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## Karnataka 2nd PUC Physics Question Bank Chapter 7 Alternating Current

### 2nd PUC Physics Alternating Current One Marks Questions and Answers

Question 1.

What is meant by alternating current?

Answer:

The current that varies simple harmonically and reverses its direction once every half cycle is called alternating or bidirectional current.

Question 2.

Mention any one advantage of ac over dc.

Answer:

AC voltage can easily and efficiently be stepped down or stepped up or transmitted over long distances with a minimum power loss.

Note : AC equipments are economical, durable and ac can easily be rectified and filtered.

Question 3.

What is the phase relation between voltage and current in a resistor connected to an ac voltage?

Answer:

The voltage across a resistor is in phase with the current in it.

Question 4.

Give an expression for the average value of a time function F(t) over a complete time period.

Answer:

<F(t)> = \(\frac{1}{\mathrm{T}} \int_{0}^{\mathrm{T}} \mathrm{F}(t) d t\)

Question 5.

What is the value of < cos 2 ωt > over a complete period?

Answer:

Zero.

Note :<cos 2 ωt> = \(\frac{1}{T} \int_{0}^{T}\) cos 2 ωt dt = 0

Question 6.

Give the average value of sin^{2} cot (<sin^{2} ωt>).

Answer:

Since < sin^{2} ωt > = \(\frac { 1 }{ 2 }\) (1 – <cos 2 ωt>) and

<cos 2 ωt> = 0, < sin^{2} ωt> = \(\frac { 1 }{ 2 }\)

Question 7.

Give the expression for average power dissipated in a resistor over a complete ac cycle.

Answer:

p_{ave} = P =<i^{2}_{m}Rsin^{2} ωt> = \(\frac { 1 }{ 2 }\)<i^{2}_{m}R

Question 8.

What will be the instantaneous power dissipation in the resistor connected to ac?

Answer:

The instantaneous power dissipated in the resistor is given by P = i^{2}_{R} = <i^{2}_{m}R sin^{2}_{ωt}

Question 9.

Draw voltage and current waveforms in a pure resistor connected to ac.

Answer:

Question 10.

Mention the average value of ac over one complete ac cycle

Answer:

Zero.

Question 11.

Relate v_{rms} and v_{peak} values of ac. (March 2014)

Answer:

v_{peak} = v_{m} = √2 v_{rms}

Question 12.

Relate v_{rms} and v_{ave} over one half of an ac cycle.

Answer:

Question 13.

Will there be any energy loss in the case of a pure inductor or a pure capacitor connected in ac circuits?

Answer:

There will be no energy loss in the case of a pure inductor or a pure capacitor in the form of heat in ac circuits.

Question 14.

What is a low pass filter?

Answer:

A circuit device that uses an inductor to allow a low frequency voltage to pass large current is known as a low pass filter.

Note : X_{L} ∝ f. For a high ‘f’ ‘X_{L}‘ c will be large, current in the circuit becomes low.

Question 15.

What is the value of inductive reactance of a pure inductor connected to DC?

Answer:

Since f = 0 for DC, and X_{L} ∝ f, X_{L} = 0.

Question 16.

Why is that a pure inductor not used to control DC?

Answer:

Pure inductor is not used to control DC because X_{L} = 0 for DC and pure inductor allows maximum current.

Question 17.

What is meant by a high pass filter?

Answer:

A circuit device that uses a qapacitor to allow high frequency current is known as a high pass filter.

Question 18.

What is the value of capacitive reactance connected to DC?

Answer:

The capacitive reactance of a capacitor connected to DC is infinity.

Question 19.

Give the expression for wattless component of current in a pure inductor or capacitor.

Answer:

The current in a pure inductor or capacitor is called wattless current and is expressed as

I_{rms} cos θ.

Question 20.

Mention the power factor of a pure inductor or a capacitor.

Answer:

Power factor (cos Φ) for a pure inductor or capacitor is zero.

Question 21.

What is the value of power factor of a pure resistor?

Answer:

Power factor (cos Φ) of a pure resistor is one.

Question 22.

