Physics Definition & Proof Based Problems

### Definition & Proof Based Problems

Q 3202556438

Define the quality factor in an ac circuit. Why should the quality factor have
high value in receiving circuits? Name the factors on which it depends.

Solution:

The quality LKtor is defined as the ratio of potential difference across an
inductor or a capacitor to potential difference across resistance in an LCR
circuit at resonance.

 Q = (omega_r L)/R

Quality factor should be high to have the current corresponding to a particular
frequency to be more and to avoid the other unwanted frequencies.
Q-factor depends on f, L, R and C.
Q 3202845738

In a series LCR circuit connected to an ae source of variable frequency and voltage
V = V_m sin omega t, draw a plot showing the variation of current (I) with angular
frequency (omega) for two different values of resistance R_1 and R_2 (R_1 > R_2).
Write the condition under which the phenomenon of resonance occurs. For
which value of the resistance out of the two curves, a sharper resonance
is produced? Define Q-factor of the circuit and give its significance.

Solution:

The following graph shown the variation of current ·with angular frequency.

When X_L = X_C, resonance occurs,

i.e ,  omega_r L = 1/( omega_r C)

 => omega_r = 1/sqrt(LC)

omega_r, is called the resonant frequency;

Lesser is the resistance, sharper is resonance.

Hence, for resistance R_ 2 sharper is the

curve. The ratio  (omega_r L)/R  is called quality

factor. Higher the quality factor (Q), better is the tuning of the circuit (i.e.
selectivity).
Q 3212845730

Define root mean square current. Also, obtain its expression.

Solution:

Rms current is that value of current which produces the same amount of
heat as it produced by alternating current when flows through the same
conductor for the same time period.
Let an instantaneous current I = I_m sin omega t is passing through a resistor R.
Let it be constant for a very short duration of time dt.
Then, very small amount of heat produced is given by

dH = I^2 R dt

Heat produced for one complete cycle is given by

 H = int_0^T dH = int_0^T I^2 Rdt

:. H = int_0^T I_m^2 ( sin^2 omega t) Rdt = I_m^2 int_0^T sin^2 omega t dt = I_m^2 int_(t = 0)^T ((1 - cos 2 omega t)/2) dt

 H = ( I_m^2 R)/2 [ int_0^T dt - int_o^T ( cos 2 omega t) dt ] = (I_m^2 R)/2 { T - [ (sin 2 omega t)/(2 omega) ]_0^T }

As omega = 2 pi // T

 H = ( I_m^2 RT)/2 - (I_m^2 R)/2 [ sin 2 xx (2 pi)/T xx T - sin 0]

 :. H = ( I_m^2 RT)/2 ............(i)

If I_(r m s) is steady current flowing through the same circuit for the same time
interval T producing the same heat amount of heat H, then the heat
produced is

H = I_(r m s)^2 - RT ........ (ii)

Equating (i) and (ii), we get

( I_m^2 RT)/2 = I_m^2 RT

:. I_m/sqrt2 = I_(r m s)  or  I_(r m s) = 0.707 I_m.
Q 3242756633

(a) Show that in an ac circuit containing a pure inductor, the voltage is ahead
of current by pi//2 in phase.
(b) A horizontal straight wire of length L extending from east to west is
falling with speed v at right angles to the horizontal component of
Earth's magnetic field B.
(i) Write the expression for the instantaneous value of the emf induced in
the wire.
(ii) What is the direction of the emf?
(iii) Which end of the wire is at the higher potential?

Solution:

(a) Inductive reactance is the opposition offered by
an inductor towards the How of current passing
through it.

X_L = 2 pi v L

Applied ac voltage, E = E_m sin omega t ... (i)

An emf induced in the inductor is given by

epsilon = - L (dI)/( dt)

In order to maintain the How of current through
the inductor, we must have

E = - epsilon

i.e.  E = L (dI)/(dt) => (dI)/(dt) = E/L = E_m/L sin omega t

:. int dI = E_m/L int sin omega t dt => I = E_m/(L omega) ( - cos omega t)

 I = E_m/(L omega) sin ( omega t - pi//2)

I = I_m , sin (omega t - pi//2) .............(ii)

where  I_m = E_m/(L omega) = E_m/X_L

From equations (i) and (ii), we conclude that voltage leads the current by a
phase angle pi//2.

