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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1201. |
The general equation for the instantneous voltage of a 50 Hz generator with peak voltage 220 V isA. `220sin50pit`B. `220sin100pit`C. `pm220sin100pit`D. `220sin25pit` |
| Answer» Correct Answer - B | |
| 1202. |
In any ac circuit, is the applied instanteneous voltage equal to the algebraic sum of the instantneous voltages across the series elements of the circuit?Is the same true for rms value? |
| Answer» Correct Answer - Yes | |
| 1203. |
A small signal voltage `V(t)=V_(0)sin omegat` is applied across an ideal capacitor `C`:A. Current `I(t)`, lags voltage `V(t)` by `90^(@)`B. Over a full cycle the capacitor `C` does not consume any energy from the voltage source.C. Current `I(t)` is in phase with voltage `V(t)`D. Current `I(t)` leads voltage `V(t)` by `180^(@)` |
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Answer» Correct Answer - B Power `=V_(rms). I_(rms) cos phi` `as cos theta=0`(Because `phi=90^(@)`) `:.` Power consumed `=0` (in one complete cycle) |
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| 1204. |
An inductor `20 mH`, a capacitor `50 muF` and a resistor `40 Omega`are connected in series across of emf `V=10 sin 340 t`. The power loss in `A.C.` circuit isA. 0.67 WB. 0.76 WC. 0.89 WD. 0.51 W |
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Answer» Correct Answer - D Given, inductance, L=20mH Capacitance, C=`50muF` Resistance, R=`40Omega` emf, V`=10sin340t` `therefore` Power loss in AC circuit will be given as `P_(av)=I_(V)^(2)R= [E_(V)^(2)/Z]^(2).R` `(10/sqrt(2))^(2). 40[1/(40^(2) + (340xx20xx10^(-3) - 1/(340 xx 50 xx 10^(-6))))]` =`100/sqrt(2) x 40 xx 1/(1600 + (6.8-58.8)^(2))` `2000/(1600+2704) ~~ 0.46W ~~ 0.51 W` |
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| 1205. |
An oscillator circuit consists of an inductance of `0.5mH` and a capacitor of `20muF`. The resonant frequency of the circuit is nearlyA. 15.92 HzB. 159.2 HzC. 1592 HzD. 15910 Hz |
| Answer» Correct Answer - c | |
| 1206. |
An oscillator circuit consists of an inductance of `0.5mH` and a capacitor of `20muF`. The resonant frequency of the circuit is nearlyA. `15.92 Hz`B. `159.2 Hz`C. `1592 Hz`D. `15910 Hz` |
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Answer» Correct Answer - C `v_(0)=1/(2pisqrt(LC))=1/(2xx3.14sqrt(5xx10^(-4)xx20xx10^(-6)))` `v_(0)=(10^(4))/6.28=1529Hz` |
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| 1207. |
A small signal voltage `V(t)=V_(0)sin omegat` is applied across an ideal capacitor `C`:A. over a full cycle the capacitor C does not consume any energy from the voltage source.B. current I(t) is in phase with voltage V(t)C. current `I(t)` leads voltage V(t) by `180^(@)`D. current I(t), lags voltage V(t) by `90^(@)`. |
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Answer» Correct Answer - A For an AC circuit containing capacitor only, the phase difference between current and voltage will be `pi/2` (i.e., `90^(@)`. In this case current is ahead of voltage by `pi/2`. Hence, power ini this case is given by `P=Vicosphi` where, `phi`= Phase difference between voltage and current) `P=VIcos90^(@)`=0 |
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| 1208. |
A `100 Omega` resistance and a capacitor of `100 Omega` reactance are connected in series across a `220` V source. When the capacitor is `50%` charged, the peak value of the displacement current isA. 2.2 AB. 0.45833333333333C. 4.4 AD. `11sqrt(2)` A |
