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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.
| 1301. |
In the given circuit the readings of the voltmeter `V_(1)` and the ammeter `A` are .A. `220 V, 2.2A`B. `110 V, 1.1A`C. `220 V, 1.1A`D. `110 V, 2.2A` |
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Answer» Correct Answer - A Since `V_(L) = V_(C)` (series resonance) `V_(1) = V= 220 V, Z =R =100 Omega` `I = (V)/(Z) = (220)/(100) =2.2A` . |
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| 1302. |
In the series L-C-R circuit , the voltmeter and ammeter readings are A. `V=100` volt, `I=2` ampB. `V=100` volt, `I=5` ampC. `V=1000` volt, `I=2` ampD. `V=300` volt, `I=1` amp |
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Answer» Correct Answer - A At resonance `(V_(C)=V_(L))` `V=I_(rms)xxR` (here `z=R`) `V=V_(rms)=100` volt & `I_(rms)=100/50=2` Amp. |
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| 1303. |
In a series LCR circuit, at resonance, power factor is …….. .A. zeroB. 1C. `1//2`D. `1//2 ohm` |
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Answer» Correct Answer - B At resonance `x_(L)=x_(C)` So, `z=R rArr cosphi=1` |
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| 1304. |
The power factor of LCR circuit at resonance is-A. 0.707B. 1C. ZeroD. 0.5 |
| Answer» Correct Answer - b | |
| 1305. |
An inductance of 1 mH a condenser of `10 mu F` and a resistance of `50 Omega` are connected in series. The reactances of inductor and condensers are same. The reactance of either of them will beA. `100 Omega`B. `30 Omega`C. `3.2 Omega`D. `10 Omega` |
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Answer» Correct Answer - D According to problem, `X_(L) = X_(C)` `omegaL=1/(omegaC) = 1/(10^(-3) xx 10 xx 10^(-6))` `omega=1/(10^(-8))=10^(4)` `therefore X_(L) = omegaL=10^(4) xx 10^(-3)=10 Omega` |
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| 1306. |
In figure `i_(1) = 10 e^(-2t) A, i_(2) = 4 A, v_(C ) = 3 e^(-2t) V` The variation of potential difference acorss `A` and `C` with time can be represented asA. B. C. D. |
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Answer» Correct Answer - B `V_(C ) - i_(2) R_(2) - i_(L) = V_(D) , V_(CD) = V_(C ) - V_(D) = i_(2) R_(2)+ i_(L)` Substitution the values we have, `V_(CD) = (4) (3) + 16 e^(-2t) , V_(CD) = (12 + 16 e^(-2t)) V` At `t = 0, V_(CD) = 28 V` , And at `t = oo, V_(CD) = 12 V` |
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| 1307. |
A choke coil has.A. low inductance and low resistanceB. high inductance and high resistanceC. low inductance and high resistanceD. high inductance and low resistance |
| Answer» Correct Answer - D | |
| 1308. |
Assertion: Average value of `AC` over a complete cycle is always zero. Reason: Average value of `AC` is always defined over half cycle.A. If both assertion ans reason are true ans reaason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion istrue but reason is false.D. If both assertion and reason are false. |
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Answer» Correct Answer - B The mean or average value of alternating current or e.m.f. during a half cycle is given by `I_(m)=0.636 I_(0)` or `E_(m) = 0.636 E_(0)` During the next half cycle, the mean value of ac will be equal in magnitude but opposite in direction. For this reason the average value of ac over a complete cycle is always zero. So the average value is always defined over a half cycle of ac. |
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| 1309. |
Assertion : The capacitive reactance limits the amplitude of the current in a purely capacitive circuit. Reason : Capacitive reactance is proportional to the frequency and the capacitance.A. If both assertion ans reason are true ans reaason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion istrue but reason is false.D. If both assertion and reason are false. |
