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Alternating Current

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Physics Part I

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Class 10 Class 12

(a) For a given a.c.,i space equals space i subscript m sin space omega t comma show that the average power dissipated in a resistor R over a complete cycle is 1 half i squared subscript m R

(b) A light bulb is rated at 100 W for a 220 V a.c. supply. Calculate the resistance of the bulb. 


a) Average power consumed in resistor R is given by, 

straight P subscript av space equals space fraction numerator 1 over denominator integral subscript 0 superscript straight T dt end fraction space integral subscript 0 superscript straight T straight i squared space straight R space dt

space space space space space equals space fraction numerator straight i squared subscript straight m space straight R over denominator straight T end fraction space integral subscript 0 superscript straight T sin squared space ωt space dt space space space space space space space space space space space space space space space space... space left parenthesis 1 right parenthesis
space space space space space space equals fraction numerator straight i squared subscript straight m space straight R over denominator 2 straight T end fraction integral subscript 0 superscript straight T left parenthesis 1 space minus space cos space 2 ωt right parenthesis space dt space space space space space space space space space space space space space space

space space space space space space equals fraction numerator straight i squared subscript straight m space straight R over denominator 2 straight T end fraction space open square brackets integral subscript 0 superscript straight T dt space minus space integral subscript 0 superscript straight T cos space 2 ωt space dt close square brackets space space space space space space space space space space space space space... space left parenthesis 2 right parenthesis thin space

space space space space space space space equals fraction numerator straight i squared subscript straight m space straight R over denominator 2 straight T end fraction space left square bracket space straight T space minus space 0 right square bracket space

space space space space space space space equals space fraction numerator straight i squared subscript straight m space straight R over denominator 2 straight T end fraction 

b) In case of ac, we have

straight P subscript av space equals space straight v squared subscript rms over straight R space equals space straight V squared subscript eff over straight R

rightwards double arrow space straight R space equals space straight V squared subscript rms over straight P subscript av

space space space space space space space space space equals space fraction numerator 220 space straight x space 220 over denominator 100 end fraction

space space space space space space space space space equals space 484 space straight capital omega

2079 Views

Draw a necessary arrangement for winding of primary and secondary coils in a step-up transformer. State its underlying principle and derive the relation between the primary and secondary voltages in terms of number of primary and secondary turns. Mention the two basic assumptions used in obtaining the above relation.

State any two causes of energy loss in actual transformers.


Underlying principle of a step-up transformer: A transformer which increases the ac voltage is known as a step up transformer.

Working of step-up transformer is based on the principle of mutual inductance and it converts the alternating low voltage to alternating high voltage. The number of turns in the secondary coil is greater than the number of turns in the primary coil.

i.e.,                                                    NS > NP

                                  


Working: When an A.C source is connected to the ends of the primary coil, the current changes continuously in the primary coil. Hence, the magnetic flux which is linked with the secondary coil changes continuously. So, the emf which is developed across the secondary coil is same as that in the primary coil. The emf is induced in the coil as per Faraday’s law.

Let, EP be the alternating emf applied to primary coil and np be the number of turns in it.

Consider straight ϕ as the electric flux associated with it.

Then, 

straight E subscript straight P space equals space minus straight n subscript straight p space dϕ over dt space space space space space space space space space space space space space space space... space left parenthesis 1 right parenthesis space

where comma space fraction numerator d ϕ over denominator d t end fraction space is space the space rate space of space change space of space flux space across space
the space primary space coil. space

Now comma space Emf space across space the space secondary space coil space
be space straight E subscript straight s space and space straight n subscript straight s space be space the space number space of space turns space in space it.

straight E subscript straight S space equals space minus straight n subscript straight s space dϕ over dt space space space space space space space space space space space space space space... space left parenthesis 2 right parenthesis

Dividing space equation space left parenthesis 2 right parenthesis space by space left parenthesis 1 right parenthesis comma space

E subscript s over E subscript p equals n subscript s over n subscript p equals straight k italic colon

Assumption: We assume that there is no leakage of flux so that, the flux linked with each turn of primary coil is same as flux linked with secondary coil.

Two sources of energy loss in the transformer:

Joule Heating – Energy is lost in resistance of primary and secondary windings in the form of heat.