How does current in a pure resistor vary in terms of phase angle with the voltage across it?

Answer:

The phase angle between voltage and current is 90°and voltage across the inductor leads the current by 90° or \(\frac{\pi}{2}\) radian.

Question 23.

How does current in a pure capacitor vary in terms of phase angle with the voltage across it?

Answer:

The phase angle between voltage and current is 90°. In this case, voltage across the capacitor lags the current by 90° or \(\frac{\pi}{2}\) radian.

Question 24.

What does an ac voltmeter or an ac ammeter in an ac circuit measure?

Answer:

AC instruments measure r.m.s values of voltage or current.

Question 25.

What does a DC voltmeter measure when connected across a load resistor in an ac circuit ?

Answer:

Zero (voltage) reading.

Question 26.

Can a capacitor of suitable capacitance replace a choke coil in an ac circuit?

Answer:

Yes. A capacitor can replace a choke coil in an ac circuit because power factor is zero for a capacitor and an inductor.

Question 27.

For circuits used for transporting electric power, a low power factor implies large power loss in transmission. Is the statement true?

Answer:

Yes. When the power factor is low, I_{rms} is large. Power is usually transmitted at fixed voltage. Hence power loss (large heat loss) I^{2}_{rms} R will be large.

Question 28.

A choke coil is connected in series with a lamp and the combination across a dc supply. What happens to the intensity of light if an iron core is used in the coil?

Answer:

Intensity of the bulb remains the same.

Note : Although L ∝ µ_{r}, X_{L} = 2πf L = 0 for DC (f= 0).

Question 29.

If ac supply is used and an iron core is introduced in a coil connected in series with the bulb, then what happens to the intensity of the bulb?

Answer:

Intensity of the lamp decreases.

Note : X_{L} ∝ L and L ∝ µ_{r}. Hence the reactance of the coil increases and voltage across the coil increases.

Question 30.

Write the expression for the electromagnetic energy of an electrical system.

Answer:

Electromagnetic energy = \(\frac{1}{2} \frac{q^{2}}{C}+\frac{1}{2} L i^{2}\)

Question 31.

Give the differential equation for an LC oscillator in the absence Of an ac source and resistor.

Answer:

The differential equation = \(\frac{d^{2} q}{d t^{2}}+\left(\frac{1}{L C}\right) q\) = 0

Question 32.

What is an alternating emf?

Answer:

Emf that varies periodically with time is called an alternating emf.

Question 33.

What is an AC waveform?

Answer:

A graph of AC voltage versus time is called an AC waveform.

Question 34.

Define the period of an AC.

Answer:

The period of an AC is the time taken by it to complete one cycle.

Question 35.

Define the frequency of an AC.

Answer:

The frequency of an AC is the number of cycles completed by that AC in one second.

Question 36.

Define the peak value of an AC.

Answer:

The peak value of an AC is the maximum value attained by that AC voltage/current in either half cycle.

Question 37.

Define the mean value of an AC.

Answer:

The mean value of an AC is the mean/average of the instantaneous values of that AC measured over a half cycle.

Question 38.

Define the rms value of an AC.

Answer:

The rms value of an AC is the square root of the mean of the squares of the instantaneous values taken over a complete cycle.

Question 39.

Give one method of generating a sinusoidal AC.

Answer:

A sinusoidal AC can be generated by rotating a coil in a uniform magnetic field at a constant speed.

Question 40.

Define instantaneous value of AC.

Answer:

Instantaneous value of AC is the value of alternating voltage/current at the given instant.

Question 41.

Define inductive reactance.

Answer:

Inductive reactance is defined as the ratio of rms value of voltage across the coil to rms value of current through it. The resistance offered by an inductor to a.c. is called the inductive reactance.

Question 42.

Define capacitive reactance.

Answer:

Capacitive reactance is defined as the ratio of rms value of voltage across the capacitor to the rms value of current through the capacitor. The resistance offered by a capacitor to ac is called the capacitive reactance.

Question 43.

Define impedance of an AC circuit.

