Bihar Board - Class 12 Physics - Chapter 7: Alternating Current Long Answer Question
Long Answer Type Questions
1. Why is an inductor an easy path for d.c. and resistive path for a.c.?
Answer: The inductive reactance (resistance offered by inductor) of an inductor is given by
XL = ωL = 2πv L
When v is the frequency of a.c. source.
For d.c. v = 0
∴ XL = 2π0.L = 0
i.e. the inductor offers no resistance to d.c.
For a.c.
v is finite
∴XL ∝ v
So larger the value of v, more will be the effective resistance offered by the inductor.
Hence an inductor is an easy path for d.c. but a resistive path for a.c.
2. Show that a capacitor is an easy path for a.c. but a block for d.c. (i.e. offers infinite resistance for d.c.).
Answer: The capacitive reactance XC (resistance offered by a capacitor) is given by
XC = 1C = 12vC
where C is the capacitance of the capacitor and v is the frequency of a.c. source.
For d.c.
v = 0
∴ XC = 10 = ∞
i.e. capacitor offers infinite resistance to d.c, in other words it blocks d.c.
For a.c., v is finite
so XC ∝ 1v
So larger the value of v, smaller will be the value of XC . Hence capacitor is an easy path for a.c.
So a capacitor is an easy path for a.c. but a block for d.c.
3. Distinguish between resistance and resistor; inductance and inductor; capacitance and capacitor.
Answer: Resistance is the property of a conductor due to which it opposes the growth of the current (or resists the flow of electrons) through it. The appliance which offers the resistance to the flow of current is called a resistor.
Inductance is the phenomenon according to which an opposing induced e.m.f. is produced in a coil as a result of change in the current or magnetic flux linked with the coil. The appliance in which this effect is produced is called inductor.
Electrical capacitance of a conductor is related to its ability to store the electric charge.
Capacitor is a device for storing large quantities of electric charge.
In brief, resistance, inductance and capacitance are the properties of devices resistor, inductor and capacitor.
4. Distinguish between resistance, inductive reactance, capacitive reactance and impedance.
Answer: Resistance is the property due to which conductor resists the flow of electrons through it. It is measured in ohm.
Reactance is the opposition offered by an inductor or a capacitor to the flow of a.c. through it. If it is due to an inductor, it is called inductive reactance and if it is due to a capacitor, it is called capacitive reactance. Both are measured in ohm.
Inductive reactance of an inductor is the measure of its effective opposition to the flow of alternating current through it.
XL = 2πv L
For d.c. v = 0, hence inductive reactance does not offer any resistance to d.c. Greater the frequency, more the inductive reactance.
Capacitive reactance of a capacitor is the measure of opposition offered by the flow of current through it.
XC = 12vC
For d.c. v = 0; XC= ∞. Thus capacitor blocks d.c. Higher the frequency, lesser the reactance.
Impedance is the total opposition offered by L, C and R to the flow of a.c.
Impedance
Z = (Resistance)2+(Reactance)2
Or Z = (R)2+(XL-XC)2
Impedance plays the same role in a.c. as resistance plays in d.c. circuit. Both are measured in ohm.
5. Describe the use of a transformer for long distance transmission of a.c.
Answer: Long distance transmission of electric power. The most important use of a transformer is in the transmission of power from a generating station to the consumer. The following difficulties arise in such cases:
1. As the wires are many km long, they have an appreciable resistance so that a good portion of the electric energy is wasted as heat produced in the wires (heat produced being proportional to I2R).
2. For carrying heavy current and also to keep the resistance of the wires low, thick wires shall have to be used. But the cost of installing very long thick wires will be very high.
3. Fall of potential (IR) is also high.
To avoid these difficulties, the voltage from the generator is increased by a step up transformer, so the current is lower. Heat produced will now be less, hence thin wires can be used and fall of potential also decreases. At the receiving end the voltage is reduced by using a step-down transformer as shown in the figure. High power transformers are immersed in oils to obtain good electric insulation and to provide a cooling medium.