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Answer» Correct Answer - C The capacitive reactance limits the amplitude of the current in a purely capacitive circuit in the same ways as the resistance limits the current in a purely resistive circuit. i.e., `I_(0)=(epsilon_(0))/(X_(C ))` But it is inversely proportional to the frequency and the capacitance. |
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| 1310. |
Assertion : The inductive reactance limits amplitude of the current in a purely inductive circuit. Reason : The inductive reactance is independent of the frequency of the current.A. If both assertion ans reason are true ans reaason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion istrue but reason is false.D. If both assertion and reason are false. |
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Answer» Correct Answer - C The inductive reactance limits the amplitude of current in a purely inductive circuit in the same ways as the resistance limits the current in a purely resistive circuit. i.e., `I_(0)=(epsilon_(0))/(X_(L))` The inductive reactance is directly proportional to the inductance and the frequency of the circuit. |
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| 1311. |
Assertion : In series LCR resonance circuit, the impedance is equal to the ohmic resistance. Reason : At resonance, the inductive reactance exceeds the capacitive reactance.A. If both assertion ans reason are true ans reaason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion istrue but reason is false.D. If both assertion and reason are false. |
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Answer» Correct Answer - C In series resonance circuit, inductive reactance is equal to acpacitive reactance. i.e., `omega L = (1)/(omega C) therefore Z = sqrt(R^(2)+(omega L-(1)/(omega C))^(2))=R` |
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| 1312. |
An Lc circuit contains a 40 mH inductor and a `25 mu F` capacitor. The resistance of the circuit is negligible.The time is measured from the instant the circuit is closed. The energy stored in the circuit is completely magnetic at time (in milliseconds)A. `0,3.14,6.28`B. `0,1.57,4.71`C. `1.57,4.71,7.85`D. `1.57,3.14,4.71` |
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Answer» Correct Answer - C Here, `L = 40 mH = 40xx10^(-3)H` `C = 25 mu F = 25xx10^(-6)F, upsilon = (1)/(2pi sqrt(LC))` Substituting the given values, we get `upsilon = (1)/(2pi sqrt(40xx10^(-3)xx25xx10^(-6)))=(10^(3))/(2pi)Hz` `therefore T = (1)/(upsilon)=(2pi)/(10^(3))s=2pixx10^(-3)s=2pi ms` Energy stored is completely electrical at times `t=0, (T)/(2), T,(3)/(2)T`, ....... Energy stored is completely magnetic at times `t=(T)/(4),(3)/(4)T,(5)/(4)T`, ......... Hence, `t=(pi)/(2)ms, (3pi)/(2)ms, (5pi)/(2)ms=1.57ms, 4.71 ms, 7.85 ms` |
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| 1313. |
The instantaneous values of current and voltage in an AC circuit are given by l = 6sin ( 100`pi`t + `pi` / 4) V = 5sin ( 100 `pi`t - `pi` / 4 ), thenA. current leads the voltage by `45^(@)`B. voltage leads the current by `90^(@)`C. current leads the voltage by `90^(@)`D. voltage leads the current by `45^(@) ` |
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Answer» Correct Answer - C The phase difference between instantaneous value of _ and V is `pi`/ 4 - (-`pi`/4) = `pi`/2 Hence current leads the voltage by `90^(@)` |
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| 1314. |
Assertion : An inductor coil normally produces more current with `DC` source compared to an AC source of same value of rms voltage. Reason : In `DC` source, applied voltage remains constant with time.A. If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.B. If both Assertion and Reason are true but Reason is not the correct explanation of AssertionC. If Assertion is true, but the Reason is false.D. If Assertion is false but the Reason is true. |