Answer:

Impedance of an AC circuit is defined as the ratio of rms value of voltage across the circuit to the rms value of current the circuit.

Question 44.

Write the formula for the impedance of a series LCR circuit.

Answer:

Question 45.

What is the phase relation between current and voltage in a series LCR circuit in which X_{L} < X_{c}?

Answer:

Question 46.

What is the power factor of a series LCR circuit in which X_{L} > X_{c}?

Answer:

Question 47.

Under what condition will the current through a series LCR circuit be in phase with voltage?

Answer:

Current through a series LCR circuit will be in phase with voltage, when inductive reactance (X_{L}) = capacitive reactance (X_{c}). The circuit in this case is said to be under resonance.

Z = Z_{min} = R; I = I_{max}

Question 48.

What is a choke?

Answer:

Choke is an inductance coil used in AC circuits to bring down the value of current without appreciable power loss.

Question 49.

What is a transformer?

Answer:

A transformer is a device used for stepping up or stepping down the AC voltages.

Question 50.

When is a transformed said to be ideal?

Answer:

A transformer is said to be ideal when it’s output power is equal to the input power.

Question 51.

What is wattless current?

Answer:

A component of current that does not allow the circuit elements like capacitors and inductors to dissipate energy in them is called as wattless current.

Question 52.

What is a step up transformer?

Answer:

A step up transformer is one that generates an output voltage which is greater than the input voltage.

Question 53.

What is a step down transformer?

Answer:

A step down transformer is one that generates an output voltage which is less than the input voltage.

### 2nd PUC Physics Alternating Current Two Marks Questions and Answers

Question 1.

Draw a phasor diagram for the ac circuit comprising of a pure resistor.

Answer:

Question 2.

Draw phasor diagram for the ac circuit comprising of a pure inductor.

Answer:

Question 3.

Draw phasor diagram for the ac circuit comprising of a pure capacitor.

Answer:

Question 4.

Draw phasor diagram for a RLC series circuit connected to ac voltage source.

Answer:

Question 5.

Based on what principle does a transformer work? Explain. (March 2014)

Answer:

A transformer is a device used for varying the AC voltages. It works on the principle of mutual induction. (It is an indispensable device in electrical power transmission and distribution.)

Question 6.

Define ‘Q’ factor and sharpness of resonant curve.

Answer:

‘Q’ factor is defined as the ratio of X_{L} or X_{c} to the DC, resistance. Q = \(\frac{X_{L} \text { or } X_{C}}{R}\)

Sharpness of resonant curve = \(\frac{f_{0}}{B . W}\)

where, f_{0} – resonant frequency,

B.W – band width.

Question 7.

What is the relation between rms and average values of a sinusoidal A.C. voltage?

Answer:

Question 8.

Give the differential equation for an LC oscillator.

Answer:

Question 9.

Give the expression for the instantaneous energy stored in’a capacitor along with the meaning of the symbols used.

Answer:

U_{E} = \(\frac{1}{2} \frac{q_{0}^{2}}{\mathrm{C}}\)cos^{2}(ωt)

where, C – electrical capacitance, cat – phase angle and U_{E} – electrical energy stored in the dielectric medium between the plates.

Question 10.

Give the expression for the instantaneous energy stored in an inductor along with the meaning of the symbols used.

Answer:

U_{m} = \(\frac { 1 }{ 2 }\)Lq_{0}^{2} ω^{2} sin^{2} (ωt)

where, U_{m} – magnetic energy, L – self inductance and ωt – phase angle.

Question 11.

If v = V_{m} sin cot represents instantaneous voltage connected across a resistor of resistance ‘R’ then write the expression for the instantaneous current in it.

Answer:

i – i_{m} sin ωt where, i = \(\frac{\mathbf{v}}{\mathbf{R}}\) and i_{m} = \(\frac{v_{m}}{R}\)

and i_{m} and v_{m} represent current amplitude and voltage amplitude.

Question 12.

Write the equation for a sinusoidal AC voltage. Give meaning of the symbols used.