Bhakra dam supplies electric power of 110 kV and then this voltage is reduced to usual 220 V with the help of a step down transformer.
6. Radio frequency choke is air cored, whereas audio frequency choke is iron cored. Explain.
Answer: Radio frequency choke :
The inductive reactance of choke is given by
XL = 2πv L
Radio frequency is very high, i.e v is large, so. XL is large. So in order to keep XL low, L should be low. This is possible only by using an air cored (µ = 1) inductance.
Audio frequency choke
In audio frequency choke, the frequency is low i.e. v is small. So to have a reasonable value of XL, L should be high hence iron should be used for which the value of µ is high (because the value of L increases with µ).
7. An aeroplane is flying horizontally from west to east with a velocity of 900 km/hour. Calculate the potential difference developed between the ends of its wings having a span of 20 m. The horizontal component of the Earth’s magnetic field is 5 x 10-4 T and the angle of dip is 30°.
Answer: Here B = 5 x 10-4 T, θ = 30°, l = 20m
v = 900 km h-1= 900 x 1018 = 250 ms-1
When aeroplane is flying horizontally from west to east its wings cross vertical component of earth’s magnetic field
∴ B1 = B sin θ = 5 x 10-4 x sin 30°
= 2.5 x 10-4 T
The potential difference developed between the ends of its wings
ε = B1 l v
= 2.5 x 10-4 x 20 x 250
= 1.25 volt.
8. Give differences between step – up and step – down transformers.
Answer: Differences between a step – up and step – down transformer :
Step – up transformer
- It increases the voltage.
- It decreases the current strength.
- The number of turns in the secondary coil is greater than that of the primary coil.
- Its transformation ratio is more than one.
Step – down transformer
- It decreases the voltage.
- It increases the current strength.
- The number of turns in the primary coil is greater than that of secondary coil.
- Its transformation ratio is less than one.
9. When a capacitor is connected in series LR circuit the alternating current flowing in the circuit increases. Explain why.
Answer:
In LR circuit, the impedance is given by
ZL = (R)2+(XL)2 ..........................(i)
In LCR circuit, the impedance is given by
Z = (R)2+(XL-XC)2 .............................(ii)
From eqns. (i) and (ii), we find that
Z < ZL
Since I = EZ
As Z decreases on connecting a capacitor in series with the LR circuit, hence the current I in the circuit increases.
10. What is Wattless current?
Answer: In a.c. circuit, when the average consumed power is zero, then that current is called wattless current.
If a pure inductance or pure capacitance is used in a.c. circuit, then the phase difference between current and e.m.f. becomes 2.
Pav = Vrms x Irms x cosϕ
= Vrms x Irms x cos2
= Vrms x Irms x o = 0
Choke coil is made up of thick copper wire. Its ohmic resistance is negligible. Hence, the phase difference between current and e.m.f. is 2. Thus, the average consumed power becomes zero. Therefore, the current is called wattless.
By using a choke coil in an alternating circuit, we can obtain a wattless current.
11. What are the reasons for energy losses in transformers? How are these losses reduced?
Answer:
Energy losses and their removal:
Copper loss:
Some part of the energy is wasted in the form of heat due to the heating effect of current in primary and secondary coils because the coils have some resistance. The amount of heat produced is I2 Rt. To reduce it, thick coils are used in the primary coil of step – up transformer and in the secondary coil of a step – down transformer.
Iron loss:
Eddy current is produced in the iron core of the transformer, causing heating. This loss is called iron loss. To minimize this loss, the core is laminated.
Magnetic flux leakage:
All the magnetic flux produced by the primary coil may not be transferred to the secondary coil. Therefore, some energy is wasted. To minimize this loss, a soft – iron core is used.
Hysteresis loss:
Some amount of energy is wasted because the iron core becomes magnetized during the first half and then gets demagnetized during the other half. This waste loss of magnetic energy is called hysteresis loss. It is minimized by taking a soft- iron core which has a thin hysteresis loop.
12. Give differences between a.c. and d.c.
Answer: Differences between a.c. and d.c. :
a.c.:
Its direction and magnitude both change.