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Answer» Correct Answer - B `I_(DC)=V_(DC)/r(r="internal resitance of inductor") ` `I_(AC)=V_(AC)/Z=V_(AC)/sqrt(r^2+X_L^2)` If `V_(DC)=V_(AC)`, then `I_(DC)gtI_(AC)`. |
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| 1315. |
An dielectric current has both DC and AC components . DC component of BA and AC component is given as I = `6sin`omega`t. So `underset(rms)(I)` value of resulatant current isA. 8.05 AB. 9.05 AC. 11.58 AD. 13.58 A |
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Answer» Correct Answer - B `underset(rms)(I)` = `sqrt (I)^(2)` = `sqrt (8 + 6 sin omegat)^(2)` `impliesunderset(rms)(I)` = `sqrt (64 + 96(sin omegat) + 36(sin^(2) omegat)` `implies` `underset(rms)(I)` = `sqrt 64 + 0 + 36 xx 0.5` = 9.05 `implies`Since, `(sin^(2)omegat)` = 0.5 |
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| 1316. |
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?A. 4ampB. `4sqrt2amp`C. `(3+4sqrt2)amp`D. 5amp |
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Answer» Correct Answer - D |
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| 1317. |
A series L-C-R circuit is connected across an AC source E = `10sin[100pit - pi/6]`. Current from the supply is I = `2sin[100pit + pi/12]`, What is the average power dissipated? |
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Answer» Phase difference between voltage and current. `phi = pi/12-((-pi)/6)=pi/4` `rArr` Power factor `=cosphi = 1/sqrt(2)` Average power dissipated, `(V_(m)I_(m))/2 cosphi = (10 xx 2)/2 xx 1/sqrt(2) = 5sqrt(2)`W |
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| 1318. |
A circuit containing a 80 mH inductor and a `60 mu F` capacitor in series is connected to a 230 V 50 Hz supply. The resistance of the circuit is negligible. What is the average power transferred to the inductor ? |
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Answer» Given Inductance `L = 80 mH = 80xx10^(-3)H` Capacitance of capacitor `C = 60 mu F = 60xx10^(-6)F` The rms value of voltage `V_("rms")=230 V` Frequency f = 50 Hz , Resistance R = 0 Average power transferred to the inductor `P = I_("rms")V_("rms").cos phi` As the phase difference is `90^(@)`, So P = 0 |
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| 1319. |
A circuit containing a 80 mH inductor and a `60 mu F` capacitor in series is connected to a 230 V 50 Hz supply. The resistance of the circuit is negligible. Obtain the rms values of protential drops across each element. |
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Answer» Given Inductance `L = 80 mH = 80xx10^(-3)H` Capacitance of capacitor `C = 60 mu F = 60xx10^(-6)F` The rms value of voltage `V_("rms")=230 V` Frequency f = 50 Hz , Resistance R = 0 Potential drop across L `V_("rms")L=I_("rms")xx X_(L)=8.29xx2xx3.14xx50xx80xx10^(-3)` `= 208.25 V` Potential drop across C `V_("rms")C=I_("rms")xx X_(C )` `= 8.29xx(1)/(2xx3.14xx50xx60xx10^(-6))` `= 440.02 V` `therefore` Applied rms voltage `= V_(C )-V_(L)` `= 440-208.25=231.75 V` |
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| 1320. |
A circuit containing a 80 mH inductor and a `60 mu F` capacitor in series is connected to a 230 V 50 Hz supply. The resistance of the circuit is negligible. Obtain the current amplitude and rms values. |
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Answer» Given Inductance `L = 80 mH = 80xx10^(-3)H` Capacitance of capacitor `C = 60 mu F = 60xx10^(-6)F` The rms value of voltage `V_("rms")=230 V` Frequency f = 50 Hz , Resistance R = 0 Impedance of circuit `Z = sqrt(R^(2)+(X_(L)-X_(C ))^(2))` `= sqrt(0+(2pi fL-(1)/(2pi fC))^(2))=(2pi fL-(1)/(2pi FC))` `= sqrt((2xx3.14xx50xx80xx10^(-3)-(1)/(2xx3.14xx50xx60xx10^(-6)))^(2))` `= 25.12-53.07=-27.95 Omega` As ZCO, means `X_(L)lt X_(C )`, emf lags by current by `90^(@)` The rms value of current `I_("rms")=(V_("rms"))/(z)=(230)/(27.95)=8.29 A`. The maximum value of current `I_(0)=sqrt(2)I_("rms")=1.414xx8.29=11.64 A` |
