RBSE 12th Physics Model Paper Set 5 with Answers in English 

Students must start practicing the questions from RBSE 12th Physics Model Papers Set 5 with Answers in English Medium provided here.

RBSE Class 12 Physics Model Paper Set 5 with Answers in English

Time: 2 Hours 45 Min.
Max Marks: 56

General instruction to the Examinees:

• Candidate must write first his/her Roll No. on the question paper compulsorily.
• All the questions are compulsory.
• Write the answer to each question in the given answer book only.
• For questions having more than one part the answers to those parts are to be written together in continuity.

Section – A

Multiple Choice Questions

Question 1.
Write the Correct Answer from multiple choice questIon 1 (Ito Ix) and write In given answer book:
(i) According to Gauss’s theorem, electric field of an infinitely long straight wire is propotional to: [1]
(a) r
(b) \(\frac{1}{r^{2}}\)
(c) \(\frac{1}{r^{3}}\)
(d) \(\frac{1}{r}\)
Answer:
(d) \(\frac{1}{r}\)

(ii) In a region of constant potential: [1]
(a) The electric field is uniform
(b) The electric field is zero
(c) There can be no charge inside the region
(d) Both (b) and (e) are correct
Answer:
(d) Both (b) and (e) are correct

(iii) 1 ampere current is equivalent to [1]
(a) 6.25 x 10¹⁸ electrons s^-1
(b) 2.25 x 10¹⁸ electrons s^-1
(e) 6.25 x 10¹⁴ electrons s^-1
(d) 2.25 x 10¹⁴ electrons s^-1
Answer:
(a) 6.25 x 10¹⁸ electrons s^-1

(iv) A strong magnetic field is applied on a stationary electron. Then the electron: [1]
(a) Moves in the direction of the field
(b) Remains stationary
(c) Moves perpendicular to the direction of the field
(d) Moves opposite to the direction of the field.
Answer:
(b) Remains stationary

(v) A solenoid is connected to a battery so that a steady current flows through it. ifan iron curt? is inserted into the solenoid, the current will [1]
(a) Increase
(b) Decrease
(c) Remains same
(d) First increase then decrease
Answer:
(b) Decrease

(vi) In an experiment on photoelectric effect the frequency f of the incident light is plotted against the stopping potential V₀. The work function of the photoelectric surface is given by (e is electronic charge): [1]

(a) OB x e in eV
(b) OB in volt
(c) OA in eV
(d) The slope of the line AB
Answer:
(a) OB x e in eV

(vii) Outside a nucleus: [1]
(a) Neutron is stable
(b) Proton and neutron both are stable
(c) Neutron is unstable
(d) Neither neutron nor proton is stable
Answer:
(c) Neutron is unstable

(viii) The approximate ratio of resistances in the forward and reverse bias of the P- Njunction diode is: [1]
(a) 10² :1
(b) 10^-2 :1
(c) 1:10^-4
(d) 1:10⁴
Answer:
(d) 1:10⁴

(ix) The following logic circuit represents: [1]
(a) NAND gate with output O = \(\bar{X}+\bar{Y} \)
(b) NOR gate with output O = \(\overline{X+Y}\)
(c) NAND gate with output O = \(\overline{X Y}\)
(d) NOR gate with out O = \(\bar{X}+\bar{Y}\)
Answer:
(b) NOR gate with output O = \(\overline{X+Y}\)

Question 2.
Fill in the blanks:

(i) The potential at each point on an ………………………………. surface is equal. [1]
Answer:
equipotential

(ii) Cell is the device which converts ……………………………. energy into electrical energy. [1]
Answer:
chemical

(iii) Eddy currents are also …………………………………. currents. [1]
Answer:
Foucault

(iv) The reverse breakdown voltages of LEDs are low generally around ……………………… . [1]
Answer:
5V.

Question 3.
Give the answer of the following question in one line 3 (i to viii)

(i) Define the term electrical conductivity of a metallic wire. Write its SI unit. [1]
Answer:
The electrical conductivity (σ) of a metallic wire is defined as the ratio of the current density to the electric field it creates. Its SI unit is mho per meter (Ω-m)^-1

(ii) Two protons of equal kinetic energies enter a region of uniform magnetic field. The first proton enters normal to the field direction while the second enters at 30° to the field direction. Name the trajectories followed by them. [1]
Answer:
First proton has circular trajectories and second has helical.