Answer:

V = V_{0} sin ωt where V → voltage induced at time t, V_{0} → maximum value of induced voltage and ω → angular velocity of the coil.

### 2nd PUC Physics Alternating Current Three Marks Questions and Answers

Question 1.

Draw current v/s frequency curve for a series RLC circuit and explain the different portions of the curve.

Answer:

Across the portion of the curve AB, the capacitive reactance will be greater than the inductive reactance, and it behaves as a capacitive circuit.

Across the portion of the curve BC, the capacitive reactance will be less than the inductive reactance and the circuit behaves as an inductive circuit.

Across the portion of the curve at B, X_{L} = X_{c}.

The impedance will be equal to the DC resistance R and the circuit behaves as a resistive circuit.

Question 2.

Distinguish between a pure inductor and a pure resistor.

Answer:

Question 3.

Distinguish between a pure inductor and a pure resistor.

Answer:

Question 4.

Define inductive reactance. On what factors does it depend?

Answer:

Inductive reactance is the effective opposition to the flow of AC and is defined as the ratio of the rms value of voltage across the coil to the rms value of current through the coil. It depends on the frequency of AC source and the self inductance of the i nductor. X_{L} ∝ f and X_{L} ∝ L; X = 2πfL.

Question 5.

Define capacitive reactance. On what factors does it depend?

Answer:

Capacitive reactance is the effective opposition to the flow of AC and is defined as the ratio of rms value of the voltage across the capacitor to the rms value of current through the capacitor. Capacitive reactance depends on the frequency of AC and the capacitance of the capacitor.

Question 6.

Define the term power factor. What is the value of power factor for a pure resistor, pure inductor and pure capacitor?

Answer: The power factor is defined as the cosine of the angle between the current and the applied voltage. For a pure capacitor and inductor, θ = 90° and hence the power factor is zero. For a pure resistor, θ = 0° and hence the power factory is one.

Question 7.

What are the uses of an induction coil?

Answer:

Induction coils are used

- to regulate the current in AC circuits,
- to separate audio and radio frequencies.
- in mercury vapour lamps to control current.
- in electronic power supply units as filters.

Question 8.

Distinguish between DC and AC.

Answer:

ipments are not easy. equipments are easy.

Question 9.

Show that the sum of electrostatic energy and magnetic energy in an LC oscillator equals \(\frac{q_{0}^{2}}{2 C}\)

Answer:

Total energy = U_{E} + U_{m}

q_{0} and C are time independent constants.

Note :

- Maximum energy of a capacitor – maximum energy of an inductor = \(\frac{1}{2} \frac{q_{0}^{2}}{\mathrm{C}}=\frac{1}{2} \mathrm{Li}_{m}^{2}\)
- At any given instant, the total energy, in an LC oscillator is the sum of energy in each of them.

Question 10.

Give an expression for an ac driven series RLC circuit.

Answer:

For a series RLC circuit and for a sustained oscillation,

Question 11.

S.T. average power over a complete cycle in a pure inductor connected to ac is zero.

Answer:

The instantaneous power supplied to an inductor

P_{L} = iv

= i_{m} sin (ω – π/2) (v_{m} sin ωt)

= -i_{m} cos ωt v_{m} sin ωt

i.e., P_{L} = \(-\frac { 1 }{ 2 }\) i_{m} v_{m} (sin 2 ωt) where sin 2 ωt = 2 sin ωt cos ωt

The average power over a complete cycle is

Thus average power supplied to an inductor over one complete cycle is zero.

### 2nd PUC Physics Alternating Current Five Marks Questions and Answers

Question 1.

Show that the current in a pure resistor is in phase with the ac voltage across it and. hence S.T. average power dissipation in a resistor is i^{2}R where i is r.m.s value of ac.

Ans:

Let i be the current in a resistor voltage drop a cross the resistor will be iR, supplying KVL to the closed loop, we write v_{m} sin ωt = IR.

Hence current is in phase with the applied voltage.

By definition of power p = i^{2}R

Question 2.