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| 1321. |
The maximum values of the alternating voltages and current are `400 V` and `20 A` respectively in a circuit connected to `50 Hz` supply and these quantities are sinusoidal. The instantaneous values of the voltage and current are `200sqrt2V` and `10 A`, respectively. At `t = 0`, both are increasing positively. (a) Write down the expression for voltage and current at time `t`. (b) Determine the power consumed in the circuit. |
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Answer» `V_(0) = 400 V. i_(0) =20A, f = 50 Hz` `omega =2pi f = 100 pi rad//sec, V =283 V, I = 10 A` `V = V_(0) sin (omegat + phi 1)` `283 =400 sin (100 phi t + phi_(1))` At `t = 0` `200sqrt2 =400 sin (0 + phi_1) implies sin phi_(1) = (1)/sqrt2` `phi_(1) (pi)/(4)` `i = i_(0) sin (omega t + phi_(2))` `10 = 20 sin (0 + phi_(2)) implies sin phi_(2) = (1)/(2) implies phi_(2) = (pi)/(2)` `V =400 sin (100 pit + pi/4)` `i =20 sin (100 pit + pi//6)` `P = (1)/(2) V_(0)i_(0) cos phi` where phi phase difference between voltage and current `phi = (pi)/(4) -(pi)/(6) = (pi)/(12) = 15^(@)` `P = (1)/ (2) (400) (20) cos 15^(@) = 4000 cos 15^(@)` ` = 3864 W` Note: `cos 15^(@) = cos (45^(@) cos 30^(@) + sin 45^(@) sin 30^(@)` . |
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| 1322. |
An AC circuit containing 800mH inductor and a `60 muF` capacitor is in series with `15 Omega` resistance. They are connected to 230 V ,50 Hz AC supply. Obtain average power transferred to each element and total power absorbed. |
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Answer» Reactance of the inductor `X_(L) = omegaL = 2pivL= 2xx 3.14 xx 50 xx 800 xx 10^(-3)=251.3Omega` Capacitive reactance, `X_(C) =1/(omegaC)=1/(2pivC) = 1/(2 xx 3.14 xx 50 xx 60 xx 10^(-3)) = 251.3 Omega` `therefore` Impedance, `Z=sqrt(R^(2)+(X_(L)-X_(C))^(2))=sqrt((15)^(2) + (251.3-53.05)^(2))` `rArr Z=198.8Omega` `therefore I_(rms) = V_(rms)/Z = 230/198.8 = 1.157`A Note that the inductance and capacitor do not consume power over a cycle. Now, average power dissipated in the resistor. `I_(rms)^(2).R=(1.157)^(2) xx 15` =20.07 W Total power consumed = `V_(rms).I_(rms).cosphi` `=V_(rms).I_(rms).R/Z` `=(230 xx 1.157 xx 15)/(198.8) = 20.07`W Note that the total power consumed always is same as that consumed in resistor. |
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| 1323. |
A circuit containing a `0.1 H` inductor and a `500 muF` capacitor in series is connected to a `230` volt `100//pi Hz` supply.The resistance of the circuit is negligible. (a)Obtain the current amplitude and `rms` values. (b)Obtain the `rms` values of potential drops across each element ( c)What is the average power transferred to the inductor?(d)What is the average power transferred to the capacitor?(e)What is the total average power absorbed by the circuit ?[Average implies average over one cycle] |
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Answer» Correct Answer - (a)`23sqrt2 A,23A` , (b)`460` volt, `230` volt , (c )zero , (d)zero , (e)zero Given that `L=0.1 H` `C=500xx10^(-6) F` `V_(rms)=230` volt , `f=100/piHz` (a)`I_(rms)=V_(rms)/Z=230/Z` where `Z=omegaL-1/(omegac)=2pi f L-1/(2pi^(f)C)=20-10=10` & `I_(@)=sqrt2xxI_(rms)=23sqrt2` (b)`V_(L)=I_(rms)(omegaL)` ( c )`lt P_(L) gt=I_(rms)V_(rms)cos phi` (d)`lt P_(C) gt=0` (e)`lt P_("net") gt=0` |
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| 1324. |