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Answer» Correct Answer - A Impedance of the R-C circuit, `Z= sqrt(R^(2)+X_(C)^(2))` Where, `R=100Omega` and `X_(C) = 100Omega` `rArr Z=sqrt((100)^(2)+(100)^(2))` = `100sqrt(2)Omega` Peak value of the current, `I_("max") = V_("max")/Z = (220sqrt(2))/(100sqrt(2))`=2.2 A |
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| 1209. |
An `AC` voltage is applied to a resistance `R` and an inductance `L` in series. If `R` and the inductive reactance are both equal to `3 Omega`, the phase difference between the applied voltage and the current in the circuit isA. `pi//4`B. `pi//2`C. zeroD. `pi//6` |
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Answer» Correct Answer - A |
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| 1210. |
A `100 Omega` resistance and a capacitor of `100 Omega` reactance are connected in series across a `220` V source. When the capacitor is `50%` charged, the peak value of the displacement current isA. `2.2A`B. `11A`C. `4.4A`D. `11sqrt2A` |
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Answer» Correct Answer - A |
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| 1211. |
The reactance of a capacitor of capacitance C is X. If both th frequency and capacitance be doubled, then new reactance will beA. `X`B. `2X`C. `4X`D. `(X)/(4)` |
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Answer» Correct Answer - D |
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| 1212. |
When `1A` is passed through three coils `A,B,C` in series the voltage drops are respectively 6,3 and 8 volt on direct current source and 7,5 and 10 volt on Alternating current source Power factor of whole circuit when atlernating current flowA. 0.6B. 0.8C. 0.78D. 1 |
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Answer» Correct Answer - C `R_(A) = 6, R_(B ) = 3, R_(C ) = 8` `Z_(A) = 7, Z_(B) = 5, Z_(C ) = 10` `R` (total) `= 6 + 3 + 8 = 17 Omega` `X_(A) = sqrt(7^(2) - 6^(2)) = 3.6 Omega` `X_(B) = 4 Omega , X_(C ) = 6 Omega` X (total) `= 3.6 + 4 + 6 = 13.6 Omega` Total impedance `= sqrt(R^(2) + X^(2)) = 21.8 m` Power factor `= (R )/(Z) = (17)/(21.8) = 0.78` |
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| 1213. |
In the given arrangement the square loop of area `10 cm^(2)` rotates with an angular velocitys `omega` about its diagonal. The loop is connected to a inductance of `L = 100 mH` and a capacitance of `10 mF` in series. The lead wires have a net resistance of `10 Omega`. Given that `B = 0.1 T` and `omega = 63 "rad"//s` Find the rms currentA. `6 xx 10^(5) A`B. `5 xx 10^(-5) A`C. `4 xx 10^(-5) A`D. `7 xx 10^(-5) A` |
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Answer» Correct Answer - C `phi = BA cos omega t` `e = - (d phi)/(dt) = BA omega sin omega t` `i_("rms") = (V_("rms"))/(Z) = (BA omega // sqrt(2))/(sqrt((omega L - (1)/(omega C))^(2) + R^(2)))` `= (0.1 xx 10^(-3) xx 63)/(sqrt(2 { (63 x 0.1 - (1)/(63 xx 0.01))^(2) + (10)^(2)}))` `= 4.0 xx 10^(-5) A` |
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| 1214. |
In an `LR`-circuit, the inductive reactance is equal to the resistance `R` of the circuit. An e.m.f `E=E_(0)cos (omegat)` applied to the circuit. The power consumed in the circuit isA. `(E_(0)^(2))/(R)`B. `(E_(0)^(2))/(2R)`C. `(E_(0)^(2))/(4R)`D. `(E_(0)^(2))/(8R)` |
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Answer» Correct Answer - C `P=E_("rms")i_("rms")cos varphi=(E_(0))/(sqrt(2))xx(i_(0))/(sqrt(2))xxR/Z` `implies (E_(0))/(sqrt(2))xx(E_(0))/(Zsqrt(2))xxR/Z implies P=(E_(0)^(2)R)/(2Z^(2))` Given `X_(L)=R`, so `Z=sqrt(2)R implies P=(E_(0)^(2))/(4R)` |
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| 1215. |
