(a) For a given a.c., show that the average power dissipated in a resistor R over a complete cycle is
(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,
b) In case of ac, we have
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 as the electric flux associated with it.
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.
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.
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 theTuning mechanism of a radio or a TV set.
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 = 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 -VL (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.
Therefore, Impedance of the circuit is given by,
This is the impedence of the LCR series circuit.
The below graph shows the variation of current with the frequency of the applied voltage.
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, , the resonance occurs and maximum current flows in the circuit; so the radio or TV gets tuned.
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.
a) Number of turns in secondary coil is given by,
b) Current in primary is given by,
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.
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 = Vo =
i.e., Vo =
Impedance, Z =
When, XL = XC , the voltage and current are in the same phase.
In such a situation, the circuit is known as non-inductive circuit.
Power factor, P1 = R/Z