂₍r₂ > q) enclosing charge 20, hnd the ratio of the electric hux through S₁ and S₂. How will the electric flux through sphere Sj change if a medium of dielectric constant K is introduced in the space inside S₂ in place Of air?  [All India 2014]

Ans.

Ans.A thin straight conducting wire will be a uniform linear charge distribution.
Let q charge be enclosed by the cylindrical surface.

11.Two charged conducting spheres of radii r_t and r₂ connected to each other by a wire. Find the ratio of electric fields at the surfaces of the two spheres. [Delhi 2011 c]
Ans.

Ans.

13.A spherical conducting shell of inner radius R₁ and outer radius R₂ has a charge Q. A charge q is placed at the centre of the shell. [All India 2010c]
(i)What is the surface charge density on the (a) inner surface, (b) outer surface of the shell?
(ii)Write the expression for the electric field at a point to x>R₂ from the centre of the shell.
Ans.Here, two points are important
(i) Charge resides on the outer surface of spherical conductor (skin effect).
(ii) Equal charge of opposite nature induces in the surface of conductor nearer to source charge.

14.Define electric flux. Write its SI unit. A charge gis enclosed by a spherical surface of radius [All India 2009]
Ans.The total electric flux linked with a surface is equal to the total number of electric lines of force passing through the surface when surface is held normal to the direction of electric field.

15.Draw the shapes of the suitable Gaussian surfaces while applying Gauss’ law to calculate the electric field due to
(i)a uniformly charged long straight wire.
(ii)a uniformly charged infinite plane sheet. [Delhi 2009 C]
Ans.The surface that we choose for application of Gauss’ theorem is called Gaussian surface. We usually choose a spherical Gaussian surface.

Ans.

Ans.

Ans.

3 Marks Questions

19.A hollow cylindrical box of lenght 1 m and area of cross-section 25 cm² is placed in a three-dimensional coordinate system as shown in the figure.

Ans.

20.State Gauss’ law in electrostatics. A cube which each side a is kept in an electric field given by E = as shown in the figure, where C is a positive dimensional constant. Find out

(i)the electric flux through the cube
(ii)the net charge inside the cube.[Foreign 2012]
Ans.

21.Using Gauss’ law, obtain the expression for the electric field due to uniformly charged spherical shell of radius R at a point outside the shell. Draw a graph showing the variation of electric field with r, for r > R and r < R.[All India 2011]
Ans.Let us consider charge +q is uniformly distributed over a spherical shell of radius R. Let £ is to be obtained at P lying outside of spherical shell.

Eat any point is radially outward (if charge q is positive) and has same magnitude at all points which lie at the same distance r from centre of spherical shell such that r> R.

Ans.

23.State Gauss’ law in electrostatics. Using this law, derive an expression for the electric field due to a uniformly charged infinite plane Sheet.       [Delhi 2009]
Ans.


Let us consider a large plane sheet of charge having surface charge density sigma

Ans.(i)

4 Marks Questions

25.Using Gauss’ law, deduce the expression for the electric field due to a uniformly charged spherical conducting shell of radius R at a point
(i)outside the shell
(ii)inside the shell
Plot a graph showing variation of electric field as a function of r > R and r < R.{r being the distance from the centre of the shell) [All India 2013]
Ans.

26.(i) Define electric flux. Write its SI unit.
(ii) A small metal sphere carrying charge + Q is located at the centre of a spherical cavity inside a large uncharged metallic spherical shell as shown in the figure.Use Gauss’ law to find the expressions for the electric field at points P_l and P₂.

(iii)Draw the pattern of electric field lines in this arrangement.[Delhi 2012 C]
Ans.

(iii) The electric field lines due to arrangement is shown as below:

Charges will be uniformly distributed on all the surfaces hence, all field lines will be uniformly separated.

