CLUSTER LEVEL WORKSHOP DECEMBER .2015

SUBJECT : PHYSICS CLASS XII

ALTERNATING CURRENT

SECTION A (1 MARKS)

  1. What is the average power loss in a pure Inductive circuit?
  2. Draw the graphs showing variation of inductive reactance and capacitive reactance with frequency of applied AC source.
  3. The power factor of an AC circuit is 0.5. What will be the phase difference between voltage and current in this circuit?
  4. What is the power factor at resonance in an LCR circuit?
  5. What is Impedance?
  6. What is the phase difference between the voltage and current in a pure capacitive circuit?
  7. Define Watt less Current.
  8. Name the device which converts high alternating voltage into low alternating voltage.
  9. The peak voltage of an ac supply is 300 V. What is the rms voltage?
  10. What is Q-factor of an AC circuit?

S.No. / Answer
1. / Zero
2. /
3. / Power factor,

Phase difference is 60°.
4. / 1
5. / It is the effective opposition to the flow of current provided by the Inductor, Capacitor and resistor.
6. / 900lagging behind
7. / It is the current passing in a pure capacitive or Inductive circuit in which the net power loss is Zero.
8. / Step down Transfer
9. / 300/(2)1/2= 212.23 volt
10. / The ratiois called the quality factor or Q-factor
Or
It can also be defined as the ratio of potential drop across either the inductor or the capacitor to the potential drop across the resistor.

SECTION B (2 MARKS)

  1. A 100 Ω resistor is connected to a 220 V, 50 Hz ac supply. (a) What is the rms value of current in the circuit? (b) What is the net power consumed over a full cycle?
  2. Calculate the average voltage in half cycle of an AC signal?
  3. What is resonance frequency? Find the expression for the resonance frequency.
  4. Draw the schematic diagram of a transformer. Distinguish between step up and step down transformer.
  5. A coil of 0.01H inductance and 1 ohm resistance is connected to 200 volt, 50Hz ac supply. Find the impedance of the circuit and phase difference of the circuit.
  6. An electric bulb B and a parallel plate capacitor C are connected in series to the a.c. Mains as shown in the given figure. The bulb glows with some brightness.

How will the glow of the bulb be affected on introducing a dielectric slab between the plates of the capacitor? Give reasons in support of your answer.

  1. In India, domestic power supply is at 220 V, 50 Hz, while in U.S.A., it is 110 V, 60 Hz. Give one advantage and one disadvantage of 220 V supply over 110 V supply.
  2. Mention various losses in a transformer.
  3. Find the resonance frequency of an AC circuit using Electric current v/s frequency graph.
  4. Show mathematically that the average power in a capacitive circuit is Zero. Draw a diagram of instantaneous power and time.

S. No. / Answer
1. / 3.11 amp, 242 J
2. / 2I0/π
3. / It is the frequency of an LCR circuit at which the current passing through the circuit is maximum.

4. /
5. / 3.28 ohm, Φ= tan-1(3) = 71.60
6. / The bulb will glow brighter.
Reasons:
The impedance of a capacitor is without dielectric.
If a dielectric is introduced inside a capacitor, then the new capacitance will be KC and the new impedance will be .
Therefore, the impedance has decreased.
This will result in higher current through the circuit and the bulb will glow brighter.
7. / Advantage:
The power loss at 220 volt supply is less than at 110 V.
Disadvantage:
It is difficult to work with 220 V supply because its peak value (311 V) is much higher than the peak value (155.5 V) of 110 V supply.
8. / Eddy current:
When a changing magnetic flux is linked with the iron core, eddy currents are set up which in turn produce heat and energy is wasted.
Eddy currents are reduced by using laminated core instead of a solid iron block because in laminated core the eddy currents are confined withinthe lamination and they do not get added up to produce larger current. In other words their paths are broken instead of continuous ones.
Hysteresis Loss:
When alternating current is passed, the iron core is magnetized and demagnetized repeatedly over the cycles and some energy is being lost in the process.
9.
10. / E = E0 sin ωt
q = CE = CE0 sin ωt
I = dq / dt
= (d / dt) [CE0 sin ωt]
I = [E0 / (1 / ωC)] ( cos ωt )
I = I0 sin (ωt + π / 2)
(where I0 = E0 / (1 / ωC) and XC = 1 / ωC = E0 / I0) XC is Capacitive Reactance.
Its SI unit is ohm.