(iii) Figure shows a current-carrying solenoid moving towards conducting loop. Find the direction of the current induced in the loop. [1]

Answer:
The direction of current in the coil is anti-clockwise. As North polarity of solenoid is moving towards the loop. So, to stop the motion of solenoid Current will induced in the coil in anticlockwise. direction (N-pole).

(iv) Write the expression for the de-Broglie wavelength associated, with a charged particle having charge q and mass m when it is accelerated by a potential. [1]
Answer:
A charged particle having charge q and mass m then kinetic energy of the particle is equal to the work done on it by the electric field. i.e.,K=qV
⇒ \(\frac{1}{2}\) mV² =qV
⇒ \(\frac{p^{2}}{2 m}\) = qV
⇒ p = \(\sqrt{2 m q V}\)
∴ de-Broglie wavelength,
λ = \(\frac{h}{p}=\frac{h}{\sqrt{2 m q V}}\)

(v) Show graphically, the variation of de-Broglie wavelength (λ) with the potential (V) through which an electron is accelerated from rest. [1]
Answer:

(vi) How is the radius of a nucleus related to its mass number? [1]
Answer:
The radius of a nucleus is proportional to the cube root of its mass number (A).
The approximate law is R = R₀A^1/3;
where R₀ = 1.2 x 10^-15m.

(vii) State the reason, why heavy water is generally used as a moderator in a nuclear reactor. [1]
Answer:
It absorbs neutrons as a fast rate via-reaction : n + P = Y + d Here, d is deuteron.

(viii) What is the difference between an n-type and a p-type extrinsic semiconductor? [1]
Answer:

n-type semiconductor	p-type semiconductor
It is formed by doping pentavalent impurities with tetravalent atoms.	It is formed by doping trivalent impurities with tetravalent atoms.
The electrons are majority carriers and holes are minority carries. (ne>>nh)	The holes are majority carriers and electrons are minority carriers. (nh>>n)

Question 4.
Define an equipotential surface. Draw equipotential surfaces
(i) In case of a single point charge.
(ii) In a constant electric field in Z-direction. Why the equipotential surface about a single charge are not equidistant?
(iii) Can electric field exist tangential to an equipotential surface? [1½]
Answer:
Any surface that has same electric potential at every point on it is called an equipotential surface.
(i) Equipotential surface in case of a single point charge

(ii) Equipotential surfaces when the electric field is in Z-direction.

The equipotential surfaces due to a single point charge is represented by concentric spherical shells of increasing radius, so they are not equidistant.

(iii) No, the electric field does not exist tangentially to an equipotential surface because no work is done in moving a charge from one point to the other on equipotential surface.

Question 5.
A 12 pF capacitor is connected to a 50 V battery. How much electrostatic energy is stored in the capacitor? If another capacitor of 6 pF is connected in series with it with the same battery connected across the combination, find the charge stored and potential difference across each capacitor. [1½]
Answer:
Energy stored in capacitor
= \(\frac{1}{2}\) C₁V² = \(\frac{1}{2}\) x 12 x 10^-12 x (50)²J
= 6 x 25 x 10^-10 J
= 15 x 10^-9J
With other capacitor 5 pF in series. total capacitance
(C) = \(\frac{C_{1} \times C_{2}}{C_{1}+C_{2}}=\frac{6 \times 12}{6+12} \mathrm{pF}\)
= \(\frac{12 \times 6}{18}\) = 4 pF
Charge stored in each capacitor is same and is given by
Q=CV=4 x 10^-12 x 50C = 2 x 10^-10C

Each of the capacitors will have charge equal to Q = 12 x 10^-10 C
Potential on capacitors with capacitance 12 pF is
= \(\frac{Q}{C_{1}}=\frac{12 \times 10^{-10}}{12 \times 10^{-12}}\) V = 16.67 V
Potential on capacitor with capacitance 6 pF is
= \(\frac{2 \times 10^{-10}}{6 \times 10^{-12}}\) V = 33.33 V

Question 6.
Twelve equal wires, each of resistance r, are joined to form a skeleton cube. The current enters at one corner and leaves at the diagonally opposite comer. Find the total resistance between these corners. [1½]

