Students must start practicing the questions from CBSE Sample Papers for Class 12 Physics with Solutions Set 6 are designed as per the revised syllabus.
CBSE Sample Papers for Class 12 Physics Set 6 with Solutions
Time Allowed: 3 Hours
Maximum Marks: 70 Marks
General Instructions:
- There are 35 questions in all. All questions are compulsory.
- This question paper has five sections: Section A, Section B, Section C, Section D and Section E. All the sections are compulsory.
- Section A contains eighteen MCQ of 1 mark each, Section B contains seven questions of two marks each, Section C contains five questions of three marks each, section D contains three long questions of five marks each and Section E contains two case study-based questions of 4 marks each.
- There is no overall choice. However, an internal choice has been provided in section B, C, D and E. You have to attempt only one of the choices in such questions.
- Use of calculators is not allowed.
Section – A (18 Marks)
Question 1.
In the given pattern of electric field lines with a proton is released from point P to Q. Which of the statement is correct for the given figure: [1]
(a) Electric field is the strongest in the case of figure 1.
(b) Protons, in figure 2, will have the greatest momentum.
(c) Protons, in figure 1, will have greater Momentum than that of figure 2.
(d) Protons, in figure 3, will have the greatest momentum.
Answer:
(d) Protons, in figure 3, wilt have the greatest momentum.
Explanation:
The lines of force near point P are closest to each other in figure 3 creating a strong electric field near the proton giving it greater momentum then the other two.
Question 2.
Considering the following figure, which of the given statement is true? [1]
(a) Current will pass from diode 1 and resistance R respectively.
(b) Current will pass from resistance R and diode 2 respectively.
(c) Current will pass from both the diodes equally following by resistance R.
(d) Current will pass from resistance R followed by diode 1.
Answer:
(b) Current wilt pass from resistance and diode 2 respectively
Explanation:
In the given circuit, diode 2 is forward biased and diode 1 is reverse bias. The path of the current will be from positive pole of the battery, through resistance, diode 2, to negative pole of the battery.
Question 3.
pn-photodiode is fabricated with band gap of 2.8eV. The energy associated with the wavelength of 600nm will be: [1]
(a) 2.07
(b) 2.80
(c) 2.0
(d) 4.14
Answer:
(a) 2.07
Explanation:
Here, λ = 600nm = 600 × 10-9m
Energy associated with the wavelength will be,
E = \(\frac{h c}{\lambda}\)
= \(\frac{6.22 \times 10^{-34} \times 3 \times 10^8}{600 \times 10^{-9}}\)
= 2.07 eV
Question 4.
Which of the following is true? [1]
(1) By using a resistor in a circuit, power factor can be increased.
(2) Appropriate capacitance of a capacitor can improve the power factor
(3) Capacitor greater than that of inductor can make the value of Z < R
(a) Both (1) and (2)
(b) Only (1)
(c) Only (2)
(d) All of the above
Answer:
(c) Only 2
Explanation:
Power factor can be improved by use of capacitor of appropriate capacitance in the circuit. By adjusting C, \(\frac{1}{\omega C}\) can be made equal to L, which will bring the value of Z, equal to R. So that power factor can become unity i.e. 1.
Question 5.
The Pupil of your eye is of 2mm and is very sensitive to light of wavelength 555nm. Then angular limit of resolution of your eyes is: [1]
(a) 3.39 × 10-9 nm
(b) 3.39 × 10-4 nm
(c) 3.39 × 10-12 nm
(d) 3.39 × 10-6 nm
Answer:
(b) 3.39 × 10-4 nm
Explanation:
Here,
λ = 555nm = 555 × 10-9m
D = 2mm = 2 × 10-3m
Angular limit of resolution of the eye is given by,
dθ = \(\frac{1.22 \lambda}{D}=\frac{1.22 \times 555 \times 10^{-9}}{2 \times 10^{-3}}\)
= 3.39 × 10-4 nm
Question 6.