In an `L-C-R` ereis circuit `R=150Omega, L=0.0750 H` and `C=0.0180muF` . The source has voltage amplitude `V=150V` and a frequencey equal to the resonacne frequency of the circuit. (a) What is the power factor ? (b) What is the average power delivered by the source? (c) The capacitor is replaced by one with `C=0.0360muF` and the source frequency is adjusted to the new resonance value. Then, what is the average power delivered by the source? |
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Answer» (a) At resonance frequency, `X_L=X_C,Z=R` and power factor `cosphi=R/Z=1.0` (b) `P=V_(rms)^2/R=((150//sqrt(2))^(2))/(150)=75W` (c) Again `P=V_("rms")^2/R=75W`. |
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| 1216. |
A resistance is connected to a capacitor in AC are the phase differece is `(pi)/(4)` between current and voltage. Whe the same resistance is connected to an inductor, phase difference becomes `tan^(-1)(2)`. Power factor of the circuit when both capacitor and inductor are connect to the resistance will beA. 1B. `(1)/(sqrt2)`C. `(1)/(sqrt3)`D. `(1)/(2)` |
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Answer» Correct Answer - B |
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| 1217. |
A `100 km` telegraph wire hasd capacity of `0.02 mu F//km`, if it carries an alternating current of frequency `5 kHZ`. The value of an inductrance required to be connected in series so that the impedance is minimum.A. `50.7 mH`B. `5.07 mH`C. `0.507 mH`D. `507 mH` |
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Answer» Correct Answer - 3 `omega = (1)/(sqrt(LC)) implies L = (1)/(omega^(2)C) = (1)/((2 pi n)^(2) C)` |
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| 1218. |
In a series L-R growth circuit, if maximum current and maximum voltage across inductor of inductane 3mH are 2A and 6V respectively, the the time constant of the circuit isA. 1msB. 2msC. 0.5msD. 0.6ms |
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Answer» Correct Answer - A |
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| 1219. |
A condenser of `10 mu F` and an inductor of `1 H` are connected in series with an `A.C` source of frequency `50 Hz`. The impedance of the combination will be (take `pi^(2) = 10)`A. zeroB. InifinityC. `44.7 Omega`D. `5.67 Omega` |
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Answer» Correct Answer - 1 `Z = (2 pi f L - (1)/(2 pi f C))` |
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| 1220. |
which of ther following is constructed on the principle of electromagnetic induction?A. GalvanometerB. Electric motorC. GeneratorD. Voltmeter |
| Answer» Correct Answer - C | |
| 1221. |
One `10V, 60W` bulb is to be connected to `100V` line. The required inductance coil has self-inductance of value `(f=50Hz)`A. 0.052 HB. 2.42 HC. 16.2 HD. 16.2 mH |
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Answer» Correct Answer - A From P = VI = `V^(2)/R` `implies I = P/V = 6A` `implies R = V^(2)/P = 100/60 = 5/3Omega` Now, `I = V/Z 6 = 100 / sqrt (5/3^(2)) + (2pifL)^(2)` Solving this equation we get, L = 0.052 H |
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| 1222. |
A virtual current of `4A` and `50Hz` flows in an `AC` circuit contaning a coil. The power consumed in the coil is `100V` its inductance will beA. `1/(3pi) H`B. `1/(5pi) H`C. `1/(7pi) H`D. `1/(9pi) H` |
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Answer» Correct Answer - B `R=P/(i_(rms)^(2))=240/16=15 Omega` `Z=V/i=100/4=25 Omega` Now `X_(L)=sqrt(Z^(2)-R^(2))=sqrt((25)^(2)-(15)^(2))=20 Omega` `:. 2pi vL =20 implies L =20/(2pixx50)=1/(5pi)Hz` |
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| 1223. |