(b) (i) E = B_H Lv
(ii) The direction of emf is from west to east.
(iii) The end of the wire towards east is at higher
potential.
Q 3232756632

(a) what do you understand by sharpness of resonance in a series LCR
circuit? Derive an expression for Q-factor of the circuit.
Three electrical circuits having ac sources of variable frequency are shown
in the figure. 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 reason for your

Solution:

The sharpness of resonance is measured by Q- factor of the LCR circuit. It
is defined as the ratio of the voltage developed across the inductance (or
capacitance) at resonance to 'the voltage developed across the resistance.

:. Q = (L omega _r I_text(max))/( R I_text(max) ) = (L omega _r)/R , here  omega_r = 1/sqrt(LC)

Also,  Q = ( 1 /( C omega_r) I_text(max) )/( R I_text(max) ) = 1/( C omega_r R)

Further,  omega_r = 1/sqrt (LC)

:. Q = L/R * 1/sqrt(LC) = 1/R sqrt (L/C)

Larger the Q-value of the circuit, sharper is the resonance curve.
(b) (i) The first is a pure resistor. Since, the resistance offered by a circuit
having a resistor only does not change with frequency, the current
will also not change with fi-equency.
(ii) In this circuit, which has an inductance L the reactance X_L = L_omega = 2 pi v L.

Thus, X_L alpha v ,i.e. as v increases X_L also increases and the current  I_V ( I_v = E_V/X_L)
decreases.

(iii) In this circuit which contains only a pure capacitance, the resistance offered by
it is given by  X_C = 1/(C omega) = 1/( 2 pi v C) . Thus as v, the frequency, increases
X_C decreases and hence, the current increases in the circuit.
Q 3252845734

A series LCR 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.

Solution:

Take the voltage of source

V = V_m sin omega t ... (i)

To determine the phase relation between current
and voltage at any instant of time, we use a phasor
technique.

As all the three components are in series, the same amount of current flows
through them at any instant of time. Let it be

I = I_m sin (omega t + phi ) ... (ii)

where phi is the phase difference between the voltage across the source and current.
We construct a phasor diagram.

On applying Pythagoras theorem, we get

V_m^ 2 = V_(Rm)^2 + ( V_(Cm) - V_(Lm) )^2

Here  V_(Rm) = I_mR , V_(Cm) = I_m X_C , V_(Lm) = I_m X_L

 V_m = I_m sqrt ( R^2 + ( X_C - X_L)^2)

 V_m = I_m Z

where, Z = sqrt ( R^2 + ( X_C - X_L)^2)

Z is called the impedance of the circuit.
The variation of current with frequency of source
is shown in the figure.

 I_m = V_m/Z

We know  X_L = omega_L  and  X_C = 1/( omega C)

Thus, on increasing the frequency, X_L increases and X_C decreases. As a result,

X_C - X_L decreases.

We know Z = sqrt ( R^2 + ( X_C - X_L)^2)

So, Z also decreases and with this current increases.

For a particular value of frequency called resonant frequency (omega _r)
we find X_L = X_C

Z = R

and  I_m = I_m^(m a x )

After this frequency, X_C - X_L again increases and I_m decreases.
Hence, we obtain bell-shaped graph.
Q 3232856732

(a) Draw a schematic sketch of an ac generator describing its basic elements.
State briefly its working principle. Show a plot of variation of
(i) Magnetic, flux and
(ii) Alternating emf versus time generated by a loop of wire rotating in a
magnetic field.
(b) Why is choke coil needed in the use of fluorescent tubes with ac mains?

Solution:

(a) An alternating current generator, designed by Nikola Tesla, is based upon
the principle of electromagnetic induction.
Construction: An AC generator consists of the
following parts:

(i) Armature: It is a rectangular coil ABCD (Fig.)
having a large number of turns of
insulated copper wire wound over a
soft-iron core. The core increases the
magnetic flux linked with the armature.
(not shown in the diagram)

(ii) Field magnet : lt is a powerful permanent magnet having concave pole-pieces
N and S'. T'he armature is rotated (say, by a water turbine) between
these pole-pieces about an axis perpendicular to the magnetic field
lines.
(iii) Slip rings: The leads from the armature coil ABCD are connected to
two copper rings R_ 1  and R_ 2 called the 'slip rings'. These rings are
concentric with the axis of the armature-coil and rotate with it.

(iv) Brushes: These are two carbon pieces B_ 1 and B_2 called 'brushes' which
remain stationary, pressing against the slip rings R_ 1 and R_2 respectively.
The brushes are connected to the external circuit in which current
is to be supplied by the generator.
Working: As the armature coil ABCD rotates,
the magnetic flux linked with it changes.
Hence, an emf is induced in the coil and
current flows in it.

