Chapter 7 Alternating Current Solutions
Question - 21 : - Obtain the resonant frequencyand Q-factor of a series LCR circuit with L =3.0 H, C = 27 μF, and R = 7.4 Ω. It isdesired to improve the sharpness of the resonance of the circuit by reducingits ‘full width at half maximum’ by a factor of 2. Suggest a suitable way.
Answer - 21 : -
Inductance, L =3.0 H
Capacitance, C =27 μF = 27 × 10−6 F
Resistance, R =7.4 Ω
At resonance, angular frequencyof the source for the given LCR series circuit is given as:
Q-factorof the series:
To improve the sharpness of the resonance by reducingits ‘full width at half maximum’ by a factor of 2 without changing
, we need to reduce R tohalf i.e.,
Resistance
Question - 22 : - Answer the following questions:
(a) In any accircuit, is the applied instantaneous voltage equal to the algebraic sum of theinstantaneous voltages across the series elements of the circuit? Is the sametrue for rms voltage?
(b) Acapacitor is used in the primary circuit of an induction coil.
(c) Anapplied voltage signal consists of a superposition of a dc voltage and an acvoltage of high frequency. The circuit consists of an inductor and a capacitorin series. Show that the dc signal will appear across C andthe ac signal across L.
(d) A chokecoil in series with a lamp is connected to a dc line. The lamp is seen to shinebrightly. Insertion of an iron core in the choke causes no change in the lamp’sbrightness. Predict the corresponding observations if the connection is to anac line.
(e) Why ischoke coil needed in the use of fluorescent tubes with ac mains? Why can we notuse an ordinary resistor instead of the choke coil?
Answer - 22 : -
(a) Yes;the statement is not true for rms voltage
It is true that in any accircuit, the applied voltage is equal to the average sum of the instantaneousvoltages across the series elements of the circuit. However, this is not truefor rms voltage because voltages across different elements may not be in phase.
(b) Highinduced voltage is used to charge the capacitor.
A capacitor is used in the primary circuit of an induction coil. This isbecause when the circuit is broken, a high induced voltage is used to chargethe capacitor to avoid sparks.
(c) Thedc signal will appear across capacitor C because for dcsignals, the impedance of an inductor (L) is negligible while theimpedance of a capacitor (C) is very high (almost infinite). Hence, a dcsignal appears across C. For an ac signal of high frequency, theimpedance of L is high and that of C is verylow. Hence, an ac signal of high frequency appears across L.
(d) Ifan iron core is inserted in the choke coil (which is in series with a lampconnected to the ac line), then the lamp will glow dimly. This is because thechoke coil and the iron core increase the impedance of the circuit.
(e) A choke coil isneeded in the use of fluorescent tubes with ac mains because it reduces thevoltage across the tube without wasting much power. An ordinary resistor cannotbe used instead of a choke coil for this purpose because it wastes power in theform of heat.
Question - 23 : - A power transmission line feedsinput power at 2300 V to a stepdown transformer with its primary windingshaving 4000 turns. What should be the number of turns in the secondary in orderto get output power at 230 V?
Answer - 23 : -
Input voltage, V1 =2300
Number of turns in primarycoil, n1 = 4000
Output voltage, V2 =230 V
Number of turns in secondary coil= n2
Voltage is related to the numberof turns as:
Hence, there are 400 turns in thesecond winding.