27.Define electric flux. Write its SI unit, (ii) Using Gauss’ law, prove that the electric field at a point due to a uniformly charged infinite plane sheet is independent of distance from it.
How is the field directed if
(a)the sheet is positively charged
(b)negatively charged? [Delhi 2012]
Ans.(i) Electric flux Electric flux over an area in an electric field represents the total number of electric lines of force crossing the area in a direction normal to the plane of the area. The SI unit of electric flux is N-m²/C
(ii)


Let us consider a large plane sheet of charge having surface charge density sigma

The field directed

• Normally away from the sheet when sheet is positively charged.
• Normally inward towards the sheet when plane sheet is negatively charged.

28.(i) State Gauss’ law. Use it to deduce the expression for the electric field due to a uniformly charged thin spherical shell at points

• inside the shell and
• outside the shell.

(ii) Two identical metallic spheres A and B having charges +40 and – 10O are kept a certain distance apart. A third identical uncharged sphere C is first placed in contact with sphere A and then with sphere B. Then, spheres A and Bare brought in contact and then separated. Find the charges on the spheres A and B.[All India 2011C]
Ans.(i) 



(ii)

29.(i) Define electric flux. Write its SI unit, (ii) The electric field components due to a charge inside the cube of side 0.1 m are shown below.


Ans.(i)Electric flux Electric flux over an area in an electric field represents the total number of electric lines of force crossing the area in a direction normal to the plane of the area. The SI unit of electric flux is N-m²/C
(ii) The electric field is directed along +X-axis. Therefore, angle between E and A for left face is 180°, whereas for right face is 0°. The angle between E and A on four non-shaded faces is 90°. Therefore, flux linked with these four faces is zero.

Ans.(i)Electric flux Electric flux over an area in an electric field represents the total number of electric lines of force crossing the area in a direction normal to the plane of the area. The SI unit of electric flux is N-m²/C

Ans.The electric lines of force emerge from the positive charge and comes into the negative charge.
(i)


Electric field lines due to positive and negative charged spherical shell are as given below in figures (a) and (b) I respectively

Important Questions for Class 12 PhysicsClass 12 PhysicsNCERT Solutions Home Page

Alternating Current Important Questions for CBSE Class 12 Physics AC Currents

1.Phasor The representation of AC current and voltage by rotating vectors is calle^ phasor and the diagram representing these phasors is known as phasor diagram. The length of the vector represents the maximum or peak value. The projection of the vectors on fixed axis gives the instantaneous value of alternating current and voltage.

2.In an AC Circuit Containing Resistance Only

Suppose a resistor of resistance R is connected to an AC source of emf with instantaneous value E is given by

3.In an AC Circuit Containing Capacitor Only

Suppose a capacitor with capacitance C is connected to an AC source with emf having instantaneous value E is given by

4. In an AC Circuit Containing Inductor Only

10 .L-C Oscillations
Consider the L-C circuit shown as below:

When the charged capacitor is connected with the inductor, current flows through the inductor and energy stored in the inductor in the form of magnetic field and capacitor discharges and vice-versa. In this way, energy oscillates between capacitor and inductor. If the circuit is not ideal oscillations finally die away.

Previous Years Examination Questions

1 Mark Questions

1.A reactive element in an AC circuit causes the current flowing

Ans

2 Marks Questions 

2.A capacitor C, a variable resistor R and a bulb B are connected in series to the AC mains in the circuit as shown. The bulb glows with some brightness. How will the glow of the bulb change if (i) a dielectric slab is introduced between the plates of the capacitor keeping resistance R to be the same (ii) the resistance R is increased keeping the same capacitance? [Delhi 2014]

Ans.

Ans.


Ans.Refer to ans3

Ans Refer to ans3.

Ans.


7. In a series L-C-R circuit, obtain the conditions under which (i) the impedance of circuit is minimum and (ii) wattless current flows in the circuit.           [Foreign 2014]

Ans.

Ans.

Ans.

Ans.

Ans.

Ans.



13.(i) The graphs (I) and (II) represent the variation of the opposition offered by the circuit element to the flow of alternating current with frequency of the applied emf. Identify the circuit element corresponding to each graph.
(ii) Write the expression for the impedance offered by the series combination of the above two elements connected across the AC sources. Which will be ahead in phase in this circuit, voltage or current?

Ans.