H = I2Rt

Flux leakage - Energy is lost due to coupling of primary and secondary coils not being perfect, i.e., whole of magnetic flux generated in primary coil is not linked with the secondary coil.

iii)

Conservation of law of energy is not violated in step-up transformer. When output voltage increases, the output current automatically decreases. Thus, there is no loss of energy. 

 

2204 Views

Derive an expression for the impedance of a series LCR circuit connected to an AC supply of variable frequency.

Plot a graph showing variation of current with the frequency of the applied voltage.

Explain briefly how the phenomenon of resonance in the circuit can be used in the

Tuning mechanism of a radio or a TV set.

a)

Expression for impedance in LCR series circuit:

Suppose a resistance R, inductance L and capacitance C are connected to series and an alternating voltage V = straight V subscript straight o space sinωt space is applied across it. 


Since L, C and R are connected in series, current flowing through them is the same. The voltage across R is VR, inductance across L is VL and across capacitance is VC. 

The voltage VR and current i are in the same phase, the voltage VL will lead the current by angle 90° while the voltage VC will lag behind the current by 90°. 

Clearly VC and VL are in opposite directions, therefore their resultant potential difference =VC -V(if VC >VC).

Thus, VR and (VC -VL) are mutually perpendicular and the phase difference between them is 90°. As seen in the fig, we can say that, as applied voltage across the circuit is V, the resultant of VR and (VC -VL) will also be V. 

So, 

V squared space equals space V subscript R squared space plus space left parenthesis V subscript c minus V subscript L right parenthesis squared italic space

rightwards double arrow space V space equals space square root of V subscript R to the power of italic 2 space italic plus space italic left parenthesis V subscript c italic minus V subscript L italic right parenthesis to the power of italic 2 end root

B u t italic comma italic space V subscript R space equals space R i comma space V subscript c space equals space capital chi subscript c i space a n d space V subscript L space equals space capital chi subscript L i

w h e r e italic comma italic space capital chi subscript c space equals space fraction numerator italic 1 over denominator omega C end fraction space a n d space capital chi subscript L space equals space omega L
S o italic comma italic space V space equals space square root of italic left parenthesis R i italic right parenthesis to the power of italic 2 italic plus italic left parenthesis capital chi subscript c i space italic minus space capital chi subscript L i space italic right parenthesis to the power of italic 2 end root italic comma italic space 

Therefore, Impedance of the circuit is given by, 

Error converting from MathML to accessible text.
This is the impedence of the LCR series circuit.

b)

The below graph shows the variation of current with the frequency of the applied voltage. 

 
c)

A radio or a TV set has a L-C circuit with capacitor of variable capacitance C. The circuit remains connected with an aerial coil through the phenomenon of mutual inductance. Suppose a radio or TV station is transmitted a programme at frequency f, then waves produce alternating voltage of frequency f, in area, due to which an emf of same frequency is induced in LC circuit, When capacitor C is in circuit is varied then for a particular value of capacitance, C italic comma italic space f space equals space fraction numerator 1 over denominator 2 pi square root of L C end root end fraction, the resonance occurs and maximum current flows in the circuit; so the radio or TV gets tuned. 

4688 Views

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

i) A transformer is a device that changes a low alternating voltage into a high alternating voltage or vice versa. 

Transformer works on the principle of mutual induction. 

A changing alternate current in primary coil produces a changing magnetic field, which induces a changing alternating current in secondary coil. 

 

Energy losses in transformer: 

Flux leakage due to poor structure of the core and air gaps in the core.

Loss of energy due to heat produced by the resistance of the windings.

Eddy currents due to alternating magnetic flux in the iron core, which leads to loss due to heating. 

Hysteresis, frequent and periodic magnetisation and demagnetization of the core, leading to loss of energy due to heat.