A series `LCR` circuit with `L=0.125//pi H,C=500//pi nF,R=23Omega` is connected to a `230 V` variable frequency supply. (a)What is the source frequency for which current amplitudes is maximum? Obtain this maximum value. (b)What is the source frequency for which average power absorbed by the circuit is maximum ?Obtain the value of this maximum power.(c )For what reactangle of the circuit, the power transferred to the circuit is half the power at resonance?What is the current amplitude at this reactangle ? (d)if `omega` is the angular frequency at which the power consumed in the circuit is half the power at resonance, write an expression for `omega` (e)What is the `Q`-factor (Quality factor) of the given circuit? |
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Answer» Correct Answer - (a)`2000 Hz,10sqrt2 A` , (b)`2000 Hz, 2300` watt , ( c)`23 Omega , 10 A` , (d)`0.125/pi omega-(1xx10^(9))/(omega(500)/(pi))+-23` , (e)`500//23` Given that `L=0.125//pi , C=500/pixx10` `R=23 , V_(rms)=230` volt (a)`f_(R)=1/(2pi) 1/sqrt(CL)=2000 Hz , I_(rms)(max)=230/R=10` Amp. (b)at Resonance `f_(R)=2000 , P_(max)=I_(rms)^(2) R=(10)xx23=2300` watt ( c)`P_(max)/2=I_(rms)^(2) R , (I_(rms)^(2) R)/ 2 , =I_(rms)^(2) R=I_(rms)=I_(rms)/sqrt2` `V_(rms)/sqrt(R^(2)+x^(2))=I_(rms)/sqrt2=2300/sqrt((23)^(2)+x^(2))=10/sqrt2 x=23 Omega`. `I_(0)=I_(rms)xxsqrt2=10` Amp. (d)`omegaL-1/(omegac)=+23 Omega rArr omega =?` (e)`Q=(omega_(r)L)/R=(2xxpixx2000xx0.125/pi)/23=50/23` |
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| 1325. |
Figure here, shows a series L-C-R circuit connected to a variable frequency 230 V source. L = 5.0H, C = `80 muF` and r =`40 Omega` (a) Determine the source frequency which drives the circuit in resonance. (b) Obtain the impedance of the circuit and the amplitude of current at the resonating frequency. (c) Determine the rms potential drops across the three elements of the circuit. show that the potential drop across the L-C combination is zero at the resonating frequency. |
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Answer» a) If `omega_(0)` be the resonant angular frequency = source frequency at resonance, then `omega_(0) = 1/(sqrt(LC)` = `1/sqrt(5 xx 80 xx 10^(-6))= 50 rads^(-1)` `f_(0) = omega_(0)/(2pi) = (50)/(2 xx 3.141) = 7.96 Hz` b) At resonance, Z=R = `40Omega` `I_(rms) = V_(rms)/Z= 230/40 = 5.75 A` and `I_(0) = V_(0)/Z = (sqrt(2)V_(rms))/Z` `(sqrt(2) xx 230)/(40) = 8.13 A` c) Tjhe rms potential drop across R is given by `V_(rms) = I_(rms) R= 5.75 xx 40 = 230 V` The rms potential drop across L is given by `V_(Lrms) = I_(rms) X_(L) = I_(rms)(omega_(0)L)` `=5.75 xx 50 xx 5 = 1437.5 V` The rms potential drop across C is given by `V_(rms) = I_(rms)X_(C)=I_(rms) xx 1/(omega_(0)C) = 5.75 xx 1/(50 xx 80 xx 10^(-6))` =1437.5 V The rms potential drop across L-C is given by `V_(LC) = V_(L) - V_(C) = 1437.5 - 1437.5=0` |
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| 1326. |
In the given figure , a series L-C-R circuit is connected to a variable frequency source of 230 V . The impedance and amplitude of the current at the resonating frequency will be A. `20 Omega` and 4.2 AB. `30 Omega` and 6.9 AC. `25 Omega` and 5.8 AD. `40 Omega` and 5.75 A |
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Answer» Correct Answer - D At resonance , `underset(L)(X)` = `underset(C )(X)` `therefore` Z = R = 40`Omega` `underset(rms)(I)` = `underset(rms)(V)` / Z = 5.75 A |
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| 1327. |
A `200Omega` resistor is connected to a 220 V, 50 Hz AC supply. Calculate rms value of current in the circuit. Also find phase difference between voltage and the current. |
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Answer» Given, `V_(rms) = 220V`, f = 50Hz, R= `200Omega` Current in the purely resistive AC circuit `I=V/R` `therefore` rms value of current `I_(rms) = (V_(rms)/R) = 220/200 = 1.1`A In purely resistive AC circuit, current and voltage are in phase. Therefore phase difference is zero. |
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| 1328. |
The figure shows variation of R, `underset(L)(X)` and `underset(C )(X)` with frequency f in a series L, C , R circuit . Then , for what frequency point, the circuit is inductive? A. AB. BC. CD. All points |
