Please Wait... While Loading Full Video### 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.

high value in receiving circuits? Name the factors on which it depends.

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.

`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.

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.

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?

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?

(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

answer.

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

answer.

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.

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.

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?

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?

(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.

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` .

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` .

(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.

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.

(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?

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?

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.

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.

(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`