Answer:
Applying Kirchhoff’s second rule (ΣE + ΣRI = 0) to the closed mesh XYMLKAX, we get
-E+[r × 2I + r × I+r × 2I]= O
or 5rI = ε ……………………….. (i)
where E is the emf of the battery of negligible internal resistance. If r is the resistance of the cube between the diagonally opposite corners A and M, then according to the Ohm’s law, we have
6RI=E ………………………. (ii)
From eqs. (i) and (ii),
6RI = 5rI
Or
R = \(\frac{5}{6} r\)
Hence, the resistance of the cube between any two diagonally opposite corners is 5/6 of the resistance of each wire.

Question 7.
(a) In a meter-bridge [figure ], the balance point is found be at 39.5 cm from the end A, when the resistance S is of 12.5 Ω. Determine the resistance of R. Why are the connections between the resistors in a Wheatstone bridge or meter bridge made of thick copper strips?[1½]

(b) Determine the balance point of the bridge if R and S are interchanged.
(c) What happens if the galvanometer and cell are interchanged at the balance point of the bridge?
Would the galvanometer show any current?[1½]
Answer:
(a) We are given that S = 12.5Ω,
l₁ = 395cm.
and l₂ = 100 – 39.5 = 60.5cm
As \(\frac{R}{S}=\frac{l_{1}}{I_{2}}, R=\frac{l_{1}}{l_{2}} \times S\)
= \(\left(\frac{39.5 \mathrm{~cm}}{60.5 \mathrm{~cm}}\right)(12.5 \Omega)\) = 8.2 Ω

Thick copper strips are used for connections between the resistors as these have negligible resistance. This is done to minimize the resistances of the connections which are not accounted for in the bridge formula. Instead, if we use wires whose resistances are not negligible, these will have to be determine and taken into account.

(b) If R and S are interchanged, the balance point will lie at a distance of 60.5 cm from the end A.
(c) Since the bridge is balanced, there will be no difference if position of cell and galvanometer are interchanged as there will be no flow of current from galvanometer.

Question 8.
Derive the expression for the magnetic energy stored in an inductor, when a current I develops in it. Hence, obtain the expression for the magnetic energy density.[1½]
Answer:
Induced emf due to inductance in an Inductor is
ε = – \(\frac{L d I}{d t}\)
Work done against the emf:
dW=εdq
dW = L\(\frac{d I}{d t} d q\)
dW=LIdI (I= \(\frac{d q}{d t}\))

Thus, the work done in increasing the current in circuit from zero to I,
W = \(\int_{0}^{I} L I d I \)
W = L \(\int_{0}^{I} I . d I\)
W = L \(\left[\frac{I^{2}}{2}\right]_{0}^{I}\)
W = \(\frac{1}{2} L I^{2}\)

This work is stored in the form of magnetic energy in inductor.
Thus, stored magnetic energy,
U_m = \(\frac{1}{2} L I^{2}\)
Magnetic energy stored per unit volume,
\(\bar{U}=\frac{U_{m}}{V}\)
If the cross-sectional area of a solenoid is A and its length is 4 then

This is the relation for energy density.

Question 9.
An air-cored solenoid with length 30 cm, area of cross-section 25 cm^2, and number of turns 500 carries a current of 2.5 A. The current is suddenly switched off in a brief time 10^-3s. How much is the average back emf induced across the ends of the open switch in the circuit? Ignore the variation in magnetic field near the ends of the solenoid.[1½]
Answer:
We are given that, length of the solenoid, l= 30 cm = 0.30m
area of cross-section of the solenoid,
A= 25cm² = 25 x 10^-4m²
Current through the solenoid,
I=2.5 A
Total number of turns in the solenoid, N = 500
Time during which the current in switched off, t = 1 ms = 10^-3 s Magnetic field in a solenoid,
B = µ₀nI = µ₀\(\left(\frac{N}{l}\right)\) I

Total flux linked with the solenoid,
Φ_B = NBA
Or Φ_B = Nµ₀\(\left(\frac{N}{l}\right)\)IA
= \(\frac{\mu_{0} N^{2} I A}{l}\)