Power dissipated in an `L-C-R` series circuit connected to an `AC` source of emf `epsilon` isA. `(epsilonR)/(R^(2)+(Lomega-1/(Comega))^(2)]`B. `sqrt(R^(2)+(Lomega-1/(Comega)^(2))/(R )`C. `[R^(2)+(Lomega-1/(Comega))^(2)]/( R)`D. `(omega^(2)R)/(sqrt(R^(2)+(Lomega+1/(Comega))^(2)` |
| Answer» Correct Answer - A | |
| 1224. |
Power dissipated in an `L-C-R` series circuit connected to an `AC` source of emf `epsilon` isA. `(epsilon^(2)sqrt(R^(2)+(omega L-(1)/(omega C))^(2)))/(R )`B. `(epsilon^(2)[R^(2)+(omega L-(1)/(omega C))^(2)])/(R )`C. `(epsilon^(2)R)/(sqrt(R^(2)+(omega L-(1)/(omega C))^(2)))`D. `(epsilon^(2)R)/([R^(2)+(omega L-(1)/(omega C))^(2)])` |
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Answer» Correct Answer - D Average power, `P = V_("rms")I_("rms")cos phi` Here, `Z = sqrt(R^(2)+(X_(L)-X_(C ))^(2)), cos phi = (R )/(Z)` But `I_("rms")=(V_("rms"))/(Z). Therefore P= V_("rms")^(2)(R )/(Z^(2))` `therefore P = V_("rms")^(2) (R )/({R^(2)+(X_(L)-X_(C ))^(2)})=(epsilon^(2)R)/([R^(2)+(omega L-(1)/(omega C))^(2)])` |
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| 1225. |
A series `LCR` circuit with `R = 20 Omega, L = 1.5 H` and `C = 35 mu F` 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 in `kW` transferred to the circuit in one complete cycle?A. 200 WB. 2000 WC. 100 WD. 4000 W |
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Answer» Correct Answer - B If the frequency of the ac equals thenatural frequency of the circuit, the impedance `Z=R=20 Omega` The average power dissipated per cycle, `P_(av)=(V_("rms")^(2))/(Z)=(V_("rms")^(2))/(R )=((200)^(2))/(20)=2000 W` |
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| 1226. |
A series LCR circuit with `R = 20 Omega, L = 1.5 H` and `C = 35 mu F` 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 ? |
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Answer» Given resistance `R = 200 Omega`, Inductance `L = 1.5 H`, capacitance `C = 35 mu F = 35xx10^(-6)F` and voltage `V_("rms")=200 V` When, the frequency of the supply equal to the natural frequency of the circuit, this is the condition of resonance. At the condition of resonance, Impedance `Z=R=20 Omega` The rms value of current in the circuit `I_("rms")=(V_("rms"))/(Z)=(200)/(20)=10A` `phi = 0^(@)` Power transferred to the circuit in one complete cycle. `P = I_("rms")V_("rms")cos phi` `= 10xx200xxcos 0^(@)=2000 omega =2K omega` |
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| 1227. |
A series LCR circuit with R=20`Omega` , L=1.5 H and C=35`muF` 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? |
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Answer» When natural frequency and supply frequency and supply frequency are equal, resonance occurs. `thereforeX_L=X_CthereforeZ=R` `P=V^2/Z=V^2/R=(200times200)/20=2000 W` |
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| 1228. |
A series `LCR` circuit with `R = 20 Omega, L = 1.5 H` and `C = 35 mu F` 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 in `kW` transferred to the circuit in one complete cycle? |
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Answer» At resonance, the frequency of the supply power equals the natural frequency of the given LCR circuit. Resistance, `R=20Omega` Inductance, L=1.5H Capacitance, `C=35muF=30xx10^(-6)F` AC supply voltage to the LCR circuit, V=200V Impedance of the circuit is given by the relation. `Z=sqrt(R^(2)=(omegaL-(1)/(omegaC))^(2)) ` At resonance, ` omegaL(1)/( omegaC)` `thereforeZ=R20Omega` Current in the circuit can be calculated as: `I=(v)/(Z)` `=(200)/(20)=10A` Hence, the average power tranferred to the circuit in one complete cycle =VI `=200xx10=2000W`. |