ii)

a)  Number of turns in secondary coil is given by, 


space space space space space space straight N subscript straight s over straight N subscript straight P space equals space n
rightwards double arrow space space N subscript S over 100 space equals space 100 space
rightwards double arrow space space space space space N subscript S space equals space 10 comma 000 

b) Current in primary is given by, 

straight I subscript straight P space straight V subscript straight P space equals space straight P

rightwards double arrow space straight I subscript straight P space equals space 1100 over 220 space equals space 5 space straight A

c) Voltage across secondary  is given by, 

space space space space space space space fraction numerator straight V subscript straight S over denominator space straight V subscript straight P end fraction space equals space N subscript S over N subscript P space equals space n

rightwards double arrow space space space straight V subscript straight S space equals space 100 space straight x space 220 space equals space 22 comma 000 space straight V

d) Current in secondary is given by, 

space space space space space straight V subscript straight S space straight I subscript straight S space equals space straight P

rightwards double arrow space straight I subscript straight S space equals space straight P over straight V subscript straight S space equals space 1100 over 22000 space equals space 0.05 space straight A

e) In an ideal transformer,

Power in secondary = Power in primary = 1,100 W
2030 Views

A group of students while coming from the school noticed a box marked “Danger H.T. 2200 V” at a substation in the main street. They did not understand the utility of such a high voltage, while they argued; the supply was only 220 V. They asked their teacher this question the next day. The teacher thought it to be an important question and therefore explained to the whole class.

Answer the following questions:

(i) What device is used to bring the high voltage down to low voltage of a.c. current and what is the principle of its working?

(ii) Is it possible to use this device for bringing down the high dc voltage to the low voltage? Explain.

iii) Write the values displayed by the students and the teacher. 


i) The device which is used to bring the high voltage down to low voltage is step down transformer. The working principle of step down transformer is mutual induction. Whenever there is change in current in one circuit, emf is induced in the neighboring circuit.

ii) Transformer cannot be used to bring down the high d.c voltage to low voltage because its working is based on electromagnetic induction, which is associated with varying magnetic flux. But, current is constant for a d.c source. So, magnetic flux will not vary. Hence, output cannot be obtained from the transformer.

iii) There is a curiosity among the students to gain knowledge. And the teachers are able to make use of their knowledge and impart it to students to explain the practical application.

2336 Views

(i) An a.c. source of voltage V = Vo sin ωt is connected to a series combination of L, C and R. Use the phasor diagram to obtain an expression for impedance of a circuit and the 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?
ii) In a series LR circuit, XL = R and the power factor of the circuit is P1. When capacitor with capacitance C, such that XL = XC is put in series, the power factor becomes P2. Calculate P1 / P2.

Voltage of the source is given by, 

V = Vo sin ωt 

                                           

Let current of the source be I = Io sin ωt

Maximum voltage across R is VR = Vo R, represented along OX

Maximum voltage across L = VL  = IO XL, represented along OY and is 90o ahead of Io.

Maximum voltage across C = VC = Io XC, represented along OC and is lagging behind Io by 90o

Hence, reactive voltage is VL - VC, represented by OB'



the vector sum of VR, VL and VC is resultant of OA and OB', represented along OK.

OK = Vosquare root of OA squared plus OB squared end root
i.e., Vosquare root of straight V subscript straight R squared plus left parenthesis straight V subscript straight L minus straight V subscript straight C right parenthesis squared end root space equals space square root of left parenthesis I subscript O R right parenthesis squared plus left parenthesis I subscript o X space minus space V subscript C right parenthesis squared end root

rightwards double arrow space straight V subscript straight o space equals space straight I subscript straight o space square root of straight R squared space plus space left parenthesis straight X subscript straight L minus straight X subscript straight c right parenthesis squared end root

Impedance, Z = straight V subscript straight o over straight I subscript straight o space equals space R square root of straight R squared space plus space left parenthesis straight X subscript straight L minus straight X subscript straight c right parenthesis squared end root

When, X= XC ,  the voltage and current are in the same phase. 

In such a situation, the circuit is known as non-inductive circuit. 

ii) 

Given,

Power factor, P1 = R/Z

rightwards double arrow space straight P subscript 1 space equals space fraction numerator straight R over denominator square root of straight R squared space plus space straight X squared end root end fraction space equals space fraction numerator straight R over denominator square root of 2 straight R squared end root end fraction space equals space fraction numerator 1 over denominator square root of 2 end fraction

space space space space space straight P subscript 2 space equals space straight R over straight Z

rightwards double arrow space straight P subscript 2 space equals space fraction numerator straight R over denominator square root of straight R squared plus left parenthesis straight X subscript straight L minus straight X subscript straight C right parenthesis squared end root end fraction space equals space 1 space
Thus, 

straight P subscript 1 over straight P subscript 2 space equals space fraction numerator 1 over denominator square root of 2 end fraction





3757 Views