Show that the voltage in an inductor leads the current by π/2 rad. for a pure inductor connected across ac. (March 2015)

Answer:

Since magnetic flux is a function of time, induced emf in the passive element ‘L’ is \(\left(-\mathrm{L} \frac{d i}{d t}\right)\)

Applying the voltage law we write.

On integrating both sides We get,

For symmetric oscillation, integration constant = 0

∴ i = i_{m} sin (ωt – π/2)

and i_{m} = v_{m}/Lω.

Inductive reactance (ac resistance offered o by an inductor) = ωL = 2πfL.

Hence, X_{L} ∝ f and X_{L} ∝ L.

We also note that the voltage across the inductor leads the current by π/2 radian.

Question 3.

S.T. the voltage across a pure capacitor lags the current by 90° or \(\frac{\pi}{2}\) rad.

Answer:

Let the applied voltage across the capacitor be v_{c} = v = v_{m} sin ωt.

X_{c} is the capacitive reactance which is the a.c resistance offered by a capacitor.

Question 4.

Obtain an expression for the im pedance of a series LCR circuit, (using phasor diagram method).

Answer:

Consider a resistance R, an inductor of self inductance L and a capacitor of capacitance C connected in series across an AC source. The applied voltage is given by,

v = v_{0} sin ωt ….. (1)

where, v is the instantaneous value, v_{0} is the peak value and ω = 2πf, f being the frequency of AC.

If i be the instantaneous current at time t, the instantaneous voltages across R, L and C are respectively iR, iX_{L} and iX_{c}. The vector sum of the voltage amplitudes across R, L, C equals the amplitude v_{0} of the voltage applied.

Let v_{R}, v_{L} and v_{c} be the voltage amplitudes across R,L and C respectively and I_{0} the current amplitude. Then

v_{R} = i_{0}R is in phase with i_{0}

v_{L} = i_{0}X_{L} = i_{0}(ωL) leads i_{0} by 90°

v_{c} = i_{0}X_{C} = i_{0} \(\left(\frac{1}{\omega C}\right)\) lags behind i_{0} by 90°

The current in a pure resistor is phase with the voltage, whereas the current in a pure inductor lags the voltage by \(\frac{\pi}{2}\) rad. The current in a pure capacitor leads the voltage by \(\frac{\pi}{2}\)rad. For v_{L} > v_{c}, phase angle Φ between the voltage and the current is positive.

From the right angled triangle OAP,

Where Z is the impedance of the circuit.

Phase angle between v & i.

Question 5.

Give the expression for the impedance of an LCR circuit and obtain the expression for the resonant frequency. What is the power factor of the resonant circuit?

Answer:

Impedance of a series LCR circuit is given by

Where ω = 2πf, f being frequency of AC

At resonance X_{L} = X_{c} and ω = ω_{0}, i.e., ω_{0}L = \(\frac{1}{\omega_{0} C}\)

Where X_{L} – inductive reactance, X_{c} – capacitive reactance,

ω – angle of frequency of ac.

i.e., ω^{2}_{0}LC = 1

Power factor = cos R/Z and Z = R at resonant frequency.

Power factor = cos R/R = 1.

Voltage is in phase with current.

Question 6.

Explain the advantages and disadvantages of AC over DC.

Answer:

Advantages of AC:

- AC can be stepped up or stepped down by using a transformer.
- AC can be easily converted into DC, using electronic power supply units.
- AC devices are more durable since power dissipation is less.
- Transmission of electrical power is more efficient and economical in the form of AC.
- Current in an AC circuit can be reduced using reactive elements like choke coils and capacitors which consume very little power.

Disadvantages of AC:

- It cannot be used for electroplating.
- The average value of alternative current over a half cycle is less than its rms value. Hence for a given value of current, wires carrying AC require better insulation.
- AC meters have a non-linear scale. As a result, accuracy of measurement is not uniform throughout the range.

Question 7.

Obtain an expression for the average power supplied to a series RLC circuit. Discuss the average power when the series RLC circuit behaves as a pure resistive, inductive or capacitive circuit.