Transformers are used in a.c.
It cannot be used for electroplating.
It is generally not used for electro – magnet.
The measuring instruments are based on its heating effect.
It is dangerous.
d.c.:
The magnitude may change but direction does not change.
Transformers cannot be used.
It can be used in electroplating.
It is used for electromagnetism.
Its measuring instruments are based on its magnetic effect.
It is less dangerous than a.c.
13. An alternating voltage given by V = 140 sin 314 t is connected across a pure resistor of 50 ohm. Find
(i) the frequency of the source.
(ii) the rms current through the resistor.
Answer:
(i) V0= 140 V, ω = 314
2πv = 314 Therefore, V = 314/2π = 50 Hz
(ii) Irms= Vrms /R (where Vrms = V/√2 )
= (V0 /√2)/R
= V0/2 R
= 140/√2 × 50
= 140/1.414 × 50
14. Explain the behavior of inductor and capacitor towards a.c. and d.c.
Answer: The reactance offered by an inductor is given by the equation
XL = ωL = 2πfL.
Where f is the frequency of the source and L is the inductance of the inductor
For a d.c. f = 0 thus XL = 0.
For an a.c. f ≠ 0 and XL ∝L.
Flence, an inductor provides no reactance to the flow of d.c. and provide reactance of the a.c. i.e. an inductor passes d.c. and blocks a.c.
The reactance offered by a capacitor is given by the equation,
XC= 1C = 12fC
Where C is the capacitance of the capacitor.
For d.c. f = 0 thus, XC = ∞
For an a.c. f ≠ 0 and XC∝ 1f
Hence, a capacitor provides infinite reactance to d.c. and with increasing frequency of an a.c. the reactance decreases, i.e. a capacitor by passes a.c. and blocks d.c.
15. (a) On which principle, are ammeters or voltmeters used for measuring alternating current or voltage based upon, and why?
(b) Generally, the divisions marked on the scale of an AC ammeter are not equally spaced. Why?
Answer: (a) Usually, AC ammeters or voltmeters utilize the heating effect of current because the heating effect depends on the square of the current and is thus, independent of the direction of current.
(b) An AC ammeter works on the principle of heating effect of current and the heating effect depends on the square of current. Consequently, in an AC ammeter, the divisions in the beginning of scale are closely marked but then separation between successive divisions goes on increasing.
16. Distinguish between resistance, impedance and reactance of an a.c. circuit.
Answer: Difference between resistance, impedance and reactance of an a.c. circuit are as follow :
S.N | Resistance | Impedance | Reactance |
1. | It is defined as the opposition offered by the resistor to the flow of current. | It is defined as the opposition offered by the circuit containing resistor,inductor and the capacitor to the flow of current. | It is defined as the opposition offered by the capacitor or inductor to the flow of current. |
2. | It is independent of the frequency of the source of current. | It depends upon the frequency of the source of current. | It depends on the frequency of the source of the current. |
3. | It is measured in ohm. | It is measured in ohm. | It is measured in ohm. |
17. Why is a choke coil needed in the use of fluorescent tubes with a.c. mains ? Why can we not use an ordinary resistor instead of the choke coil ?
Answer: For choke coil, φ = π/2
Average power consumed
Pav =lV EV cosφ = lV EV cosπ/2
Pav = O and XL = ωL = 2πvL
Hence ωL can control a.c. with consuming power.
For ordinary resistor φ = 0
∴ P = lV EV cosφ = lV EV
and impedance = R
Therefore, the resistor consumes maximum power. Hence a choke is used in fluorescent tubes with a.c. main than an ordinary resistor.
18. Explain the different energy losses in a transformer. Does the Step-up transformer violate the law of conservation of energy, if not, give reason.
Answer: Two sources for energy loss in a transformer are:
(1) The primary and secondary are made up of copper wires and some of the energy will be lost due to the heating of these wires.
This power loss (= I2R) can be decreased by using thick copper wires of low resistance.
(2) Eddy current loss: Eddy currents are induced in the iron ore due to the alternating magnetic flux which leads to some energy loss in the form of heat. We can reduce these types of loss by laminating the iron core.