Let ε be the induced emf set up across the ends of the open switch in the circuit.
Clearly, ε = \(\frac{\phi_{B}}{t}=\frac{\frac{\mu_{0} N^{2} I A}{\lambda}}{10^{-3}}\)
Or

Question 10.
A convex lens of focal length f₁ is kept in contact with a concave lens of focal length f₂.
Find the focal length of the combination.[1½]
Answer:
Focal length for convex lens = f₁
Focal length for concave lens = – f₂
The equivalent focal length of a combination of convex lens and concave lens is given by
\(\frac{1}{F}=\frac{1}{f_{1}}+\frac{1}{-f_{2}}\)
⇒ F = \(\frac{f_{1} f_{2}}{f_{2}-f_{1}}\)

Question 11.
State the conditions for the phenomenon of total internal reflection to occur.[1½]
Answer:
Two essential conditions for total internal reflection are:

1. Light should travel from an. optically denser medium to an optically rarer medium.
2. The angle of incidence in the denser medium must be greater than the critical angle for the two media.

Question 12.
A convex lens of focal length 20 cm is placed coaxially with a concave mirror of focal length 10 cm at a distance of 50 cm apart from each other. A beam of light coming parallel to the principal axis is incident on the convex lens. Find the position of the final image formed by this combination. Draw the ray diagram showing the formation of the image. [1½]
Answer:
Image formed by the lens will be f at focus.

For mirror, u = -30, f = -10
According to lens formula,
\(\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\)
⇒ \(\frac{1}{-10}=\frac{1}{v}-\frac{1}{30}\)
⇒ \(\frac{1}{v}=\frac{1}{30}-\frac{1}{10}\)
⇒ \(\frac{1}{v}=\frac{1-3}{30}\)
⇒ υ = -15cm

Question 13.
At night, it is difficult to see objects outside from the glass window of a lighted room easily. Explain.[1½]
Answer:
The intensity of reflected light inside the room is large while the light coming from outside the room is negligible so it is difficult to see objects outside the room. If the light is off, then we see lights only coming from outside the room therefore we see outside objects easily.

Question 14.
State the law of radioactive decay. [1½]
Answer:
According to this law:
(i) Radioactivity is a nuclear phenomenon and the rate of emission of radioactive ray can not be controlled by physical or chemical process that mean neither can it be extended nor can it be reduced.

(ii) The nature of the disintegration of radioactive substance is statistical, this is, it is very difficult to say which nucleus will be disintegrated and which particle will emit α, β, and γ With the emission of α, β, and y rays in the process of disintegration one element change into another new element, its chemical and radioactive qualities are completely new.

(iii) At any time the rate of decay of radioactive atom is proportional to the number of atoms present at that time.

Question 15.
Prove that the rate of activity of a radioactive substance is inversely proportional of the square of its half-life.[1½]
Answer:
Activity of radioactive substance
R = \(\frac{d N}{d t}=-\lambda N \)
Or \(|R|=\frac{d N}{d t}=\lambda N\)
∴ Rate of activity

λ = \(\frac{0.6932}{T}\) ; where T is half-age.
∴ \(\frac{d R}{d t}=\frac{(0.6932)^{2}}{T^{2}} N\)
∴ \(\frac{d R}{d t} \propto \frac{1}{T^{2}}\)
i.e., the rate of activity of a radioactive substance is inversely proportional to the square of the half-life.

Section – C

Question 16.
(i) State Ampere’s circuital law. Use this law to obtain the expression for the magnetic field inside an air-cored toroid of average radius r, having n turns per unit length and carrying a steady current I.
(ii) An observation to the left of a solenoid of N turns each of cross-section areas A observes that a steady current I flows in the clockwise direction. Depict the magnetic field lines due to the solenoid specifying its polarity and show that it acts as a bar magnet of magnetic moment m = NIA,[3]

Or
With the help of a labelled diagram, state the underlying principle of a cyclotron. Explain clearly how it works to accelerate the charged particles? Show that cyclotron frequency is independent of energy of the particle. Is there an upper limit on the energy acquired by the particle? Give reason. [3]
Answer:
(i) Ampere’s circuital law states that the line integral of the magnetic field \(\vec{B}\) around any closed
equal to µ₀ (permeability constant) times the total current through this closed Mathematically,
\(\oint \vec{B} \cdot \overline{d l}=\mu_{0} I\) …………………………… (1)

Suppose there are n turns per unit length of toroid and I is the current flowing through it. Due to flow of current, a magnetic field is produced inside the turns of toroid. Magnetic field inside the toroid Suppose, there is a circular path of radius r inside the toroid.