SECTION C ( 3 MARKS)

  1. Define resonance frequency. In an LCR circuit a capacitor 15microfarad, an inductor 100 mH are connected in series with an AC voltage source of 100V. Find the resonance frequency of the circuit.
  2. A sinusoidal A.C. has a maximum value of 15 A. What are its rms values? If the time is recorded from the instant the current is zero and is becoming positive, what is the instantaneous value of the current after 1/300s given the frequency is 50 Hz.
  3. Write the working and principle of a Transformer by drawing a schematic diagram.
  4. An inductor ‘L” of inductive reactance XL is connected in series with a bulb B to an ac source as shown in fig. Briefly explain how does the brightness of the bulb change when (i) number of turns of the inductor is reduced(ii) and a capacitor of reactance XC=XL is included in series in the same circuit.
  1. Fig shows an inductor L a resistor R connected in parallel to a battery through a switch. The resistance of R is same as that of the coil that makes L. Two identical bulbs are put in each arm of the circuit.

(i)Which of the bulbs lights up earlier when S is closed ?

(ii)(ii) Will the bulbs be equally bright after some time ?

  1. Write the difference between step up and step down transformer. A transformer is giving output current at 110 amp to 22 amp. What is ratio of the number of coils in primary and secondary coils.
  2. A resistor of 15 ohm, an inductor of 200 mH and a capacitor of 800 μC are connected in series with 220 volt, 50 Hz in an LCR circuit. Find the Impedance and resonance frequency of the circuit.
  3. What is the significance of Vrms? Find the expression of Vrms of an AC signal?
  4. What is relation between Vrms and Vdc? Why AC is preferred over DC.
  5. In an LCR circuit, a resistor of 50 ohmand an Inductance of 30 mH in series with 220 volts, 50 Hz. Calculate the quality factor and the capacitance of the capacitor.

S.No. / Answer
1. / It is the frequency of an LCR circuit at which the current passing through the circuit is maximum.
Resonance frequency = 816.5 Hz
2. / Irms = 21.21 amp, I = 12.99 amp
3. /
In a transformer with Ns secondary turns and Npprimary turns, induced emf or voltage Es is:

Back emf = Ep =
EP = VP
Es = Vs
Thus, Vs = … (i)

Dividing equations (i) and (ii), we obtain

If the transformer is 100% efficient, then

Thus, combining the above equations,

If Ns > Np, then the transformer is said to be step-up transformer because the voltage is stepped up in the secondary coil.
No, the transformer does not violate the principal of conservation of energies. This can be easily observed by the following equation:

Power consumed in both the coils is the same as even if the voltage increases or current increases, their product at any instant remains the same.
4. / (i)With the reduction in the number of turns of the coil, the inductive reactance XL will decrease and more current will flow through the bulb. The bulb glows brighter.
(ii)When XL= XC in the LCR series , the circuit behaves as a resistive circuit only so circuit current will increase and the bulb glows brighter.
5. / (i)When switch is closed induced emf in inductor, called back emf delays the glowing of lampB1 so lamp B2 lights up earlier.
(ii)Yes. In this case at steady state inductive effect becomes meaningless so both bulbs become equally bright after some time.
6. / 5 : 1
7. / 25 ohm, 7.9 Hz
8. /
9. / Vrms= Vdc.
Advantages:
1. Production of AC is less expensive
2. can be transmitted over a long distance without much a much power loss
3. It can be easily rectified.
10. / Q = 0.188, C= 33.8 μF

SECTION –E (5 MARKS)

Q1. (a) State the principle on which AC generator works. Draw a labelled diagram and explain its working.

(b) A conducting rod held horizontally along East-West direction is dropped from rest from a certain height near the Earth’s surface. Why should there be an induced emf across the ends of the rod? Draw a plot showing the instantaneous variations of emf as a function of time from the instant it begins to fall.

Ans

(a Principle of AC generator-It works on the principle of electromagnetic induction.

In an A.C. generator, mechanical energy is converted to electrical energy by virtue of electromagnetic induction.

* Rotation of rectangular coil in a magnetic field causes change in flux (Φ = NBACosωt).

* Change in flux induces emf in the coil which is given by

ε= -dΦ/dt = NBAωSinωt ε = ε0Sinωt

* Current induced in the coil I = ε/R = ε0Sinωt/R = I0Sinωt

b.) Since the earth’s magnetic field lines are cut by the falling rod, the change in magnetic flux takes place this change in flux induces an emf across the ends of rod.

