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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.
1251. |
A 44 mH inductor is connected to 220 V, 50 Hz ac supply. The rms value of the current in the circuit is |
Answer» As As `I_(rms) = V_(rms)/X_(L)` Where, `X_(L) = omegaL=2pifL` is the reactance of the inductor Here, f=50 Hz, L= 44mH = `44 xx 10^(-3)= 13.82Omega` `therefore` rms value of current in the circuit `I_(rms) = 220/13.82 = 15.9A` |
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1252. |
A 44 mH inductor is connected to 220 V, 50 Hz a.c. supply. Determine rms value of current in the circuit. |
Answer» Inductance of inductor, `L=44mH=44xx10^(-3)H` Supply voltage, V=220V Frequency, v=50V Angular frequenc, `omega=2piv` Inductive reactivance, `x_(L)= 2pivL=2pixx50xx44xx10^(-3)Omega` Rms value of current is given as: `=(220)/(2pixx50xx44xx10^(-3))=15.92A` Hence, the rms value of current in the circuit is 15.92A. |
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1253. |
A coil of inductive reactance `31 Omega` has a resistance of `8 omega`. It is placed in series with a condenser of capacitive reactance `25 Omega`. The combination is connected to an `ac` source of `110 V`. The power factor of the circuit isA. `0.56`B. `0.64`C. `0.80`D. `0.33` |
Answer» Correct Answer - C Power factor `(cos phi)` is a ratio fo resistance and impedence of `AC` circuit. Power factor of `AC` circuit is given by `cos phi=R/Z......(i)` Where `R` is resistance employed and `Z` the impedence of the circuit. `Z=sqrt(R^(2)+(X_(L)-X_(C)^(2)))........(ii)` Eqs (i) and (ii) meet to give, `cos phi=R/(sqrt(R^(2)+(X_(L)-X_(C))^(2)))........(iii)` Given, `R=8 Omega, X_(L)=21Omega, X_(C)=25 Omega` `cos phi=8/(sqrt((8)^(2)+(31-25)^(2)))=8/(sqrt(64+36))` `cos phi=0.80` |
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1254. |
A coil of inductive reactance `31 Omega` has a resistance of `8 ohm`. It is placed in series with a condenser of capacitive reactance `25 Omega`. The combination is connected to an `ac` source of `110 V`. The power factor of the circuit isA. `0.56`B. `0.64`C. `0.80`D. `0.33` |
Answer» Correct Answer - C |
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1255. |
Power loss in `AC` circuit will be minimum whenA. Inductance is high, resistance is highB. Inductance is low, resistance is highC. Inductance is low, resistance is lowD. Inductance is high, resistance is low |
Answer» Correct Answer - D `l = (V)/(sqrt(R^(2) + (omega L)^(2)))` power loss `= l^(2) R = (V^(2) R)/(R^(2) + (omega L)^(2)) = (V^(2))/(R + (omega^(2) L^(2))/(R ))` For minimum power loss, resistance should be low and inductance should be high. |
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1256. |
A `750 Hz, 20 V` source is connected to as resistance of `100 Omega` an inductance of `0.1803 H` and a capacitance of `10 muF` all in sereis.Calculate the time in which the resistance (thermalcapacity `2J//.^(@)C`) wil get heated by `10^(@)C`. |
Answer» Here, `v=750 Hz, E_(v)=20V, R=100 Omega` `L=0.1803 H, C=10 muF=10^(-5) F, t=?` `Delta theta=10^(@)C`, thermal capacity `=2J//.^(@)C` `X_(L)= omegaL=2pi vL=2xx3.14xx750xx0.1803` `X_(C)=1/(omega C)=(1)/(2pi xx750xx10^(-5))=21.2 Omega` `Z=sqrt(R^(2)+(X_(L)-X_(C))^(2))` `=sqrt(100^(2)+(850-21.2)^(2))=835 Omega` Power dissipated `=E_(v)I_(v) cos phi` `=E_(v) (E_(v)/Z)(R/Z)=(20^(2)xx100)/((835)^(2))=0.0574 W` Heat produced in resistance `=2xx10=20 J` If t is the required time, then `Pxxt=20 rArr t=20/P=20/0.0574=348 s` |
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1257. |
A sinusoidal voltage of peak value 283 V and frequency 50 Hz is applied to a series LCR circuit in which `R = 3 Omega. L = 25.48mH`. And `C = 796mu F`. The power factor. |
Answer» Power factor `= cos phi = cos 53.1^(@)=0.6` . | |
1258. |
A sinusoidal voltage of peak value 283 V and frequency 50 Hz is applied to a series LCR circuit in which `R = 3 Omega. L = 25.48mH`. And `C = 796mu F`. The phase difference between the voltage across the source and the current |
Answer» Phase difference, `phi=tan^(-1)(X_(C )-X_(L))/(R )` `= tan^(-1)((4-8)/(3))=-53.1^(@)` Since `phi` is negative, the current in the circuit lags the voltage across the source. |