14. An AC source of voltage V = V_m sin wt is applied across a series L-C-R Draw the phasor diagram for this circuit when the

• capacitative impedance exceeds the inductive impedance.
• inductive impedance exceeds the capacitative impedance. [All India 2008C]

Ans.

15. An AC source of voltage V = V_m sin wt is applied across a

• series R-C circuit in which the capacitative impedance is a times the resistance in the circuit.
• series R – L circuit in which the inductive impedance is b times the resistance in the circuit.

Calculate the value of the power factor of the circuit in each case. [All India 2008C]

Ans.

3 Marks Questions

16. A voltage V = V₀ sin wt is applied to a series L-C-R Derive the expression for the average power dissipated over a cycle.
Under what condition is (i) no power dissipated even though the current flows through the circuit (ii) maximum power dissipated in the circuit? [All India 2014]

Ans.

Ans.

Ans.

19.A series L-C-R circuit is connected to an AC source. Using the phasor diagram, derive the expression for the impedance of the circuit. Plot a graph to show the variation of current with frequency of the source, explaining the nature of its variation. [All India 2012]

Ans.

Ans.


21. An AC voltage V = V₀ sin wt is applied across a pure inductor Obtain an expression for the current / in the circuit and hence obtain the

• inductive reactance of the circuit and
• the phase of the current flowing with respect to the applied voltage. [All India 2010C]

Ans.

Ans.




Observe the graphs carefully.

(i) State the relation between L and C values of the two circuits when the current in the two circuits is maximum.

(ii) Indicate the circuit for which

• power factor is higher
• quality factor Q is larger.

Give the reasons for each case. [Delhi 2009C]

Ans.

Ans.

Ans.

Ans.



Ans. Refer Q.no 26

28. When a given coil is connected to a 200 V DC source, 2 A current flows and when the same coil is connected to 200 V, 50 Hz AC source, only 1 A current flows in the circuit.

• Explain why the current decreases in the latter case?
• Calculate the self-inductance of the coil used.                                     [Foreign 2008]

Ans.

Ans.

Ans.



31. What does the term ‘phasors’ in AC circuit analysis mean? An AC source of voltage, V = V_m sin wt is applied across a pure inductor of inductance L. Obtain an expression for the current I flowing in the circuit. Also draw the

• phasor diagram.
• graphs of V and I versus wt for this circuit.          [Delhi 2008C]

Ans.

Ans.

5 Marks Questions

33. (i) A series L-C-R circuit is connected to an AC source of variable frequency. Draw a suitable phasor diagram to deduce the expressions for the amplitude of the current and phase angle.

(ii) Obtain the condition at resonance. Draw a plot showing the variation of current with the frequency of AC source for two resistances R₁ and R2 (R₁ > R2). Hence, define the quality factor Q and write its role in the tuning of the circuit. [Delhi 2014C]

Ans.


34. Derive an expression for the impedance of a series L-C-Rcircuit connected to an AC supply of variable frequency.

Plot a graph showing variation of current with the frequency of the applied voltage. Explain briefly how the phenomenon of resonance in the circuit can be used in the tuning mechanism of a radio or a TV set? [Delhi 2011]

Ans.Refer to ans.19 and for resonance condition refer to ans. 30(b)
35. (i) What do you understand by sharpness of resonance in a series L-C-R circuit? Derive an expression for O-factor of the circuit.

(ii) Three electrical circuits having AC sources of variable frequency are shown in the figures. Initially, the current flowing in each of these is same. If the frequency of the applied AC source is increased, how will the current flowing in these circuits be affected? Give the reason for your answer.

Ans.







36.  A series L-C-Rcircuit is connected to an AC source having voltage V = V_m sin wt Derive the expression for the instantaneous current I and its phase relationship to the applied voltage. Obtain the condition for resonance to occur. Define power factor. State the conditions under which it is

• maximum and 
• minimum.       [All India 2010]

Ans.

Ans.

Ans. (i) Refer to ans 12(i)

(ii) Refer to ans. 35 (i)

Quality factor depends on ohmic resistance of L-C-R Series.Ac and Inductance and Capacitance of L-C-R Series AC Circuit.

Ans.

Ans.


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