Applying Ampere’s circuital law for this loop,

\(\oint \vec{B} \cdot \vec{d} l=\mu_{0} \Sigma I\) ……………………….. (1)
Total current flowing through closed loop,
ΣI = n × 2πr x I
= 2πrnl

Therefore from equation (1)
\(\oint \vec{B} \cdot \vec{d} l=\mu_{0} 2 \pi r n I\) ……………………………….. (2)
Here \(\vec{B}\) and \(\vec{d} l\) are in same direction.
∴ \(\oint \vec{B} \cdot \vec{d} l=\oint B d l \cos 0^{\circ}=\oint B d l\)
= \(=B \oint d l=B 2 \pi r\)

Therefore from equation (2)
B2πr = µ₀ In 2 πr
Or B = µ₀nI

(ii) Since, it is given that the current flows in the clockwise direction for an observer on the left side of the solenoid. It means that the left face of the solenoid acts as South pole and right face acts as North pole. south to North. Therefore, the magnetic field lines are directed from left to right in the solenoid.

Magnetic moment of a single current-carrying loop is given by m’ = IA. So, magnetic moment of the whole solenoid is given by
m = Nm’- N(IA)

Question 17.
(a) Draw a ray diagram for a convex mirror showing the image formation of an object placed anywhere in front of the mirror. [3]
(b) Use this ray diagram to obtain the expression for its linear magnification.
(c) Why are convex mirrors used as side-view mirrors in vehicles?
Or
Deduce with the help of ray diagram, the expression for the mirror equation in the case of convex mirror. [3]
Answer:

(b) As shown in figure
ΔA’B’P and ¿ABP are similar.
∴ \(\frac{A^{\prime} B^{\prime}}{A B}=\frac{P B^{\prime}}{P B}\)
Applying the new cartesian sign convention, we get
A’B’ = + h₂, AB = + h₁
BP’= – u, PB’ = + υ
∴ \(\frac{h_{2}}{h_{1}}=\frac{v}{-u}\)
Magnification, m = \(\frac{h_{2}}{h_{1}}=\frac{v}{-u}=\frac{-v}{u}\)

(c) A convex mirror is used as side-view mirror in vehicles because it has larger field of view as compared to other mirrors. The image so formed is small and erect.

Question 18.
(a) Define the term intensity in photon picture of electromagnetic radiation,
(b) A deuteron and an a-particle are accelerated with the same accelerating potential. Which one of the two has :
(i) greater value of de-Broglie wavelength, associated with it, and
(ii) less kinetic energy? Explain. [3]
Or
(a) Every metal has a definite work function why do photoelectrons not come out all with same energy if incident radiation is monochromatic? Why is there an energy distribution of photoelectrons?
(b) The energy and momentum of an electron are related to the frequency and wavelength of the associated matter wave by the relation :
E = hυ, p = \(\frac{\boldsymbol{h}}{\lambda}\)
But while the value of λ is physically significant the value of υ (and, therefore, the value of the phase speed υλ) has no physical significance. Why? [3]
Answer:
(a) Intensity: It is the number of photons passing through an area in a given interval of time. Its SI unit is watt/metre²
(b)
(i) de-Broglie wavelength is given by
λ = \(\frac{h}{\sqrt{2 m V_{0} q}}\)
⇒ λ ∝ \(\frac{1}{\sqrt{m q}}\)
\(\frac{\lambda_{d}}{\lambda_{\alpha}}=\frac{1 / \sqrt{2 m e}}{1 / \sqrt{4 m 2 e}}=\frac{2}{1}\)
Wavelength of deuteron is two times the wavelength of α-particle.

(ii) \(\frac{\mathrm{KE}_{d}}{\mathrm{KE}_{\alpha}}=\frac{v_{0} e}{v_{0} 2 e}=\frac{1}{2}\)
KE of deuteron is half of KE of α-particle.