Since the rod is falling under gravity,

v = gtu = 0

e.m.fε = Blv = Blgtε∝ t

Q2. a) State the principle of a step-up transformer. Explain its working with the help of a labelled

diagram,

b) Describe briefly any two energy losses, giving the reasons for their occurrence in actual

Transformers.

a Principle, It works on the principle of mutual induction.

In a transformer with Ns secondary turns and Npprimary turns, induced emf or voltage Es is:

Back emf = Ep =

EP = VP

Es = Vs

Thus, Vs = … (i)

Dividing equations (i) and (ii), we obtain

If the transformer is 100% efficient, then

Thus, combining the above equations,

If Ns > Np, then the transformer is said to be step-up transformer because the voltage is stepped up in the secondary coil.

No, the transformer does not violate the principal of conservation of energies. This can be easily observed by the following equation:

Power consumed in both the coils is the same as even if the voltage increases or current increases, their product at any instant remains the same.

b two losses and their causes

Eddy current:

When a changing magnetic flux is linked with the iron core, eddy currents are set up which in turn produce heat and energy is wasted.

Eddy currents are reduced by using laminated core instead of a solid iron block because in laminated core the eddy currents are confined withinthe lamination and they do not get added up to produce larger current. In other words their paths are broken instead of continuous ones.

Hysteresis Loss:

When alternating current is passed, the iron core is magnetized and demagnetized repeatedly over the cycles and some energy is being lost in the process.

Q3.A series LCR circuit is connected to a source having voltage v = vmsin ωt. 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 (i) maximum and (ii) minimum.

v=vmsin ωt

Let the current in the circuit be led the applied voltage by an angleΦ.

The Kirchhoff’s voltage law gives.

It is given thatv=vmsin ωt(applied voltage)

On solving the equation, we obtain

On substituting these values in equation (1), we obtain

Letand

This gives

On substituting this in equation (2), we obtain

On comparing the two sides, we obtain

Or

And

The condition for resonance to occur

For resonance to occur, the value ofimhas to be the maximum.

The value ofimwill be the maximum when

Power factor = cos Φ

Where,

(i) Conditions for maximum power factor (i.e., cos Φ = 1)

  1. XC=XL

Or

  1. R= 0

(ii) Conditions for minimum power factor

  1. When the circuit is purely inductive
  2. When the circuit is purely capacitive

Q 4. An a.c source generating a voltageis connected to a capacitor of capacitanceC. Find the expression for the currenti, flowing through it, plot a graph ofvandiversusωtto show that the current is, ahead of the voltage.

A resistor of 200Ωand a capacitor of 15μF are connected in series to a 220 V, 50 Hz a.c source. Calculate the current in the circuit and thermsvoltage across the resistor and the capacitor. Is the algebraic sum of these voltages more than the source voltage? If yes, resolve the paradox.

Ans

Solution:

A.C source containing capacitor:

Alternatingemfsupplied is:

Potential difference across the plates of capacitor

At every instant, the potential differenceVmust be equal to theemfapplied i.e.,

Or,q=C

ItIis instantaneous value of current in the circuit at instantt, then

Numerical:

Here,r= 200Ω,C= 15μF = 15 × 10−6F

E= 220 V,f= 50 Hz,I=?

This is because these voltages are not in same phase and they cannot be added like ordinary numbers.

Q5.Define the term 'mutual inductance' between the two coils.
Obtain the expression for mutual inductance of a pair of long coaxial solenoids each of lengthland radiir1andr2(r2r1). Total number of turns in the two solenoids are N1and N2,respectively.

Solution:

The ratio of magnetic flux passing through one coil to the current passing through the other is known as mutual inductance between the two coils.

Suppose a current i is passed through the inner solenoid S1.
A magnetic fieldB=μ0n1iis produced inside S1, whereas the field outside it is zero.
The flux through each turn S2isBπr12=μ0n1iπr12
The total flux through all the turns in a lengthlof S2is
ϕ=(μ0n1iπr12)n2l=(μ0n1n2πr12l)i⇒M=μ0n1n2πr12l

ELECTROMAGNETIC INDUCTION

SECTION A (1 MARKS )

1.What is thebasic cause of electromagnetic induction?