A Slit of width a is illuminated by monochromatic light of wavelength 400nm at normal incidence. The value of width, when the first minimum and maximum falls at an angle of diffraction of 30°. [1]
(a) 12 × 10-7 m and 9 × 10-7 m
(b) 8 × 10-7 m and 12 × 10-7 m
(c) 1.3 × 10-7 m and 1.9 × 10-7 m
(d) 4 × 10-7 m and 6 × 10-7 m
Answer:
(b) 8 × 10-7 m and 12 × 10-7 m
Explanation:
For the first minumum,
sin θ1 = \(\frac{\lambda}{a}\)
amin = \(\frac{\lambda}{\sin \theta_1}=\frac{400 \times 10^{-9}}{\sin 30^{\circ}}\)
= \(\frac{400 \times 10^{-9}}{0.5}\)
= 8 × 10-7 m
For the first maximum,
sin θ2 = \(\frac{3\lambda}{2a}\)
amax = \(\frac{3 \lambda}{2 \sin \theta_1}=\frac{400 \times 10^{-9}}{\sin 30^{\circ}}\)
= \(\frac{3 \times 400 \times 10^{-9}}{2 \times 0.5}\)
= 12 × 10-7 m
Question 7.
A steel wire of Length l and magnetic moment m is bent into a circuLar arc. The new magnetic moment will be: [1]
(a) \(\frac{m}{\pi}\)
(b) \(\frac{2m}{\pi}\)
(c) \(\frac{2 \pi}{m}\)
(d) \(\frac{2 r}{\pi}\)
Answer:
(b) \(\frac{2m}{\pi}\)
Explanation:
When the wire is bent into a semicircular arc, the separation between the poles changes from I to 2r.
I = πr or r = \(\frac{I}{\pi}\)
New magnetic moment will be, \(\frac{m}{I} \times \frac{2 I}{\pi}=\frac{2 m}{\pi}\)
Question 8.
A Ray of light travelling from medium 1 to medium 2 makes an angle of incidence in medium 1 and angle of refraction in medium 2. [1]
(a) Medium 2 is optically denser
(b) Medium 2 is optically rarer
(c) Optical density of both mediums is same
(d) None of the above
Answer:
(a) Medium 2 is optically denser
Explanation:
Angle of refraction in medium 2 is less than the angle of incidence in medium 1. The ray bends towards the normal and medium 2, so medium to is optically denser than medium 1.
Question 9.
Identify the given image: [1]
(a) An astronomical telescope when final image is formed at infinity.
(b) A microscope when final image is formed at the least distance of distinct vision.
(c) A simple microscope.
(d) A terrestrial telescope.
Answer:
(d) A terrestrial telescope
Explanation:
A refracting type of telescope used to see erect, distant, earthly objects is known as terrestrial telescope. It uses an additional convex lens between object eyepiece for obtaining an erect image.
Question 10.
Which of the given statement is incorrect about sign convention for applying loop rule? [1]
(a) We can take any direction as a direction of the transversal.
(b) The emf of a cell is taken positive if the direction of transfer cell is from positive to negative terminal.
(c) Total incoming current is less than total outgoing current.
(d) Both (b) and (c).
Answer:
(b) The emf of a cell is taken positive if the direction of transfer cell is from positive to negative terminal.
Explanation:
The emf of a cell is taken negative if the direction of transfer cell is from positive to negative terminal. The emf of a cell is taken positive if the direction of transfer cell is from negative to positive terminal.
Question 11.
Which of the given statement is correct? [1]
(a) Kirchhoffs law also explains conservation of momentum.
(b) Kirchhoffs law is an extension of Ohm’s Law.
(c) Algebraic sum of currents in Kirchhoffs law is always 1.
(d) The currents going towards a junction are always taken negative.
Answer:
(b) Kirchhoffs law is an extension of Ohm’s law.
Explanation:
Kirchhoffs law only explains conservation of charge and conservation of energy.
Related Theory
As per Kirchhoffs first law, the algebraic sum of the currents at any junction is zero and the currents flowing towards the junction are always taken positive while the currents flowing away from the junction are taken negative.
Question 12.
A capacitor is disconnected after charging to a DC source in an ideal parallel LC circuit. [1]
Then:
(a) the current in the circuit becomes zero instantaneously. .
(b) the current in the circuit grows monotonically.
(c) the current in the circuit oscillates instantaneously.
(d) None of the above
Answer:
(c) the current in the circuit oscillates instantaneously.
Explanation:
When a charged capacitor is allowed to get discharge through a non- resistive inductor, the energy of the system continuously surges back and forth between the electrostatic and magnetic forms.
Question 13.