No, a step-up transformer does not violate the law of conservation of energy because whatever we gain through the voltage ratio is lost through the current ratio and vice-versa. This kind of transformer steps up the voltage while it steps down the current.
19. At a hydroelectric power plant, the water pressure head is at a height of 300 m and the water flow available is 100 m3 s−1. If the turbine generator efficiency is 60%, estimate the electric power available from the plant (g= 9.8 m s-2).
Answer:
Height of water pressure head, h = 300 m
Volume of water flow per second, V = 100 m3/s
Efficiency of turbine generator, n = 60% = 0.6
Acceleration due to gravity, g = 9.8 m/s2
Density of water, ρ = 103 kg/m3
Electric power available from the plant = η × hρgV
= 0.6 × 300 × 103 × 9.8 × 100
= 176.4 × 106 W
= 176.4 MW
20. Answer the following questions:
(a) In any ac circuit, is the applied instantaneous voltage equal to the algebraic sum of the instantaneous voltages across the series elements of the circuit? Is the same true for rms voltage?
(b) A capacitor is used in the primary circuit of an induction coil.
(c) An applied voltage signal consists of a superposition of a dc voltage and an ac voltage of high frequency. The circuit consists of an inductor and a capacitor in series. Show that the dc signal will appear across C and the ac signal across L.
(d) A choke coil in series with a lamp is connected to a dc line. The lamp is seen to shine brightly. Insertion of an iron core in the choke causes no change in the lamp’s brightness. Predict the corresponding observations if the connection is to an ac line.
(e) Why is choke coil needed in the use of fluorescent tubes with ac mains? Why can we not use an ordinary resistor instead of the choke coil?
Answer:
(a) Yes; the statement is not true for rms voltage
It is true that in any ac circuit, the applied voltage is equal to the average sum of the instantaneous voltages across the series elements of the circuit. However, this is not true for rms voltage because voltages across different elements may not be in phase.
(b) High induced voltage is used to charge the capacitor.
A capacitor is used in the primary circuit of an induction coil. This is because when the circuit is broken, a high induced voltage is used to charge the capacitor to avoid sparks.
(c) The dc signal will appear across capacitor C because for dc signals, the impedance of an inductor (L) is negligible while the impedance of a capacitor (C) is very high (almost infinite). Hence, a dc signal appears across C. For an ac signal of high frequency, the impedance of L is high and that of C is very low. Hence, an ac signal of high frequency appears across L.
(d) If an iron core is inserted in the choke coil (which is in series with a lamp connected to the ac line), then the lamp will glow dimly. This is because the choke coil and the iron core increase the impedance of the circuit.
(e) A choke coil is needed in the use of fluorescent tubes with ac mains because it reduces the voltage across the tube without wasting much power. An ordinary resistor cannot be used instead of a choke coil for this purpose because it wastes power in the form of heat.
21. A power transmission line feeds input power at 2300 V 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 output power at 230 V?
Answer:
Input voltage, V1 = 2300
Number of turns in primary coil, n1 = 4000
Output voltage, V2 = 230 V
Number of turns in secondary coil = n2
Voltage is related to the number of turns as:
V1V2 = n1n2
2300230 = 4000n2
n2 = 40002302300 = 400
Hence, there are 400 turns in the second winding.
22. A small town with a demand of 800 kW of electric power at 220 V is situated 15 km away from an electric plant generating power at 440 V. The resistance of the two wire lines carrying power is 0.5 Ω per km. The town gets power from the line through a 4000-220 V step-down transformer at a sub-station in the town.
(a) Estimate the line power loss in the form of heat.
(b) How much power must the plant supply, assuming there is negligible power loss due to leakage?
(c) Characterize the step up transformer at the plant.
Answer:
Total electric power required, P = 800 kW = 800 × 103 W
Supply voltage, V = 220 V
Voltage at which electric plant is generating power, V' = 440 V
Distance between the town and power generating station, d = 15 km
Resistance of the two wire lines carrying power = 0.5 Ω/km
Total resistance of the wires, R = (15 + 15)0.5 = 15 Ω
A step-down transformer of rating 4000 − 220 V is used in the sub-station.