Answer:

For a series RLC circuit, instantaneous power supplied is given by

P = vi

i.e., P = (v_{m} sin ωt) (i_{m} sin (ωt + Φ))

The average power supplied,

Case (i):

For a series RLC to behave as a resistive circuit:

X_{L} = X_{c} so that Z = Z_{min} = R ; X_{L} – X_{c} = 0

Average power dissipation at resonance = \(\bar{p}\) = v_{rms} i_{rms}

Case (ii):

For a series RLC to behave as an inductive circuit:

X_{L} > X_{c}

Case (iii):

For a series RLC to behave as a capacitive circuit

Question 8.

S.T an LC oscillator executes SHM and hence obtain an expression for the angular frequency of oscillation.

Answer:

Applying KVL we get,

Question 9.

Write a note on transformers.

Answer:

A transformer is a device used for varying AC voltages. It works on the principle of mutual induction.

When electrical power is transmitted over long distances, it is economical to transmit power at a high voltage and low current in order to minimise the loss due to heating effect of the electric current. So the voltage is stepped up at a generating station. At the receiving end the AC must be at a low voltage so as to operate various electrical appliances. Hence voltage is stepped down at the receiving end.

A transformer consists of two coils called primary P and secondary S, wound on the same laminated soft iron core. The alternating emf to be stepped up or stepped down is applied to the primary coil and the output of the altered emf is obtained across the secondary.

Working: As the current through the primary varies, the magnetic flux linked with the secondary also varies. As a result, alternating emf of the same frequency is induced across the secondary. The magnitude of the induced voltage in the secondary, depends on the voltage across primary.

If V_{p} and V_{s} are the input and output voltages respectively and N_{p} and N_{s} are the number of turns in the primary and secondary respectively then

Sources of energy loss in transformers are due to

- copper loss
- magnetic flux linkage loss,
- hysteresis loss and
- eddy current loss.

Question 10.

Give the qualitative explanation of the action of an LC oscillator in the absence of any external ac voltage source. Assume that the capacitor is charged initially.

Answer:

(i) At t = 0, the capacitor has energy of \(\frac{1}{2} \frac{q_{m}^{2}}{\mathrm{C}}\). Energy associated with the inductor is zero. When the switch is closed, capacitor discharges, magnetic energy gets stored in the inductor. When the current reaches its maximum value i_{m} (at t = \(\frac{T}{4}\)) all the electrostatic energy is converted into magnetic energy in the inductor. Magnetic energy = \(\frac { 1 }{ 2 }\)Li^{2}_{m}

Energy in the capacitor becomes zero.

(ii) During the time interval from t = \(\frac{T}{4}\) to t = \(\frac{T}{2}\), the current starts charging the capacitor until it gets fully charged (at t = \(\frac{T}{2}\)). The polarity of the capacitor will be opposite.

Energy in the inductor become zero.

(iii) During the time interval from t = \(\frac{T}{2}\) to t = \(\frac{3}{4}\)T, the inductor gets the current from the capacitor, until the inductor gets the maximum current and maximum energy (\(\frac { 1 }{ 2 }\)Li^{2}_{m}) is stored in the inductor.

(iv) During the time interval from t = \(\frac{3}{4}\) T to t = T, the initial polarity of capacitor gets

restored and charging the capacitor to acquire energy of \(\frac{1}{2} \frac{q_{m}^{2}}{\mathrm{C}}\)

This process of conversion of electrostatic energy to magnetic energy takes place without any external source voltage as long as the oscillating system does not lose energy in the form of heat or electromagnetic energy.

(i) at t = 0

(ii) at t = \(\frac{T}{4}\)

(iii)

\(\frac{T}{4}<t<\frac{T}{2}\)

(iv) at t = \(\frac{T}{2}\)

Question 11.

Obtain an expression for quality factor of series RLC circuit connected to ac source.

Answer:

We know that for maximum current

From the current frequency curve, the band width ω_{1} – ω_{2} = 2Δω.

The quantity \(\) is regarded as a measure of the sharpness of resonance. .