In the given figure, apply the Kirchhoffs law in the loop ABEFA; [1]
(a) 3I1 + 5I3 = 7
(b) 3I1 + 5I3 = 12
(c) 5I3 – 5I1 = 7
(d) 2I3 + 5I1 = 12
Answer:
(a) 2I3 + 5I1 = 12
Explanation:
Mathematically, ΣE = ΣIR Starting from BE, the product of current I3 and a resistor 2Ω and the sum of product of current I1 and the register 5Ω in AF will be equal applied voltage 12V.
Question 14.
If a conductor is moved in a direction parallel to the magnetic field, with no induced current be developed in the conductor? [1]
(a) False
(b) True
(c) Can’t say
(d) None of these
Answer:
(b) True
Explanation:
Induced current will not be developed in a conductor if it is moved in the direction parallel to the magnetic field. Such as, Lorentz Force on the free electrons in the conductor is zero and consequently no potential difference is produced across the two ends of the conductor.
Question 15.
Red colour is used for danger signals but they are useful in foggy conditions. The given statement is: [1]
(a) True
(b) False
(c) True in some cases
(d) None of these
Two statements are given-one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below: [1]
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true and R is NOT the correct explanation of A.
(c) A is true but R is false.
(d) A is false and R is also false.
Answer:
(a) True
Explanation:
According to Rayleigh’s law, the intensity of scattered light is inversely proportional to the fourth power of the wavelength. In the visible spectrum, red colour has the largest wavelength, it is scattered the least. Even in foggy conditions, set a signal covers large distances without any appreciable loss of intensity due to scattering. Therefore, red coloured danger signal can work perfectly in foggy conditions.
Questions 16.
Assertion (A): Electromagnetic waves can be deflected by magnetic or electric field.
Reason (R): Electromagnetic waves consists of charged particles. [1]
Answer:
(d) A is false and R is also false.
Explanation:
An electromagnetic wave cannot be deflected by magnetic or electric field. It does not consist of charged particles but they merely propagate in the form of electric and magnetic fields varying both in space and time.
Question 17.
Assertion (A): The phenomenon of lagging of magnetic induction behind the magnetising field is called hysteresis. [1]
Reason (R): Hysteresis loop provides us information about retentivity, coercivity and hysteresis loss of a magnetic material.
Answer:
(b) Both A and R are true and R is NOT the correct explanation of A.
Explanation:
Hysteresis loop, is a cycle, explains that the magnetic field the lags behind the magnetising field intensity. The value of magnetic field, when magnetic field intensity is decreasing, is always more than when magnetic intensity is increasing.
Related Theory
Practical importance and study of hysteresis loop helps us in proper selection of magnetic materials for designing cores of transformers and electromagnets
Question 18.
Assertion (A): Protons and neutrons are considered as particles having orbit and spin motion to explain hyperfine structure of the spectral lines. [1]
Reason (R): When a particle like proton moves in a closed path, it possess angular momentum and at the same time it produces magnetic field.
Answer:
(a) Both A and R are true and R is the correct explanation of A.
Explanation:
To explain hyperfine structure of the spectral lines, Pauli suggested angular momentum for the nucleus. The magnetic field at large distances may be described due to magnetic dipole located at the centre of the current loop.
Section – B (14 Marks)
Question 19.
Why are the electric field always normal to the potential surface at every point? [2]
Answer:
If the field were not normal to the equipotential surface, it would have a non-zero component along the surface. So, to move a test charge against this component, a work would have to be done. But there is no potential difference between any two points on an equipotential surface and consequently in the work is required to move a test charge on the surface.
Hence, the electric field must be normal to the equipotential surface at every point. As the name suggests, the body has equal potential at every point of the surface making potential difference is zero.
Question 20.
What is a depletion region and potential barrier? [2]
OR
A student colours the lower part of a convex lens while experimenting with convex lens. By performing the experiment with half black lens, what will the student observe?
Answer:
The region in the vicinity of p-n junction which is depleted of free charge carriers and has only immobile ions is called depletion region.
The potential difference set up across a p-n junction due to accumulation of opposite charges on its two sides is called potential barrier. It opposes the further diffusion of electrons and holes across the junction.
OR
Initial image formed by the lens was formed right but the new image after colour in the bottom half of the lens will decrease the intensity of the image. The light gathering power of the lens depends on its diameter without altering the size of the image.
Question 21.