Ans: Change in magnetic flux

2. Does the change in magnetic flux induce emf or current?

Ans: Induces emf , current is induced when circuit is closed.

3.Induced emf is also called back emf? Why.

Ans: It opposes any change in magnetic flux.

4.A wire kept along the north-south direction is allowed to fall freely. Will an emf be induced in the wire ?

Ans: No because there is no change in the magnetic flux linked with the wire.

5.Why the inductance coils are made of copper?

Ans: Because copper has small resistance.

6. A solenoid with an iron core and a bulb are connected to a d.c. source. How does the brightness of the bulb change when the iron core is removed from the solenoid?

Ans: It will not change because of reactance of inductor is zero for d.c. source.

SECTION B( 2 MARKS)

1. Define self inductance. How does the self inductance of an coil change when an iron rod is introduced in it ?

Ans: The magnetic flux linked with a coil when unit current is passed through it.

The self inductance increases.

2. Why is spark produced in the switch of a fan, when it is switched off?

Ans: The break of circuit is sudden. A large induced emf set up across the gap of switch due to which sparking occurs.

3. State Faraday’s laws of electromagnetic induction.

Ans Faraday’s first law: whenever there is change in the magnetic flux associated with a coil , an emf is induced in it.

Faraday’s second law: The magnitude of emf induced is directly proportional to rate of change of magnetic flux

4. State Lenz’s law.A coil of metal wire is held stationary in a non-uniform magnetic field. Is any emf induced in the coil?

Ans: The emf induced opposes the cause that produces it.

Yes, there will be an induced current in the coil.

5.Define coefficient of mutual induction and give its SI unit.

Ans: It is the magnetic flux linked with a coil when unit current is passed though the neighbouring coil.

SI unit of coefficient of mutual induction is Henry.

6. What are eddy currents? Mention any two useful application of eddy current.

Ans: The current induced in a metallic sheet or plate when the magnetic flux linked with it changes.

Applications: (i) In electric furnace to melt metal

(ii) Dead beat galvanometer

7. Write three factors on which self inductance of a coil depend.

Ans: (i) Number of turns

(ii) Its area of cross-section

(iii) The permeability of the core material

SECTION C (3 MARKS)

1. A rectangular loop of area 20cm x 30cm is placed in a magnetic field of 0.3 T with its plane

(i)Normal to the field

(ii)Inclined 300 to the field

(iii)Parallel to the field

Find flux linked with the coil in each case.

Ans((i) 1.8 x 10-2wb,(ii) 0.9 x 10-2wb, (iii) zero)

2. Derive an expression for the induced emf in a conductor of length l, moving with a velocity v in a uniform magnetic field B.

A magnetic flux threading a coil changes from 12 x 10 -3Wb to 6x10-3Wb in 0.01 sec. Calculate the induced emf.

E =dΦ/dt (0.6 v)

3. (a) Define self inductance. Write its S.I. units.

(b) Derive an expression for self inductance of a long solenoid of lengthl, cross-sectional area A having N number of turns.

Ans

(a)The phenomenon in whichemfis induced in a single isolated coil due to change of flux through the coil by means of varying the current through the same coil is called self inductance. S.I unit of inductance is Henry.

(b)Magnetic fieldBinside a solenoid carrying a currentiis.

B=

Letnbe the number of turns per unit length.

Where,

Nis total number of turns

lis the length of the solenoid

Inductance,

Substituting, we obtain

Substituting the value ofB, we obtain

InductanceLof a solenoid is:

4.ELECTROMAGNETIC INDUCTION AND ALTERNATING CURRENTS

GIST

1 / The phenomenon in which electric current is generated by varying magnetic fields is called electromagnetic induction.
2 / Magnetic flux through a surface of area A placed in a uniform magnetic field B is defined as
ΦB = B.A = BACosθ where θ is the angle between B and A.
3 / Magnetic flux is a scalar quantity and its SI unit is weber (Wb). Its dimensional formula is [Φ] = ML2T-2A-1.
4 / Faraday’s laws of induction states that the magnitude of the induced e.m.f in a circuit is equal to the time rate of change of magnitude flux through the circuit.
ε
5 / According to Lenz law, the direction of induced current or the polarity of the induced e.m.f is such that it tends to oppose the change in magnetic flux that produces it. (The negative sign in Faraday’s law indicates this fact.)
6 / Lenz law obeys the principle of energy conservation.
7 / The induced e.m.f can be produced by changing the (i) magnitude of B (ii) area A (iii) angle θ between the direction of B and normal to the surface area A.
8 / When a metal rod of length l is placed normal to a uniform magnetic field B and moved with a velocity v perpendicular to the field, the induced e.m.f is called motional e.m.f produced across the ends of the rod which is given by ε = Blv.
9 / Changing magnetic fields can setup current loops in nearby metal bodies (any conductor). Such currents are called eddy currents. They dissipate energy as heat which can be minimized by laminating the conductor.
10 / Inductance is the ratio of the flux linkage to current.
11 / When a current in a coil changes it induces a back e.m.f in the same coil. The self induced e.m.f is given by ε where L is the self-inductance of the coil. It is a measure of inertia of the coil against the change of current through it. Its S.I unit is henry (H).
12 / A changing current in a coil can induce an e.m.f in a nearby coil. This relation,
ε , shows that Mutual inductance of coil 1 with respect to coil 2 (M12) is due to change of current in coil 2. (M12 = M21).
13 / The self-inductance of a long solenoid is given by L = µ0n2Al where A is the area of cross-section of the solenoid, l is its length and n is the number of turns per unit length.
14 / The mutual inductance of two co-axial coils is given by M12 = M21 = µ0 n1n2Al where n1& n2 are the number of turns per unit length of coils 1 & 2. A is the area of cross-section and l is the length of the solenoids.
15 / Energy stored in an inductor in the form of magnetic field is and
Magnetic energy density
16 / In an A.C. generator, mechanical energy is converted to electrical energy by virtue of electromagnetic induction.
* Rotation of rectangular coil in a magnetic field causes change in flux (Φ = NBACosωt).
* Change in flux induces e.m.f in the coil which is given by
ε= -dΦ/dt = NBAωSinωt ε = ε0Sinωt
* Current induced in the coil I = ε/R = ε0Sinωt/R = I0Sinωt
17 / An alternating voltage ε=ε0Sinωt, applied to a resistor R drives a current I = I0Sinωt in the resistor, I0 = ε0 /R where ε0& I0 are the peak values of voltage and current. (also represented by Vm & Im)
18 / The root mean square value of a.c. may be defined as that value of steady current which would generate the same amount of heat in a given resistance in a given time as is done by the a.c. when passed through the same resistance during the same time.
Irms = I0/√2 = 0.707i0
Similarly, vrms = v0/√2 = 0.707v0.
For an a.c. ε = εm Sin ωt applied to a resistor, current and voltage are in phase.
19 / In case of an a.c. circuit having pure inductance current lags behind e.m.f by a phase angle 90°. ε = εm Sin ωt and i = im Sin (ωt-Π/2)
Im = εm/XL; XL = ωL is called inductive reactance.
20 / In case of an a.c. circuit having pure capacitance, current leads e.m.f by a phase angle of 90°.
ε = εmSinωt and I= ImSin(ωt+π/2) where
Im = εm/XC and XC = 1/ωC is called capacitive reactance.
21 / In case of an a.c. circuit having R, L and C, the total or effective
resistance of the circuit is called impedance (Z).
Z = εm / Im =
tanΦ = where φ is the phase difference
between current and voltage.
ε = εmSinωt, I= ImSin(ωt+Φ)
23 / Average power loss over a complete cycle in an LCR circuit is
P = εrmsIrmsCosΦ
* In a purely resistive circuit Φ = 0; P = VRMSIRMS.
* In a purely inductive circuit Φ = Π/2; P = 0.
* In a purely capacitive circuit Φ = Π/2; P = 0.
24 / In an LCR circuit, the circuit admits maximum current if XC = XL, so that Z = R and resonant frequency
25 / Q factor of series resonant circuit is defined as the ratio of voltage developed across the inductance or capacitance at resonance to the applied voltage across ‘R’,
Q= also where is bandwidth.
26 / for a transformer,
In an ideal transformer, εPIP = εSIS. i.e
If NS>NP; εS>εP& IS<IP – step up. If NP>NS; εP>εS & IP<IS – step down.
27 / A circuit containing an inductor L and a capacitor C (initially charged) with no a.c. source and no resistors exhibits free oscillations of energy between the capacitor and inductor. The charge q satisfies the equation

CONCEPT MAP