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Answer» Correct Answer - C At point A , at point B, `X_(C)` = `X_(L)` at point C, `X_(C) lt X_(L)` So at C,circuit is inductive. |
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| 1329. |
A coil of `200Omega` resistance and `1.0H` inductance is conneted to an `AC` source of frequency `200//2pi Hz`. Phase angle between potential and current will beA. `30^(@)`B. `90^(@)`C. `45^(@)`D. `0^(@)` |
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Answer» Correct Answer - C `tan phi=(X_(L))/R=(2pivL)/R` `=(2pixx200/(2pi)xx1)/200=1 implies phi=45^(@)` |
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| 1330. |
A coil of inductance 8.4 mH and resistance `6 (Omega)` is connected to a 12 V battery. The current in the coil is 1.0 A at approximately the timeA. `500 sec`B. `20`secC. `35` milli secD. `1`milli sec |
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Answer» Correct Answer - D Peak current in the circuit `i_(0)=12/6=2A` Current decreases from `2A` to `1A`, i.e. becomes half in time `t=0.693 L/R=0.693xx(8.4xx10^(-3))/6=i millisec` |
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| 1331. |
For the series `LCR` circuit shown in the figure, what is the resonance frequency and the amplitude of the current at the resonating frequency A. `2500 rad s^(-1) and sqrt5A`B. `2500 rad s^(-1) and 5A`C. `2500 rad s^(-1) and (5)/(sqrt2)A`D. `2500 rad s^(-1) and 5sqrt2 A` |
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Answer» Correct Answer - B |
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| 1332. |
For the series `LCR` circuit shown in the figure, what is the resonance frequency and the amplitude of the current at the resonating frequency A. 2500 `rad s^(-1)` and `5`sqrt 2AB. 2500 `rad s^(-1)` and 5 AC. 2500 `rad s^(-1)` and `5/sqrt 2`AD. 25 `rad s^(-1)` and 5`sqrt 2`A |
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Answer» Correct Answer - B Resonance frequency, `omega=1/(sqrt(LC))` `1/sqrt(8xx 106^(-3)xx20xx10^(-6))` `=2500 rads^(-1)` Resonance current, `I=V/R =220/44 = 5A` |
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| 1333. |
For the series `LCR` circuit shown in the figure, what is the resonance frequency and the amplitude of the current at the resonating frequency A. `2500 rad s^(-1)` and `5sqrt(2) A`B. `2500 rad s^(-1)` and `5 A`C. `2500 rad s^(-1)` and `5/(sqrt(2)) A`D. `25 rad s^(-1)` and `5sqrt(2) A` |
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Answer» Correct Answer - B Resonance frequency `omega=1/(sqrt(LC))=1/(sqrt(8xx10^(-3)xx20xx10^(-6)))` `=2500rad//s` Resonance current `=V/R=220/44=5A` |
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| 1334. |
An AC source is connected to a capacitor. The current in the current is I. Now a dielectric slab is inserted into the capacitor , then the new current isA. equal IB. more than IC. less than ID. may be more than or less than I |
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Answer» Correct Answer - B By introducing the slab, C will increase . Therefore , `underset(C )(X)` will decrease or I will increase. |
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| 1335. |
Assertion : Resonance frequency will decrease in `L-C-R` series circuit if a dielectric slab is inserted in between the plates of the capacitor. Reason : By doing so, capacity of capacitor will increase.A. If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.B. If both Assertion and Reason are true but Reason is not the correct explanation of AssertionC. If Assertion is true, but the Reason is false.D. If Assertion is false but the Reason is true. |
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Answer» Correct Answer - A `omega=1/sqrt((LC))` by inserting a slab, `C` will increase so `omega` will decrease. |
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