Draw the circuit diagram consisting of resistance and forward and reverse biased diode respectively. [2]
Answer:
Question 22.
An AC generator consists of coil with 500 turns, rotating at constant angular speed of 20 Radian per second in a uniform magnetic field of 0.02T. If the cross-sectional area of coil is 2m2 and resistance is 100 Ohm. Then [2]
(A) Calculate the maximum current drawn from the generator.
(B) Calculate the maximum power dissipation in the coil.
Answer:
Here, n = 500
A = 2m2, R = 100Ω
ω = 20 rad s-1
B = 0.02T
(A) Maximum emf produced in coil,
ε0 = n BAw = 500 × 0.02 × 2 × 20
= 400V
The resistance of the coil is 100 Ω ,
The maximum current draw from the coil will be,
I0 = \(\frac{\varepsilon_0}{R}=\frac{400}{100}\) = 4A
(B) Maximum power dissipation in the coil,
ε0l0 = 400 × 4 = 800 W
Question 23.
The daily use sunglasses have curved lenses. So why do we see normally through sunglasses? [2]
Answer:
The two surfaces of the sunglasses possess same radii of curvature. Since one surface is convex, the other is equally concave, the powers of the two surfaces are equal but of opposite signs cancelling out. Therefore, sunglasses do not possess any power.
Question 24.
An early type of x-Ray tube, used by Roentgen, was gas filled. Due to which defects it cannot be used now? [2]
OR
Explain the formation of secondary rainbow.
Answer:
The quality of x-rays produced changes with the pressure of the gas inside the tube. During the operation of the tube, of the walls of the tube absorb the gas and it results into a fall in pressure. When the pressure decreases below a limit, the discharge stops passing through the gas.
Very high potential difference had to be applied between electrodes of the x-ray tubes but the main defect of the gas filled in x-ray tube is that intensity and the quality of x-Ray produced is not be controlled separately.
OR
It is the result of few-step process. Refraction with dispersion followed by internal reflection and finally reflection.
It is found that red light emerges at an angle of 50°, related to the incoming sunlight and violet light emerges at an angle of 53°. Thus, the observer sees secondary rainbow with violet colour on top and rec colour on the bottom. The intensity of light is reduced at second internal reflection and hence it is fainter than primary rainbow.
Thus, in the case of plastic lens the thickness of the lens should be increased to keep the same focal length as that of the glass lens to give the same power.
Question 25.
Give a brief account of yous factors on which the stability of the nucleus is determined. [2]
Answer:
The stability of a nucleus is determined by the value of its binding energy per nucleon. In general, higher the value of the binding energy per nucleon more stable the nucleus is.
The stability of the nucleus is also determined by the neutron and proton ratio. For stable nuclei, neutron Proton ratio should be greater than 1.
Apart from the high value of binding energy per nucleon and proper neutron Proton ratio, the stability of A nucleus determined by the consideration, whether it is an even-even nucleus, even-odd nucleus, odd-even nucleus or odd-odd nucleus depending on the value of Z and A.
Section – C (15 Marks)
Question 26.
Draw a circuit diagram of full wave rectifier with its input and output characteristics. [3]
Answer:
Question 27.
Derive a relation between speed of light, permeability constant and permittivity constant. [3]
OR
An electric motor is operating at full speed drawing 2A when 200V is applied. If the resistance of the armature is 5 Ω, find the emf and effciency of the motor.
Answer:
Relation between µ0, ε0 and c.
We know that,
OR
I = \(\frac{\mathrm{E}-e}{\mathrm{R}}\)
2 = \(\frac{200-e}{5}\)
10 = 200 – e
e = 190 v
Input power EI = 200 × 2 = 400 W
Output power = EI – I2R
= 200 × 2(2)2(5)
= 380 W
Efficiency, η = \(\frac{e}{E}\) × 100
= \(\frac{190}{200}\) × 100
= 95%
Question 28.
During an interference experiment, two sources are infinitely close to each other. How will it affect the interference pattern? What if two sources are far apart from each other? [3]
Answer:
When two coherent sources are placed infinitely close to each other, the fringe width becomes very large. Even a single fringe may occupy the entire screen. Due to this, the interference pattern may not be observable. As the distance between the two sources is increased, the fringe width goes on increasing. At a very large separation, it becomes too small to be detected. The interference pattern again cannot be observed.
Question 29.