Section-D

Question 19.
Using Gauss’s law, obtain the expression for the electric field due to uniformly charged spherical shell of radius R at a point P.
When point P is situated (i) inside the spherical shell, (ii) at surface (iii) outside the spherical shell.
Draw a graph showing the variation of electric field with r. [4]
Or
(i) Derive an expression for the electric field at any point on the equatorial line of an electric dipole.
(ii) Two identical point charges, q each are kept 2 m apart in air. A third point charge Q of unknown magnitude and sign is placed on the line joining the charges such that the system remains in equilibrium. Find the position and nature of Q. [4]
Answer:
Electric Field Intensity due to a Uniformly Charged
Spherical Shell:
Suppose the radius of the spherical shell is R and a charge +q is given to it.

(i) When point P is situated outside the spherical shell (i.e., r> R): To apply the Gauss’s theorem, a Gaussian surface is required. For simplicity taking O as centre and r as radius we draw a spherical Gaussian surface. The whole charged sphere is enclosed by this Gaussian spherical surface. Therefore the enclosed charge,
Σq = q
Therefore, from Gauss’s law,
Φ_E = \(\frac{\Sigma q}{\varepsilon_{0}}\)
Or Φ_E = \(\frac{q}{\varepsilon_{0}}\) ……………………………. (1)

Now, from the definition of electric flux

Equating equations (1) and (2), we have
E4πr² = \(\frac{q}{\varepsilon_{0}}\)
Or E = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r^{2}}\) …………………….. (3)
In vector form
\(\vec{E}=\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r^{2}} \hat{r}\) …………………….. (4)

(ii) When point P is situated inside the spherical shell: In this case also we draw a spherical Gaussian surface of radius r < R.
This spherical is completely inside the charged spherical shell as shown in the figure Here we see
Σq = 0
Therefore, by Gauss’s law
Φ_E = \(\frac{\Sigma q}{\varepsilon_{0}}\) = 0
or Φ_E = 0

From definition,
Φ_E = \(\oint_{S} \vec{E} \overrightarrow{d s}=\oint_{S} E d s \cos 0^{\circ}\)
(for spherical surface θ = 0⁰)
or Φ_E = \(E \oint_{S} d s=E 4 \pi r^{2}\)
Equating these values of Φ_E , we have
E x 4πr² = 0
∴ E = 0 ……………………….. (5)

(iii) When point P is situated at the surface of the spherical shell: In this case we can consider the surface of the spherical shell as Gaussian surface. Therefore to get the electric field at this surface we have to put r = R is equation (3),
E = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{R^{2}}\) ………………………. (6)
In vector form,
\(\vec{E}=\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{R^{2}} \hat{r}\) ………………………….. (7)
Where \(\hat{r} \) is unit vector in direction OP i.e., outwards.

Question 20.
Draw a labelled diagram of a half-wave rectifier circuit. State its working principle. Show the input-output waveforms. [4]
Or
(i) State briefly the processes involved in the formation of p-n junction, explaining clearly how the depletion region is formed.
(ii) Using the necessary circuit diagrams, show how the V-I characteristics of a p-n junction are obtained in (a) forward biasing (b) reverse biasing. How are these characteristics made use of in rectification? [4]
Answer:
Figure shows the circuit of a half-wave rectifier. In this an input alternating voltage source is connected to the primary coil of a transformer. In the second coil of the transformer, a diode and load resistor R_L are connected in series.

At the middle point of A and B from the secondary coil of the transformer, a desirable alternating V_s is obtained. This alternating potential can be more or less than the input alternating potential, if the used transformer is of step-up nature then V_s the alternating potential between A and B will be the more than the input potential V_in and if it is of step down nature than potential V_s will be less than V_in.

For positive half cycle of alternating potential V_i (or V_s) the end A of the secondary coil is connected to a positive potential and end B is connected to negative potential due to which the P end of the diode is at a higher potential than N end. Due to this diode will be in a state of forward biasing and will flow current.

In this state there will be a potential drop that the load resistor R_L. In the negative half cycle of input ac end A is connected to negative and end B is connected to positive potential, due to which diode is in reverse biasing state. In reverse biasing the current is negligible hence diode is in a non-conducting position. Hence there will be no current in the circuit.

In load resistor R_L the current flows only in one direction that is from A to B. Hence, potential drop at the ends of R_L is of d.c. nature.