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1259. |
A sinusoidal voltage of peak value 283 V and frequency 50 Hz is applied to a series LCR circuit in which `R = 3 Omega. L = 25.48mH`. And `C = 796mu F`. The power dissipated in the circuit |
Answer» The power dissipated in the circuit is `P = I^(2)R` but `I=(i_(m))/(sqrt(2))` and `i_(m)=(upsilon_(m))/(Z)` hence `I=(i_(m))/(sqrt(2))=(upsilon_(m))/(sqrt(2)Z)` given `upsilon m = 283 V` `I = (1)/(sqrt(2))(283)/(5)=40A` Therefore, `P=(40A)^(2)xx3Omega =4800W` `= 4.8 kW`. |
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1260. |
A sinusoidal voltage of peak value 283 V and frequency 50 Hz is applied to a series LCR circuit in which `R = 3 Omega. L = 25.48mH`. And `C = 796mu F`. The impedance of the circuit |
Answer» To find the impedance of the circuit, we first calculate `X_(L)` and `X_(C )`. `X_(L)=2pi vL` `= 2xx3.14xx50xx25.48xx10^(-3)Omega=8Omega` `X_(C )=(1)/(2pivC)` `= (1)/(2xx3.14xx50xx796xx10^(-6))=4Omega` Therefore, `z=sqrt(R^(2)+(X_(L)-X_(C ))^(2))=sqrt(3^(2)+(8-4)^(2))` |
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1261. |
A series combination of an inductor of self-inductance L, capacitor of capacitance C and resistor R is connected to an alternating voltage source of `V=V_0sinomegat`. The current through the circuit is `I=I_0sin(omegat-theta)`,where `I_0=V_0/(sqrt(R^2+(omega-1/(omegaC))^2))`and`theta=tan^-1 ""1/R(omegaL-1/(omegaC))`. Now that, the frequency of both voltage and current is `f=omega/(2pi)` . The rms value of these parameters during one complete cycles are `V_(rms)=V_0/sqrt2`and`I_(rms)=I_0/sqrt2` respectively. These values are shown in alternating voltmeter and ammeter. The power consumed by the circuit P=VI. The mean value i.e., the effective power of the circuit in a complete cycle is `overlineP=V_(rms)I_(rms)costheta`. This `costheta` is termed the power factor. The power factor of the circuit in question (iii) isA. -1B. zeroC. between zero and 1D. 1 |
Answer» Correct Answer - C | |
1262. |
In the circuit shown in the figure , the alternating currents through inductor and capacitor are 1.2 and 1.0 A respectively . The current drawn from the generator is A. 0.4 AB. 0.2 AC. 1.0 AD. 1.2 A |
Answer» Correct Answer - B | |
1263. |
In the circuit shown in figureure the `AC` source gives a voltage `V=20cos(2000t)`. Neglecting source resistance, the voltmeter and and ammeter readings will be .A. `0 V, 0.47 A`B. `1.68 V, 0.47 A`C. `0 V, 1.4 A`D. `5.6 V, 1.4 A` |
Answer» Correct Answer - D Ammeter reading, `i = (E_("rms"))/(sqrt(R^(2) + (omega L - (1)/(omega C))^(2)))` and `R = 4 + 6 = 10 Omega` voltmeter reading, `V = i sqrt(R_(1)^(2) + (omega L - (1)/(omega C))^(2))` and `R_(1) = 4 Omega` |
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1264. |
In the circuit shown in figureure the `AC` source gives a voltage `V=20cos(2000t)`. Neglecting source resistance, the voltmeter and and ammeter readings will be .A. 0 V, 0.47 AB. 1.68 V, 0.47 AC. 0 V, 1.4 AD. 5.6 V, 1.4 A |
Answer» Correct Answer - C::D Z = `sqrt ((R)^(2) + (X_L - X_C)^(2))` R = `10Omega`, `X_L` = `omega`L = `2000 xx 5 xx 10^(-3)` = `10Omega` `X_C` = `1/omega`C = `1 / 2000 xx 50 xx 10^(-6)` = 10` Omega` i.e., Z = 10`Omega` Maximum current, `i_o` = `V_o`/Z = 20 / 10 = 2A Hence, `i_rms` = 2/`sqrt 2`= 1.4A and `V_rms` = `4 xx 141` = 5.64V |
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1265. |
In the circuit shown below, what will be the reading of the voltmeter and ammeter? A. 200 V , 1 AB. 800 V and 2 AC. 220 V , 2 AD. 220 V , 2.2 A |
Answer» Correct Answer - D Taking , `V^(2)` = `` + `(underset(L)(V) - underset(C )(V))^(2))` According to the figure , we conclude that `underset(L)(V)` = `underset(C )(V)` `therefore` `underset(R )(V)` = V = 220 V Reading of voltmeter = 220 V Reading of ammeter , I = 220/100 = 2.2 A |
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1266. |
The following series L-C-R circuit , when driven by an emf source of angular frequency 70 kilo-radians per second , the circuit effectively behaves like A. purely resistive circuitB. series R-L circuitC. series R-C circuitD. series L-C circuit with R = 0 |