(i) The variation of magnetic flux with time is
given by

 phi = NBA cos omega t

(ii) The variation of alternating emf with time is given by

 (d phi)/(dt) = - BA omega sin omega t

 epsilon = (- N d phi)/(dt)

 epsilon = NBA omega sin omega t

(b) Not in syllabus.
Q 3212845730

Define root mean square current. Also, obtain its expression.

Solution:

Rms current is that value of current which produces the same amount of
heat as it produced by alternating current when flows through the same
conductor for the same time period.
Let an instantaneous current I = I_m sin omega t is passing through a resistor R.
Let it be constant for a very short duration of time dt.
Then, very small amount of heat produced is given by

dH = I^2 R dt

Heat produced for one complete cycle is given by

 H = int_0^T dH = int_0^T I^2 Rdt

:. H = int_0^T I_m^2 ( sin^2 omega t) Rdt = I_m^2 int_0^T sin^2 omega t dt = I_m^2 int_(t = 0)^T ((1 - cos 2 omega t)/2) dt

 H = ( I_m^2 R)/2 [ int_0^T dt - int_o^T ( cos 2 omega t) dt ] = (I_m^2 R)/2 { T - [ (sin 2 omega t)/(2 omega) ]_0^T }

As omega = 2 pi // T

 H = ( I_m^2 RT)/2 - (I_m^2 R)/2 [ sin 2 xx (2 pi)/T xx T - sin 0]

 :. H = ( I_m^2 RT)/2 ............(i)

If I_(r m s) is steady current flowing through the same circuit for the same time
interval T producing the same heat amount of heat H, then the heat
produced is

H = I_(r m s)^2 - RT ........ (ii)

Equating (i) and (ii), we get

( I_m^2 RT)/2 = I_m^2 RT

:. I_m/sqrt2 = I_(r m s)  or  I_(r m s) = 0.707 I_m.
Q 3242856733

(a) An a.c. source of voltage V = V_0 sin omega t is connected to a series combination
of L , C and R. Use the phasor diagram to obtain expressions for
impedance of the circuit and phase angle between voltage and current.
Find the condition when current will be in phase with the voltage.
What is the circuit in this condition called?

(b) In a series LR circuit X_L = R and power factor of the circuit is P_1. When
capacitor with capacitance C such that X_L = X_C is put in series the
power factor becomes P_ 2 . Calculate P_1/P_2 .

Solution:

(a) Take the voltage of source

V = V_m sin omega t ........ (i)

To determine the phase relation between current and voltage at any instant
of time, we use a phasor technique.

As all the three components are in series, the same amount of current flows

through them at any instant of time. Let it be

I = I_m sin ( omega t + phi ) ...............(ii)

where phi is the phase difference between the voltage across the source and
current.

We construct a phasor diagram.

On applying Pythagoras theorem, we get

V_m^ 2 = V_(Rm)^2 + ( V_(Cm) - V_(Lm) )^2

Here  V_( Rm) = I_m R , V_(Cm) = I_m X_C , V_(Lm) = I_m X_L

V_m = I_ m sqrt( R^2 + (X_C - X_L)^2) = I_m Z

where Z = sqrt( R^2 + (X_C - X_L)^2)

Z is called the impedance of the circuit.

(ii) Phase angle between voltage and current is given by phi = tan^(-1) ( ( V_(Cm) - V_(Lm) )/V_(Rm) )

The voltage and current will be in phase, if phi = 0 or V_(Cm) = V_(Lm)  or X_L =X_C
This condition is called resonance condition of the circuit and the nature of
circuit is purely resistive.

(b) Power factor ( P ) = R/Z

In a RL circuit, Z = sqrt(R^2 + X_L^2)

:. P_1 = R/sqrt( R^2 + R^2) = 1/sqrt2

When a capacitor of capacitance C is connected, such that X_L = X_C, then

 I = sqrt( R^2 + (X_L - X_C)^2) = R

Power factor  = P_2 = R/R = 1

:. P_1/P_2 = 1/sqrt2
Q 3282756637

(a) With the help of a labelled diagram, describe briefly the underlying
principle and working of a step-up transformer.
(b) Write any two sources of energy loss in a transformer.
(c) A step-up transformer converts a low input voltage into a high output voltage.
Does it violate law of conservation of energy? Explain.