A light of wavelength 6000A falls on the reflecting plane surface. Find the wavelength and frequency of the reflected light and for what angle of incidence is the reflected ray normal to the incident ray? [3]
OR
Why does a concave lens produce a virtual and diminished image independent of the location of object?
Answer:
The speed of light is fixed in a given medium and frequency of light does not change during its reflection of the surface. Consequently, the wavelength should also remain unchanged.
As we know,
Speed of reflected light is equal to speed of light in air
c = 3 × 108 ms-1
Wavelength of reflected light is equal to wavelength of incident light,
= 6000A = 6 × 10-7m
Frequency of reflected light,
ν = \(\frac{c}{\lambda}\)
= \(\frac{3 \times 10^8}{6 \times 10^{-7}}\)
= 5 × 1014Hz
By the law of reflection, angle of incidence is equal to angle of reflection.
Given that angle of incidence + angle of reflection = 90°.
angle of reflection = 45°
OR
From the Lens formula, we have,
\(-\frac{1}{u}+\frac{1}{v}=\frac{1}{f}\)
or v = \(\frac{u f}{u+f}\)
The focal length of a concave lens is negative. Since distance of a real object from the tens is also negative, it follows that the position formed of the image will also be negative. Therefore, a concave lens forms the image on the same side of the object.
m = \(\frac{f}{u+f}\)
Since both focal length and position of the object or negative, the value of magnification is positive and always less than one. Therefore, the image formed is always diminished.
Question 30.
Side AB of the given mirror is silver. Find the focal length of the given arrangement. [3]
Answer:
The arrangement acts as a concave mirror of focal length F, formed by a Plano convex lens of focal length f, a plane mirror of focal length oo and again a Plano convex lens of focal length f.
\(\frac{1}{F}=\frac{1}{f}+\frac{1}{f_m}+\frac{1}{f}=\frac{2}{f}+\frac{1}{\infty}=\frac{2}{f}\)
From the Lens Maker’s Formula,
\(\frac{1}{f}=(\mu-1)\left[\frac{1}{R}-\frac{1}{\infty}\right]=\frac{\mu-1}{R}\)
F = \(\frac{f}{2}=\frac{R}{2(\mu-1)}\)
Section – D (15 Marks)
Question 31.
Define the term excitation, ionization energy, and excitation, ionization potential. [5]
OR
Light intensity of 10-5 Wm-2 falls on a sodium photocell of surface area 2cm2. Assuming the top 5 layers of sodium absorb the incident energy, estimate the time required for photoelectric emission in the wave picture of radiation. The work function of the metal is given to be about 2eV.
Answer:
Excitation energy of an atom is defined as the energy required by electron to jump from the ground state to any one of the excited states. First excited energy of hydrogen,
E2 – E1 = -3.4 – (-13.6) = 10.2eV
Second excited energy of the hydrogen,
E3 – E2 = -1.5 – (-13.6) = 12.09eV
Ionization energy is defined as the energy required to knock and electron completely out of atom. The energy required to take an electron from its ground state to its outermost orbit. After the removal of the electron the atom is left with positive charge and is said to be ionized.
Ionization energy of hydrogen,
E∞ – E1 = 0 = -(-13.6) = 13.6eV
Excitation potential is the accelerating potential which gives to a bombarding electron, suffcient e nergy to excite t he target atom by raising one of its electrons from an inner of its outer orbit.
First excitation potential of hydrogen,
= -3.4 -(-13.6) = 10.2V
Second excitation potential of hydrogen,
-1.5 -(-13.6) = 12.09V
Ionisation potential is the accelerating potential which gives you a bombarding electron, suffcient e nergy t o i onize t he target atom by knocking one of its electrons completely out of atom.
Ionisation potential of hydrogen,
0 – (-13.6) = 13.6eV
OR
Support sodium has one conduction electron available per atom
Effective atomic area = 10-20
Number of conduction electron in 5 layers,
n = \(\frac{5 \times \text { Area of } 1 \text { layer }}{\text { Effective atomic area }}\)
n = \(\frac{5 \times 2 \times 10^{-4}}{10^{-20}}\) = 1017
Intensity = \(\frac{\text { Energy }}{\text { Area } \times \text { Time }}=\frac{\text { Power }}{\text { Area }}\)
Incident power = incident intensity × area
P = 10-5 × 2 × 10-4
P = 2 × 10-9W
In terms of wave picture, incident power is uniformly absorbed by all the electrons continuously.