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| 1229. |
A radio can tune over the frequency range of a portion of MW broadcasr band: 800kHz to 1200kHz. IF its LC circuit has an effective inductance of 200`muH`, what must be the range of the variable capacitor? [Hint: for tuning, the natural frequency of the LC circuit should be equal to the frequency of the radio wave.] |
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Answer» `f=1/(2pisqrt(LC))`or,`C=1/(4pi^2f^2L)` when L=`200muH` and f=800 kHz, `C_1=1/(4pi^2times800times800times10^8times200times10^-6)` =197.8pF when f=1200 kHz, `C_2=1/(4pi^2times1200times1200times10^6times200times10^-6)` `=87.9pF` `therefore` The range of the capacitance should be between 87.9 pF ad 197.8 pF, i.e., between 88pF and 198 pF. |
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| 1230. |
A radio can tune over the frequency range of a portion of MW broadcast band (800 kHz to 1200 kHz). If its LC circuit has an effective inductance of `200 mu H`, what must be the range of its variable capacitor ? |
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Answer» The range of frequency (v) of a radio is 800KHz to 1200 KHz. Lower tuning frequency, `v_(1)=800KHz=800xx10^(3)Hz` Upeer tuning frequency, `v_(2)=1200KHz=1200xx10^(3)Hz` Effective inductance of circuitL=200`muH=200xx10^(-6)H` Capacitance of variable capacitor for `v_(1)` is given as, `=(1)/(omega_(1)^(2)L)` Where, `omega_(1)=` Angular frequency for capacitor `C_(1)` `=2piv_(1)=2pixx800xx10^(-3) "rad"s^(-1)` `therefore C_(1)=(1)/((2pixx800xx10^(3))^(2)xx200xx10^(-6))` `=1.9809xx10^(-16)F=198.1pF` Capacitanc e of variable capacitor for `v_(2)`. ` =(1)/(omega_(2)^(2)L)` Whre, `omega_(2)=` angular frequency for capacitor `C_(2)` `=2piv_(2)=2pixx1200xx10^(3)"rad"s^(-1)` `thereforeC_(2)=(1)/((2pixx1200xx10^(3))^(2)xx200xx10^(-6))` =`88.04 pF` Hence, the range of the variable capacitor is from 88.04pF to 198.1 pF. |
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| 1231. |
A charged `30 mu F` capacitor is connected to a 27 mH inductor. What is the angular frequency of free oscillations of the circuit ? |
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Answer» Capacitance of capacitor `C = 30 mu F = 30xx10^(-6)F` Inductance `L = 27mH = 27xx10^(-3)H` For free oscillations, the angular frequency should be resonant frequency. Resonant angular frequency of oscillation of the circuit `omega_(r )=(1)/(sqrt(LC))` `= (1)/(sqrt(27xx10^(-3)xx30xx10^(-6)))=(10^(4))/(9)` `= 1.1xx10^(3)` rad/s |
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| 1232. |
Obtain the answers to (a) and (b) in Q .15, if the circuit is connected to 110 V, 12 kHz supply. Hence explain the statement that a capacitor is a conductor at very high frequencies. Compare this behaviour with that of a capacitor in d.c. after the steady state. |
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Answer» Capacitance of the capacitor, `C=100muF=100xx10^(-6)F` Resistance of the resistor, R=40`Omega` Supply voltage, V=110V Frequency of the supply. V=12KHz=`12xx10^(3)Hz` `= omega=2nv=2xxnxx12xx10^(3)03` =24nxx `10^(3)` rad/s Peak voltage, `V_(@)=V sqrt(2)=110sqrt(2)V` `I_(@)=(V_(@))/ (sqrt(R^(2)+(1)/(omega^(2)C^(2)))` Maximum current. `=(110sqrt(2))/(sqrt((40)^(2)+(24pixx10^(3)xx1 00 xx10^(-6))^(2))` For an d RC circuit the voltage lags behind the current by a phase angle of `phi` given as: `tan phi=((1)/( omegaC))/(R)=(1)/(omegaCR)` `=(0.2pi)/(180)`rad `therefore` Time lag `=(phi)/(omega)` `=(0.2pi)/(180xx24pixx10^(3))=1.55x x10^(-3)s=0.04mus` Hence `phi` tends to become zero at high frequencies. At a high frequency capacitor C acts as a conductor. In a dc circuit, after the steady state is acheived `omega=0` . Hence, capacitor C amounts to an open circuit. |