Solution:

(a) A step-up transformer is based on the
principle of mutual induction.
An alternating potential (V_p) when
applied to the primary coil is induced
an emf in it.

 epsilon_p = - N_p (d phi)/(dt)

If resistance of primary coil is low, then V_p = epsilon_p => V_p = - N_p (d phi)/(dt)

As same flux is linked with the secondary coil with the help of soft iron core
due to the mutual induction, an emf is induced in it.

 epsilon_s =- N_s (d phi)/(dt)

If an output circuit is opened, V_s = epsilon_s => V_s =- N_s (d phi)/(dt)

Thus,  V_s/V_p = N_s/N_p

For step-up transformer N_s/N_p > 1

In case of dc voltage, flux does not change. Thus, no emf is induced in the
circuit.

(b) Two sources of energy loss are:

(i) flux leakage, and (ii) resistance of the windings.

(c) For a given power supply, a high output voltage means a low output current. As
there is no gain in power, the law of conservation of energy is not violated.
Q 3242556433

A voltage V = V_0 sin omega t is applied to a series LCR circuit. 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?

Solution:

Let an alternating current of I = I_m sin omega t be passing through a network of L,

C and R creating a potential difference of V = V_m sin (omega t pm phi) where phi is

the phase difference. Then the power consumed is given by

P = VI = V_m I_ m sin ( omega t ± phi ) sin omega t

P = V_m I_m (sin omega t cos phi ± cos omega t sin phi ) sin omega t

P = V_m I_ m (sin^2 omega t cos phi pm 1/2 sin omega t sin phi )

 P_(av) = ( int_0^T Pdt )/( int_0^T dt) P_(av) = (V_m I _m)/T [ int_0^T sin^2 omega t cos phi dt + 1/2 int _0^T sin phi sin 2 omega t dt]

 P_(av) = (V_m I _m)/T [ T/2 cos phi + 0]

 P_(av) = (V_m I _m)/2 cos phi = V_(r m s) I_(r m s) cos phi

(i) No power is dissipated if (a) resistance in the circu;t is zero and (b) phase
angle between voltage and current is pi//2.
(ii) Maximum power is dissipated if (a) resistance in the circuit is maximum
and (b) phase angle between voltage and current is zero.
Q 3252856734

(a) Write the function of a transformer. State its principle of working with
the help of a diagram. Mention various energy losses in this device.
(b) The primary coil of an ideal step-up transformer has 100 turns and
transformation ratio is also 100. The input voltage and power are
respectively 220 V and 1100 W. Calculate
(i) number of turns in secondary, (ii) current in primary,
(iii) voltage across secondary, (iv) current in secondary.
(v) power in secondary.

Solution:

(a) Function: It is a device which transforms electrical power at low voltage
(and high current) to electrical power at high voltage (and low current)
and vice versa.
It works on the principle of mutual induction.

Working: A step-up transformer is based on the
principle of mutual induction.
An alternating potential (V_p) when
applied to the primary coil is induced
an emf in it.

 epsilon_p = - N_p (d phi)/(dt)

If resistance of primary coil is low, then V_p = epsilon_p => V_p = - N_p (d phi)/(dt)

As same flux is linked with the secondary coil with the help of soft iron core
due to the mutual induction, an emf is induced in it.

 epsilon_s =- N_s (d phi)/(dt)

If an output circuit is opened, V_s = epsilon_s => V_s =- N_s (d phi)/(dt)

Thus,  V_s/V_p = N_s/N_p

For step-up transformer N_s/N_p > 1

In case of dc voltage, flux does not change. Thus, no emf is induced in the
circuit.

In case of de voltage, flux does not change. Thus, no emf is induced in the
circuit.
The various energy losses in the transformer are:
(i) Joule's heating of the primary and secondary windings
(ii) I-Ieating of the core due to eddy currents.
(iii) Hysteresis loss and
(iv) Flux leakage or incomplete flux linkage.

(b) Given: N_P = 100, k = 100, V_P = 220 V, P_P = 1100 W

(i) N_S/N_P = k => N_S = 100 xx 100 = 10^4  turns

(ii)  P_P => V_P I_P => I_P = (1100)/(220) = 5 A

(iii) V_S/V_P = 100 => V_S = 100 xx 220 = 2.2 xx 10^4 V

(iv)  I_P/I_S = k => I_S = 5/(100) = 0.05 A

(v)  ∵ P_text (output ( primary)) = P_text( input (Secondary) )

:. P_S = 1100 W