Energy absorbed per second per electron,
E = \(\frac{2 \times 10^{-9}}{10^{17}}\) = 2 × 10-26W
T = \(\frac{\text { Energy required for electron }}{\text { Energy absorbed per second per electron }}\)
T = \(\frac{2 \times 10^{-26}}{2}\) = 1.6 × 10-7s = 0.5 years
Question 32.
Determine the electric field on equatorial line of an electric dipole and identify the electric field if the dipole has a very small length. [5]
OR
By defining the instantaneous power of the circuit, derive an expression for average power of an AC circuit.
Answer:
Consider an electric dipole consisting of two equal and opposite charges +q and -q, separated by a distance of 2 and placed in a free space. Let point P be an equatorial line of the dipole at distance r from the centre of a dipole.
Electric field due to both charges will have equal magnitude by two adjacent sides of PL and PM of parallelogram. PN will be the diagonal of the parallelogram representing the resultant electric field due to the dipole. Using the Triangle Law of addition of vectors, we get
q(2a) = p, magnitude to the electric dipole movement of the dipole.
E = \(\frac{1}{4 \pi \varepsilon_0} \cdot \frac{p}{\left(r^2+a^2\right)^{3 / 2}}\) (along PX’)
It is noted that the direction of electric field at a point on the equatorial line of the dipole is from positive charge to negative charge. Therefore, in vector notation
\(\overrightarrow{\mathrm{E}}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\vec{p}}{\left(r^2+a^2\right)^{3 / 2}}\)
When dipole is of very small length, a << r, then a2 can be neglected as compared to r2. Therefore, for an electric dipole of very small length,
E = \(\frac{1}{4 \pi \varepsilon_0} \cdot \frac{p}{r^3}\) (along PX’)
OR
Instantaneous power of an AC circuit is defined as the product of instantaneous EMF and the instantaneous current unit.
Suppose in an a.c. circuit, the voltage and current at any instant are given by,
ε = ε0 sin ωt and I = I0 sin (ωt – Φ)
where, Φ is the phase angle by which the voltage s leads the current I.
The instantaneous power is given by,
P = εI = ε0I0 sin ωt. sin (ωt – Φ)
= \(\frac{\varepsilon_0 I_0}{2}\) [2 sin ωt. sin (ωt – Φ)]
= \(\frac{\varepsilon_0 I_0}{2}\) [cos Φ – cos (2ωt – Φ)]
[∵ 2 sin A sin B = cos(A – B) – cos (A + B)]
Average power dissipated per cycle,
= Average of \(\frac{\varepsilon_0 I_0}{2}\) [cos Φ – cos (2ωt – Φ)]
The second cosine term [cos (2ωt – Φ)] is time dependent. Its average over a cycle is zero.
∴ Pav = \(\frac{\varepsilon_0 I_0}{2}\) cosΦ
or Pav = \(\frac{\varepsilon_0}{\sqrt{2}} \cdot \frac{I_0}{\sqrt{2}}\). cosΦ
or Pav = εrms Irms cosΦ = εrms Irms. \(\frac{\mathrm{R}}{\mathrm{z}}\)
Question 33.
Prove that an ideal capacitor connected to an AC source does not dissipate any power. [5]
OR
(A) Obtain the expression for the angle of incidence of a ray of light which is incident on the face of a prism of refracting angle A. So that it suffers total internal reflection at the other face. (Given the refractive index of the glass of the prism is µ).
(B) Draw a labelled ray diagram of an astronomical telescope in the near point adjustment position.
Answer:
When an AC voltage is applied to a capacitor, the current leads the voltage by \(\frac{\pi}{2}\)
V = V0 sin ωt
and I = I0 sin (ωt + \(\frac{\pi}{2}\)) = I0 cos ωt
Work done in the circuit in small time dt will be,
dW = Pdt =VI dt =V0I0 sin ωt cos ωt dt
= \(\frac{V_0 I_0}{2}\) sin 2 ωt dt
The average power dissipated per cycle in the capacitor is,
Thus the overage power dissipated per cycle in a capacitor is zero.