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| 1233. |
Obtain the answers (a) to (b) in Exercise 13 if the circuit is connected to a high frequency supply (240 V, 10 kHz). Hence, explain the statement that at very high frequency, an inductor in a circuit nearly amount to an open circuit. How does an inductor behave in a dc circuit after the steady state ? |
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Answer» Given frequency `f=10kHz = 10^(4)Hz` the rms value of voltage `V_("rms")=240 v` from Exercise 13 Resistance `R = 100 Omega`, Inductance `L = 0.5 H` Inductance `Z = sqrt(R^(2)+X_(L)^(2))` `= sqrt(R^(2)+(2pi fL)^(2))` `= sqrt((100)^(2)+(2xx3.14xx10^(4)xx0.5)^(2))` `= 31400.15 Omega` The rms value of current `I_(0)=sqrt(2)I_("rms")=1.414xx0.00764` `= 0.01080 A` and `tan phi = (X_(L))/(R )=(2pi xx1000xx0.5)/(100)` `tan phi = 100 pi` (very large). |
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| 1234. |
Obtain the answers (a) to (b) in Exercise 15 if the circuit is connected to a 110 V, 12 kHz supply ? Hence, explain the statement that a capacitor is a conductor at very high frequencies. Compare this behaviour with that of a capacitor in a dc circuit after the steady state. |
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Answer» Given, the ems value of voltage, `V_("rms")=110 V` The frequency of capacitor `f=12kHz = 12000 Hz`. Capacitance of conductor `C = 10^(-4)F`. Tesistance `R = 40 Omega` Capacitive Resistance `X_(C )=(1)/(2pi fC)=(1)/(2xx3.14xx12000xx10^(-4))` `= 0.133 Omega` The rms value of current `I_("rms")=(V_("rms"))/(sqrt(X_(C )^(2)+R^(2)))=(110)/(sqrt((40)^(2)+(0.133)^(2)))` `= 2.75 A`. The maximum value of current, `I_(0)=sqrt(2) I_("rms")` `= 1.414xx2.75` `= 3.9 A` Here, the value of `X_(C )` is very small, so term containing C is negligible. `tan phi = (1)/(omega CR)` `= (1)/(2xx3.14xx12000xx10^(-4)xx40)` `= (1)/(96pi)` It is very small. In DC circuits, `omega = 0` `X_(C )=(1)/(omega C)=oo` So, it behaves like an open circuit. |
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| 1235. |
A current is made of two components a `dc` component `i_(1) = 3A` and an `ac` component `i_(2) = 4 sqrt(2) sin omega t`. Find the reading of hot wire ammeter? |
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Answer» `i= i_(1) + i_(2) = 3 + 4 sqrt(2) sin * t` `i_("rms")^(2) = (int_(0)^(T) i^(2) dt)/(int_(0)^(T) dt) = (int_(0)^(T) (3 + 4 sqrt(2) sin * t)^(2) dt)/(T)` `i_("rms")^(2) = (1)/(T) int_(0)^(T) (9 + 24 sqrt(2) sin * t + 32 sin^(2) * t) dt` `:. i_("rms") = 5 A` |
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| 1236. |
A small town with a demand of 800 kW of power at 220 V is situated 15 km away from an electric plant generating power at 440 V. The resistance of the two wire line carrying power is `0.5Omega`.`km^-1`. The line gets power from the line through a 4000-220 V step-down transformer at a substation in the town. Estimate the line power loss in the form of heat? |
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Answer» Generating power of Electric plant = 800 kW at V = 220 V Distance = 15 km, Generating voltage = 440 V Resistance/length `= 0.5 Omega//km` Primary voltage VP = 4000 V Secondary voltage Vs = 220 V Power `= I_(P).V_(P)` `800xx1000=I_(P)xx4000` `I_(P)=200A` Line power loss in form of heat `= (I_(P))^(2)xx` Resistance of line `= (I_(P))^(2)xx0.5xx15xx2` `= (200)^(2)xx0.5xx15xx2` `= 60xx10^(4)W` `= 600 KW` |
