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.

---

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).

---

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.`

---

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.

---

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.

---

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.

---

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.

---

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.`

---

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`

---

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.

---

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.

---

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`

---