State the two Kirchhoff's laws. Explain briefly how these rules are justified.
Kirchhoff’s First Law or Junction Rule states that “The sum of the currents flowing towards a junction is equal to the sum of currents leaving the junction.”
This is in accordance with the conservation of charge which is the basis of Kirchhoff’s current rule.
Here, I_{1}, I_{2} I_{3}, and I_{4} are the currents flowing through the respective wires.
Convention: The current flowing towards the junction is taken as positive and the current flowing away from the junction is taken as negative.
I_{3 }+ (− I_{1}) + (− I_{2}) + (− I_{4}) = 0
Kirchhoff’s Second Law or Loop Rule states that In a closed loop, the algebraic sum of the emf is equal to the algebraic sum of the products of the resistances and the currents flowing through them.
OR
“The algebraic sum of all the potential drops and EMFs along any closed path in a network is zero.”
For the closed loop BACB:
E_{1} − E_{2} = I_{1}R_{1} + I_{2}R_{2} − I_{3}R_{3}
For the closed loop CADC:
E_{2} = I_{3}R_{3} + I_{4}R_{4} + I_{5}R_{5}
This law is based on the law of conservation of energy.
In a beam of unpolarized light, the vibrations of light vectors are in all directions in a plane perpendicular to the direction of propagation.
In the linearly polarised light, the vibrations of light take place in a particular direction, perpendicular to the direction of wave motion.
Polarized light can be distinguished, from unpolarized light, when it is allowed to pass through a Polaroid. Polarized light does can show the change in its intensity, on passing through a Polaroid; intensity remains same in case of unpolarized light.
When the unpolarised light wave is incident on a polaroid, then the electric vectors along the direction of its aligned molecules, get absorbed; the electric vector, oscillating along a direction perpendicular to the aligned molecules, pass through. This light is called linearly polarized light.
(a) Draw a labelled diagram of a step-up transformer. Obtain the ratio of secondary to primary voltage in terms of number of turn and currents in the two coils.
(b) A power transmission line feeds input power at 2200 V to a step-down transformer with its primary windings having 300 turns. Find the number of turns in the secondary to get the power output at 220 V.
The current is drawn from a cell of emf E and internal resistance r connected to the network of resistors each of resistance r as shown in the figure. Obtain the expression for (i) the current draw from the cell and (ii) the power consumed in the network.
From the above figure, we can use the horizontal symmetry to deduce the current.
As both resistance in the circuit, as well as the internal resistance in the circuit, have the same resistance 'r'
The resistance of the loop I will be
Because the circuit I and II are same we can say
R_{II} = 2r/3
Combined resistance will be
R = R_{I} + R_{II}
Now this circuit is series with internal resistance 'r'
Resultant resistance will be
The current drawn from the cell will be
I=3E/4r
And power consumed by the network
Derive an expression for drift velocity of electrons in a conductor. Hence deduce Ohm's law.
Let an electric field E be applied the conductor. Acceleration of each electron is
Velocity gained by the electron
Let the conductor contain n electrons per unit volume. The average value of time't', between their successive collisions, is the relaxation time, 't'.
Hence average drift velocity
The amount of charge, crossing area A, in time Δt is
=neAv_{d}Δt = IΔt
Substituting the value of v_{d}, we get
But I = JA, where J is the current density
(a) Define the term 'self-inductance' and write its S.I. unit.
(b) Obtain the expression for the mutual inductance of two long co-axial solenoids S_{1} and S_{2} wound one over the other, each of length L and radii r_{1} and r_{2} and n_{1} and n_{2} number of turns per unit length when a current I is set up in the outer solenoid S_{2}.
a) Mutual inductance of two coils is equal to the e.m.f induced in one coil when the rate of change of current through the other coil is unity.
SI unit of mutual inductance is Henry.
b) Consider two long solenoids S_{1} and S_{2} of same length ‘l’ such that S_{2} surrounds S_{1 }completely.
Let,
n_{1} = Number of turns per unit length of S_{1}
n_{2} = Number of turns per unit length of S_{2}
I_{1} = Current passing through solenoid S_{1}
= Flux linked with S2 due to current flowing in S1
is the coefficient of the mutual inductance of two solenoids.
When current is passed through S_{1}, emf is induced in S_{2} .
Magnetic field inside solenoid S_{1} is given by,
Magnetic flux linked with each turn of the solenoid =
Total magnetic flux linked with S_{2} is given by,
Similarly, mutual inductance between two solenoids, when current is passed through S_{2} and emf induced in solenoid S_{1} is given by,
Hence, coefficient of mutual induction between the two long solenoids is given by,
(i) A ray of light incident on face AB of an equilateral glass prism shows the minimum deviation of 30°. Calculate the speed of light through the prism.
(ii) Find the angle of incidence at face AB so that the emergent ray grazes along the face AC.
At the minimum deviation, the refracted ray inside the prism becomes parallel to its base.
Angle of minimum deviation is given as Dm = 30°
Since, the prism is equilateral, So, A = 60°
Refractive index of the prism
We know that μ = v_{1}/v_{2}
Hence the speed of light in prism would be 1/√2 times the speed of light in air i.e = 3 x10^{8} /1.414 = 2.121 x10^{8} m/s
(ii)
From Snell's law, we know that
For the emergent ray to graze at the face AC, the angle of refraction should be 90
So, applying snell's law at face AC, we get
From figure, we can see that angle of refraction at face AB is 15
So applying Snell's law we get
Sin i_{AB} = sin r_{AB} x μ_{12}
or i_{AB} = sin^{-1} (√sin 15°)
Construction
Main parts of an ac generator:
Armature − Rectangular coil ABCD
Filed Magnets − Two pole pieces of a strong electromagnet
Slip Rings − The ends of coil ABCD are connected to two hollow metallic rings R_{1} and R_{2}.
Brushes − B_{1} and B_{2} are two flexible metal plates or carbon rods. They are fixed and are kept in tight contact with R_{1} and R_{2} respectively.
Theory and Working − As the armature coil is rotated in the magnetic field, angle θ between the field and normal to the coil changes continuously. Therefore, magnetic flux linked with the coil changes. An emf is induced in the coil. According to Fleming’s right-hand rule, current induced in AB is from A to B and it is from C to D in CD. In the external circuit, current flows from B_{2} to B_{1}.
To calculate the magnitude of emf induced:
Suppose
A → Area of each turn of the coil
N → Number of turns in the coil
→ Strength of magnetic field
θ → Angle which normal to the coil makes with at any instant t
∴ Magnetic flux linked with the coil in this position:
= NBA cosθ= NBA cosωt …(i)
Where, ‘ω’ is angular velocity of the coil
As the coil rotates, angle θchanges. Therefore, magnetic flux Φ linked with the coil changes and hence, an emf is induced in the coil. At this instant t, if e is the emf induced in the coil, then
Moving coil galvanometer is an instrument used for the detection and measurement of small currents.
Principle: The working of moving coil galvanometer is that when a current carrying coil is placed in a varying magnetic field, it experiences torque.
Consider a rectangular coil for which no. of turns = N
Are of cross-section is A = lb
Intensity of the uniform magnetic field = B
Current through the coil = I
Therefore,
Deflecting torque is given by,
BIL x b = BIA
For N number of turns,
= NBIA
Restoring torque in the spring = k
Therefore,
i) The soft iron coil in a galvanometer will make the field radial. Also, it increases the strength of the magnetic field.
ii) Current sensitivity in the galvanometer is given by,
Voltage sensitivity in the galvanometer is given by,
Yes a galvanometer can be used for measuring the current.