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| 1237. |
If `i= t^(2), 0 lt t lt T` then `r.m.s.` value of current isA. `(T^(2))/(sqrt(2))`B. `(T^(2))/2`C. `(T^(2))/(sqrt(5))`D. None of these |
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Answer» Correct Answer - C `i_(rms)=sqrt(1/Tint_(0)^(T)i^(2)dt)=(T^(2))/(sqrt(5))` |
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| 1238. |
The variation of the instantaneous current `(I)` and the instantaneous e.m.f `(E)` in a circuit is as shown in figure. Which of the following statement is correct? A. The voltage lags behind the current by `pi//2`B. The voltage leads the current by `pi//2`C. The voltage and the current are in phaseD. The voltage leads the current by `pi` |
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Answer» Correct Answer - B At `t=0`, phase of the voltage is zero, while phase of the current is `-(pi)/2` i.e, voltage leads by `(pi)/2` |
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| 1239. |
The output current versus time curve of a rectifire is shown in the figure. The voltage value of output current in this case is A. `0`B. `(I_(0))/2`C. `(2I_(0))/(pi)`D. `I_(0)` |
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Answer» Correct Answer - C `I_(av)=(int_(0)^(T//2) idt)/(int_(0)^(T//2) dt)=(int_(0)^(T//2)I_(0)sin (omegat)dt)/(T//2)` `=(2I_(0))/T[(-cos omegat)/(omega)]_(0)^(T//2)=(2I_(0))/T[-(cos((omegaT)/2))/(omega)+(cos0^(@))/(omega)]` `=(2I_(0))/(omegaT)[-cospi+cos^(@)]=(2I_(0))/(2pi)[1+1]=(2I_(0))/(pi)` |
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| 1240. |
The output current versus time curve of a rectifire is shown in the figure. The voltage value of output current in this case is |
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Answer» Correct Answer - C c) Average value of output current is given by `I_(av)` = `I_(av) = (int_(0)^(T//2)Idt)/(int_(0)^(T//2)dt)= (int_(0)^(T//2)I_(0)sinomegatdt)/(T//2)` `=(2I_(0))/T[-(cosomegat)/(omega)]_(0)^(T//2)` `=(2I_(0))/T[-(cos(omegaT//2)/(omega)+(cos0^(@))/(omega))]` `=(2I_(0))/(omegaT)[(-cospi+cos0^(@))]=(2I_(0))/(2pi)[1+1]=(2I_(0))/(pi)` |
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| 1241. |
The variation of induced emf `(E)` with time `(t)` in a coil if a short bar magnet is moved along its axis with a constant velocity is best represent as A. B. C. D. |
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Answer» Correct Answer - B |
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| 1242. |
The electric and the magnetic field, associated with a electromagnetic, wave, propagating along the `+z-"axis"`, can be represented byA. `[E=E_(@)hatk,B=B_(@)hati]`B. `[E=E_(@)hatj,B=B_(@)hatj]`C. `[E=E_(@)hatj,B=B_(@)hatk]`D. `[E=E_(@)hati,B=B_(@)hatj]` |
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Answer» Correct Answer - D |
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| 1243. |
In an `AC` circuit, `V` and `I` are given by `V=100sin(100t)volts, I=100sin(100t+(pi)/3)mA`. The power dissipated in circuit isA. `10^(4)W`B. `10W`C. `2.5W`D. `5W` |
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Answer» Correct Answer - C `V_(0) = 100 V i_(0) = 100 mA = 100 xx 10^(-3) = 0.1 A phi = 60^(@)` `P = (1)/(2) V_(0)i_(0) cos phi = (1)/(2) xx 100 xx 0.1 xx (1)/(2) = 2.5 W` . |
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| 1244. |
In an `AC` circuit, `V` and `I` are given by `V=100sin(100t)volts, I=100sin(100t+(pi)/3)mA`. The power dissipated in circuit isA. `10^(4)`WB. 10 WC. 2.5 WD. 5 W |