Hence, ω_{1} = ω_{0} + Δω and ω_{2} = ω_{0} – Δω

we note that ω_{1} and ω_{2} are obtained by drawing perpendicular to the frequency axis corresponding to the \(\frac{1}{\sqrt{2}} i_{m}\)

Equation (2) may be simplified as

Question 12.

Give analytical solution to a series RLC circuit connected to ac.

Answer:

### 2nd PUC Physics Alternating Current List of Formulae

### 2nd PUC Physics Alternating Current Current Numericals and Solutions

Question 1.

A 220 AC voltage source is connected across a pure inductor of inductance 0.5 H. Find the current through the inductor, if the frequency of the source is

(i) 100 Hz

(ii) 150 kHz

(iii) What is the power lost in the inductor?

Answer:

Given, v_{rms} = 220V,L = 0.5H

Question 2.

If a coil is to have a reactance of 16 Ω for a frequency 2 kHz., then what must be its inductance? What will be its reactance at 10 Hz?

Answer:

Required reactance of the coil X_{L} = 16 Ω

Frequency of AC f = 2 kHz = 2 × 10^{3} Hz

Let L be the inductance of the coil then, X_{L} = 2πfL

Reactance of the coil at a frequency f = 10 Hz is X_{L} = 2πfL

= 2 × 3.14 × 10 × 10 × 1.27 × 10^{-3}

= 0.0791 Ω or 79.7 mΩ

Question 3.

A pair of adjacent coils have a mutual inductance of 0.25 H. If the current in the primary changes from zero to 2A in 0.05 s., then find the average induced emf in the secondary?

Answer:

Induced emf in the secondary = 10 V.

Question 4.

A sinusoidal current is represented by the equation 1 = 2 sin 100 πt. Calculate

(i) Period

(ii) RMS value

(iii) Current at t = \(\frac{1}{16}\)s and

(iv) Time required by the current to change its value from zero to the RMS value.

Answer:

Given, I = 2 sin 100 πt …. (1)

This equation is of the form I = I_{0} sin ωt …. (2)

Comparing equations (1) and (2), we note that

Peak value I_{0} = 2A and 2πf = 100 π

Question 5.

An A.C. Source of 250 V, 50 Hz is connected to a circuit consisting of an electric lamp rated 100 W, 50 V and a capacitor in series. What should be value of the capacitance of the capacitor to make the lamp work at the rated value?

Answer:

Given : v_{rmf} = 250 V, f = 50 Hz, P = 100 W; v = 50 volt.

Question 6.

A coil of resistance 100 Ω, an inductor of 0.5 H and a capacitor of 15 pF are connected in series with a 200 V – 50 Hz AC source. Calculate the current in the circuit.

Answer:

Given : R = 100 Ω, L = 0.5 H, C = 15 µF = 15 × 10^{-6}F, V_{rms} = 200V, f = 50 Hz

Induced reactance is given by,

X_{L} = Lω = 2πfL = 2 × 3.14 × 50 × 0.5

X_{L} = 157.1 Ω

Impedance of a series LCR circuit is given by,

Current through the circuit is given by,

Question 7.

In the circuit as shown in figure, find the impedance of the circuit. What is the phase angle between voltage and current? (given V_{rms} = 220V, f = 50 Hz)

Answer:

V_{rms} = 220 V, f = 50 Hz, L = 2 × 10^{-2}H, C = 2000 µF = 2 × 10^{-3} F, R = 20Ω

Inductive reactance, X_{L} = 2πfL

= 2 × 3.14 × 50 × 2 × 10^{-2}

= 6.28Ω

Capacitive reactance,

Inpedance of the circuit is given by,

Phase angle between voltage and current is given by

Question 8.

A loo resistor ¡s connected to a 220 V, 50 Hz ac supply.

(a) What is the rms value of the current in the circuit?

(b) What is the net power consumed over a full cycle?

Answer:

R = 100Ω, v_{rms} = 220V, f = 50Hz

Question 9.

A 44 mH inductor is connected to a 220 V, 50hz ac supply. Determine the r.m.s value and peak value of the current in the circuit. (JuIy 2014)

Answer:

Given: L = 44 × 1o^{-3} H, v_{rms} = 220 V, f = 50 Hz

Question 10.