OR
(A) Let C is the critical angle,
sin C = \(\frac{1}{\mu}\)
(B)
r2 = C for total internal reflection
r1 + r2 = A
r1 + C = A
r1 = A – C
µ = \(\frac{\sin i}{\sin r_1}\)
⇒ sin i = µ sin r1
sin i = µ sin(A – C)
i = sin-1 µ sin (A- C)
Section – E (8 Marks)
Question 34.
In a half cycle, no net power is consumed by the capacitor. The external source has to supply an energy of \(\frac{1}{2}\) CV2 to charged capacitor to a potential difference V but this energy is returned back during the discharging process. When the capacitor is connected to an AC source, it absorbs energy from the source for a quarter cycle and is charged. It returns energy to the source in the next quarter cycle as it is discharged.
(A) In an only resistor AC circuit, register is replaced by a capacitor. How the power of *he AC circuit will be affected? What if an inductor replaces the capacitor? [1]
(B) A perfect self-inductance connected to an AC source reduces current but does not produce any heating effect. Why? [1]
(C) No matter how much power is consumed by an inductor connected to an AC source, why the average value is always zero?
OR
No power is consumed by an ideal inductor in an AC circuit. Same goes for a capacitor. Why? [2]
Answer:
(A) When an AC circuit contains resistor only, its power is maximum but if it is replaced by an inductor or a capacitor, the power of the AC circuit always decreases
(B) As the average power consumed per cycle in an inductive AC circuit is zero, there is no heating in the circuit but the inductive reactance plays the same role in in an AC circuit as the resistance in DC circuit So, an inductance reduces current in an AC circuit.
(C) the average power consumed per cycle in an inductor connected to an AC source is always zero. During the first quarter of each cycle, magnetic flux through the inductor builds up and energy is stored in the inductor from the external source as the current in the circuit increases. In the next quarter of cycle, as the current decreases, the flux decreases and the Stored energy is returned to the source.
OR
In an AC circuit containing inductor only, the voltage leads the current by phase angle of \(\frac{\pi}{2}\), of show the average power consumed per cycle is zero.
Pav = Vrms – Irms cos \(\frac{\pi}{2}\) = 0
In an AC circuit containing capacitor only, lags lacks behind the current by the phase angle of \(\frac{\pi}{2}\), so average power consumed per cycle is zero
Pav = Vrms – Irms cos (-\(\frac{\pi}{2}\)) = 0
Question 35.
Kirchhoff’s circuit rules constitute two equalities in about current and potential difference. A German scientist, Kirchhoff, was the first to describe them. This extended Georg Ohm’s work and arrived earlier James Clerk Maxwell’s. Kirchhoff’s rules, often known as Kirchhoff’s laws, are widely utilized in electrical engineering.
These laws apply in both time and frequency levels and serve as a basis for network theory. In the low-frequency limit, both of Kirchhoff’s laws can be interpreted as corollaries of Maxwell’s equations. They are accurate for DC circuits and AC circuits at frequencies when electromagnetic radiation wavelengths are very big in comparison to the circuits.
(A) If an AC circuit shifted to a DC circuit, how will it affect the Kirchhoff’s laws? [1]
(B) What is the significance of Kirchhoff’s laws? [1]
(C) Potentiometer is based on Kirchhoff’s law on to measure the potential difference, but why is potentiometer preferred over voltmeter?
OR
State the sign conventions used for applying the loop rule. [2]
Answer:
(A) There will be no change in either of the Kirchhoffs laws the as it can be applied to both AC and DC circuit.
(B) Ohm’s law can be used to analyse simple electrical circuits but when the circuit contains more than one source of EMF and the distribution of current in various branches of the circuit, it is quite complicated to analyse. In such cases, KirchhofPs laws to analyse such complicated circuits.
(C) When a voltmeter is connected across the two poles of a cell, the voltmeter measures The Terminal potential difference between the two poles of the cell, which is always less than the emf of the cell But when a potentiometer is used for the measurement of emf of a cell, it does not draw any current from the cell, measuring the emf of the cell only. Due to this reason, potentiometer is preferred over voltmeter for measuring potential difference.
OR
Sign conventions use for applying the loop rule also known as Kirchhoff’s second law are:
- The emf of the cell is taken as positive if direction of transversal is from its negative to the positive terminal, and negative if the direction of transversal is from its positive to negative terminal.
- The current resistance product is taken as positive if the resistor is traversed in the same direction of assumed current. And negative if the resistor is traversed in the opposite direction of the assumed current.