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Answer» Correct Answer - C Voltage amplitude m `underset(o)(V)` = 100 V, current amplitude `underset(o)(i)` = 100 mA = `100 xx 10^(-3)` A and phase difference between I and V is `phi` = `pi`/3 = `60^(@)` . Now power dissipated is given by P = `underset(o)(V)``underset(o)(i)`/2 cos`phi` = `100 xx 100 xx 10^(-3) / 2 xx cos 60^(@)` = 2.5 W |
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| 1245. |
In an `AC` circuit, `V` and `I` are given by `V=100sin(100t)volts, I=100sin(100t+(pi)/3)mA`. The power dissipated in circuit isA. 100 WB. 10 WC. 5 WD. 2.5 W |
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Answer» Correct Answer - D Average power, `P_(av) = E_("rms")I_("rms") cos phi` `=E_(0)/sqrt(2) .I_(0)/sqrt(2) cos phi = (100V)/sqrt(2) xx (100 xx 10^(-3)A)/sqrt(2)cos (pi/3)` `=(100 xx 100)/2 xx 10^(-3) xx 1/2 = 2.5` W |
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| 1246. |
An electric bulb and a capacitor are connected in series with an `AC` source.On increasing the freqency of the source, the brightness of the bulbA. increaseB. decreaseC. remains unchargedD. sometime increases and sometimes decreases |
| Answer» Correct Answer - A | |
| 1247. |
A bulb and a capacitor are connected in series to an a.c. source of varialbe frequency. How will the brightness of the bulb change on increasing the frequency of a.c. source ? |
| Answer» As the frequency of the `a.c.` source increase, the capacitive reactance decrease `(X_(C) prop 1//f)` So the bulb glows with more brightness. | |
| 1248. |
In the a.c. circuit shown in the figure. The supply voltage has a constant r.m.s value `V` but varialbe frequency `f`. Resonance frequency in hertz is A. 10B. 100C. 1000D. 200 |
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Answer» Correct Answer - C `f_(0) = (1)/(2pi sqrt(LC)) = (1)/(2 pii xx sqrt((1)/(pi) xx (1)/(4 pi) xx 10^(-6))) = 1000 Hz` |
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| 1249. |
A capacitor has a resistance of `1200 M Omega` and capacitance of `22 mu F`. When connected to an a.c. supply of frequency 80 hertz, then the alternating voltage supply requried to drive a current of 10 virtual ampere isA. `904 sqrt(2) V`B. `904 V`C. `904 sqrt(2) V`D. `452 V` |
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Answer» Correct Answer - 2 `f = 80 Hz, I_(V) = 10 A` Current through `R, I_(R ) = (E_(V))/(R ) = (E_(V))/(12 xx 10^(8))` current through `C I_(C) = (E_(V))/(X_(C)) = 2 pi fC xx E_(V)` `2 pi xx 80 xx 22 10^(-6) xx E_(V)` `= 352 pi xx 10^(-5) xx E_(V)` `I_(V)^(2) = I_(R )^(2) + I_(C )^(2)` `(10)^(2) (E_(V)^(2))/((12 xx 10^(8))^(2)) + (352 xx 10^(-5) xx E_(V))^(2)` `= E_(V)^(2) ((1)/(144 xx 10^(16)) + 1.2 xx 10^(-4))` `E_(V)^(2) = (100 xx 10^(4))/(1.2) E_(V) ~~ 904` volt |
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| 1250. |
The core of any transformaer is laminated so as toA. energy losses due to Eddy currents may be minimisedB. the weight of the transformer may be reducedC. rusting of the core may be preventedD. ratio of voltage in primary and secondary may be increased |
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Answer» Correct Answer - A When magnetic flux linked with a coil changes, induced emf is produced in it and the induced current flows through the wire forming the coil. These induced currents are set up in the conductor in the from closed loops. These currents look like eddies or whirlpools and likewise are known as Eddy currents. These currents oppose the cause of their origin, therefore, due to Eddy currents, a great amount of energy is wasted in form of heat energy. if core of transformer is laminated, then their effect can be minimised. |
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