A 60 (J.F capacitor is connected to a 110V, 60Hz, ac supply. Determine the r.m.s value of the current in the circuit.

Answer:

Given : C = 60 × 1o^{-6} F, v_{rms} = 110V, f = 60 Hz

Question 11.

Obtain the resonant frequency ω_{r} of a series LCR circuit with L = 2.0 H. C = 32µF, and R = 10Ω. What is the Q – value of this circuit?

Answer:

Question 12.

A charged 30 µF capacitor is connected in series to a 27mH inductor. What is the angular frequency of free oscillations of the circuit?

Answer:

Given : C = 30 × 10^{-6} F, L = 27 × 10^{-3}H

Question 13.

A series LCR circuit with R = 20Ω, L = 1.5 H, and C = 35 µF is connected to a variable frequency 200 V ac supply. When the frequency of the supply equals the natural frequency of the circuit, what is the average power transferred to the circuit in one complete cycle?

Answer:

Given : R = 20 Ω, L = 1.5H, C = 35 × 10^{-6} F, v_{rms} = 200 V, f = f_{0}, \(\overline{\mathbf{P}}\) = ?

At resonance Z = R

Power factor of a series RLC at resonance cos Φ = 1

Average power transferred to the circuit in one complete cycle,

= i_{rms} v_{rms} cos Φ

= 10 × 200 × 1

= 2000W

= 2.0 kW

Question 14.

A coil of inductance 0.50 H, and resistance 100Ω is connected to a 240V, 50Hz ac supply.

(a) What is the maximum current in the coil?

(b) What is the time lag between the voltage maximum and the current maximum?

Answer:

(a) Given: L = 0.50 H, R = 100 W, v_{rms} = 240 V, f = 50Hz.

We know that X_{L} = 2πfL = 2 × 3.142 × 50 × 0.5 = 157.1 Ω

Question 15.

A circuit containing an inductor of 80mH inductance and a capacitor of 60 pF capacitance in series, is connected to a 230V, 50Hz supply. The resistance of the circuit is negligible.

(a) Obtain the current amplitude and r.m.s value.

(b) Obtain the r.m.s values of potential drop across each element.

(c) What is the average power transferred to the inductor?

(d) What is the average power transferred to the capacitor?

(e) What is the total average power absorbed by the circuit (averaged over one complete cycle)?

Answer:

Given : L = 80 × 10^{-3}H, C = 60 × 10^{-6} F, v_{rms} = 230 V, f = 50 Hz, R = 0.

Using, XL = 2a/L= 2 x 3.142 x 50 x 8 x 10~2

we get,

Question 16.

A power transmission line feeds input power at 2300V to a step down transformer with its primary windings having 4000 turns. What should be the number of turns in the secondary in order to get the output power at 230V? For an efficiency of 0.70, calculate the rms value of the primary current when rms value of the secondary current is 5A.

Answer:

Question 17.

At a hydroelectric power plant, the water pressure head ¡s at a height of 300 m and the water flow available is loo m3s1. If the turbine generator efficiency is 60%, estimate the electric power available from the plant (g = 9.8 ms^{-2}).

Answer:

Question 18.

A series LRC circuit comprising of. an inductor of inductance 3H, a resistor of rcsistance 7.4 and a capacitór of capacitance 27 μE When connected tò an a.c source, the. circuit resonates at a particular frequency. Calculate angular resonant frequency and quality factór of the resonant circuit. (March 2014)

Answer:

Question 19.

A sinusoidal voltage of peak vale 23V and frequency. 50Hz ‘s applied to a series LCR circuit ¡n which R = 3Ω, L = 25.48 mH and C = 786 μF.

Find: (a) Impedance of the circuit.

(b) The phase difference between the voltage across the source and the çurrent.

(c) The power factor. (March 2015)

Answer:

Hence reactance of the circuit = 4Ω, impedance of the circuit = 5Ω, phase angle between V and I is 53°6′ and power factor of the circuit = 0.6.