Input voltage, V1 = 4000 V
Output voltage, V2 = 220 V
Rms current in the wire lines is given as:
I = PV1
= 800 1034000 = 200A
(a) Line power loss = I2R
= (200)2× 15
= 600 × 103 W
= 600 kW
(b) Assuming that the power loss is negligible due to the leakage of the current:
Total power supplied by the plant = 800 kW + 600 kW = 1400 kW
(c) Voltage drop in the power line = IR = 200 × 15 = 3000 V
Hence, total voltage transmitted from the plant = 3000 + 4000 = 7000 V
Also, the power generated is 440 V.
Hence, the rating of the step-up transformer situated at the power plant is 440 V − 7000 V.
23.What is a Transformer ? How many types of transformers ?
Answer: Transformer: A Transformer is used to convert low voltage (or high current) to high voltage (or low current) and vice versa. It is based on the principle of electromagnetic induction.
One of the coils called the primary coil has Np turns. The other coil is called the secondary coil; it has Ns turns. Often the primary coil is the input coil and the secondary coil is the output coil of the transformer.
When an alternating voltage is applied to the primary, the resulting current produces an alternating magnetic flux which links the secondary and induces an emf in it. The value of this emf depends on the number of turns in the secondary.
In a transformer, if all the magnetic flux linked with the secondary then,
NS/NP = VS/VP =k
Here, k is called the transformation ratio. Now, NP and NS are the number of turns in the primary and the secondary respectively and VP and VS are the rms voltages across the primary and secondary respectively.
The primary coil of a transformer is connected to an AC source and load is connected to the secondary coil of the transformer.
Types of Transformer:
There are two types of transformer and they are:
• Step-Up Transformer
• Step-Down Transformer
Step-Up Transformer:
The transformers which are used to convert low voltages into higher ones are called ‘step-up’ transformers.
In step-up transformer VS > VP for which NS > NP or k > 1.
Step- Down Transformer:
The transformers which are used to convert high voltages into lower ones are called ‘step-up’ transformers.
In step-down transformer VS < VP for which NS < NP or k < 1.
24. (a) Though voltage and current in an AC circuit are represented by phasors, yet they are scalars. Explain how.
(b) Although there is no direct electrical connection between the two coils of a transformer, energy is being transferred from primary coil to secondary coil. How?
Answer:(a) Yes, although voltage and current in an AC circuit are represented by phasors (the rotating vectors), yet they are scalar quantities. The concept of phasors is being used in AC circuits because the amplitudes and phases of harmonically varying scalars combine mathematically in the same way as do the projections of rotating vectors of corresponding magnitudes and directions.
(b) Although there is no electrical connection between the primary and secondary coils of a transformer, still energy is transferred from the primary circuit to secondary circuit. Owing to an AC voltage applied across the primary coil an alternating current flows in primary coil and an alternating magnetic flux per unit turn is created. Through the iron core, this alternating magnetic flux is also linked with each turn of secondary coil, due to which, an induced emf is set up in the secondary coil. In this manner, electrical energy is continuously being transferred from primary to secondary coil circuit.
25. A series LCR circuit which is connected to a variable frequency with a 200 V source with L = 50 mH, C = 80 µF and R = 40 Ω. Determine
(i) The frequency of the source which derives the circuit in resonance.
(ii) the quality factor (Q) of the circuit.
Answer:
(i) 0 = 1/√LC
= 1/√(50 × 10-3 ) (80 × 10-6 )
= 1/√4000 × 10-9
= 1/√4 × 10-6
= 103 /2= 1000/2 = 500 rad S-1
f =0/2r = 500/2r = 80 hertz
(ii) Q = 0L/R = (500 × (50 × 10-3 ))/40 = 0.625हिंदी के सभी अध्याय के महत्वपूर्ण प्रशन उत्तर के लिए अभी Download करें Vidyakul App - Free Download Click Here