Answer» Correct Answer - B Resonance frequency, `underset(o)(omega)= 1/sqrt LC= 1/sqrt 100 xx 10^(-6) xx 1 xx 10^(-6) = 10^(5) rads^(-1)` Now, given `omega` = 70 kilo-`rads^(-1)` = `70000 rads^(-1)` `impliesunderset(o)(omega)gtomega` As `underset(L)(X ) propto omega` and `underset(C )(X)propto1/omega` `therefore` `underset(L)(X) gt underset(C )(X)` The circuit will be R-L circuit. |
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1267. |
The diagram shows a capacitor `C` and a resistor `R` connected in series to an `AC` source. `V_(1)` and `V_(2)` are voltmeters and `A` is ammeter Now, consider the following statemensts : (I) Reading in `A` and `V_(2)` are always in phase. (II) Reading in `V_(1)` is ahead in phase with reading in `V_(2)`, (III) Reading in `A` and `V_(1)` are always in phase. Which of these statements are/is correctA. `I` onlyB. `II` onlyC. `I` and `II` onlyD. `II` and `III` only |
Answer» Correct Answer - A In `RC` series circuit voltage across the capacitance leads the voltage across the resistance by `(pi)/2` |
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1268. |
In the circuit shown in figureure the `AC` source gives a voltage `V=20cos(2000t)`. Neglecting source resistance, the voltmeter and and ammeter readings will be .A. `0V, 0.47 A`B. `1.68V,0.47 A`C. `0V,1.4A`D. `5.6 V, 1.4 A` |
Answer» Correct Answer - D `Z=sqrt((R)^(2)+(X_(L)-X_(C))^(2))`, `R=10 Omega, X_(L)=omegaL=2000xx5xx10^(-3)=10 Omega` `X_(C)=1/(omegaC)=1/(2000xx50xx10^(-6))=10 Omega i.e., Z=10 Omega` Maximum current `i_(0)=(V_(0))/Z=20/10=2A` Hence `i_(rms)=2/(sqrt(2))=1.4A` and `V_(rms)=4xx1.41=5.46 V` |
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1269. |
In the circuit shown in figure neglecting source resistance the voltmeter and ammeter reading will respectively, will be A. `0V,3A`B. `150V,3A`C. `150 V,6A`D. `0V, 8A` |
Answer» Correct Answer - D The voltage `V_(L)` and `V_(C)` are equal and opposites so voltmeter reading will be zero. Also `R=30 Omega, X_(L)=X_(C)=25 Omega` So, `i=V/(sqrt(R^(2)+(X_(L)-X_(C))^(2)))=V/R=240/30=8A` |
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1270. |
In the circuit shown in figure neglecting source resistance the voltmeter and ammeter reading will respectively, will be A. 0 V , 3 AB. 150 V , 3 AC. 150 V, 6 AD. 0 V, 8 A |
Answer» Correct Answer - D The voltages `underset(L)(V)` and `underset(C )(V)` are equal and opposite so, voltmeter reading will be zero. Also, R = `30Omega` , `underset(L)(X)` = `underset(C )(X)` = `25Omega` So, `I = V/sqrt R^(2) + (underset(L)(X) - underset(C )(X))^(2))` `V/R = 240/30 = 8 A` |
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1271. |
In the circuit shown in fig. if both the bulbs `(B_1) and (B_2)` are identical A. (A) their brightness will be the sameB. (B) `(B_2)` will be brighter than `(B_1)`C. (C) `(B_1)` will be brighter than `(B_2)`D. (D) only `(B_2)` will glow because the capacitor has infinite impedance |
Answer» Correct Answer - B `(X_C)=(1)/(2 pi fC) =(10^6)/(2 pi 50 xx 500) =(20)/(11) Omega` `(X_L) = 2 pi f L =2 pi xx 50 xx 10 xx10 ^(-3) = pi Omega` Since `(X_L) lt (X_C)`, so inductive brach has less impeedance, and so more current. Hence `(B_2)` will be brighter. |
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1272. |
In the adjoining `AC` circuit the voltmeter whose reading will be zero at resonance is A. `V_(1)`B. `V_(2)`C. `V_(3)`D. `V_(4)` |
Answer» Correct Answer - d | |
1273. |
In the circuit shown in figure neglecting source resistance the voltmeter and ammeter reading will respectively, will be A. 0 V, 3AB. 150 V, 3AC. 150 V, 6AD. 0 V, 8A |
Answer» Correct Answer - d | |
1274. |
In the circuit given below, what will be the reading of the voltmeter? A. 300 VB. 900 VC. 200 VD. 400 V |
Answer» Correct Answer - c | |
1275. |
In the circuit shown below, what will be the reading of the voltmeter and ammeter? A. 800 V, 2AB. 300 V, 2AC. 220 V, 2.2 AD. 100 V, 2A |
Answer» Correct Answer - c | |
1276. |
In the circuit shown in fig. if both the bulbs `(B_1) and (B_2)` are identical A. their brightness will be the sameB. `B_(2)` will be brighter than `B_(1)`C. As frequency and that of `B_(2)` will decreasesD. Only `B_(2)` will glow because the capacitor has infinite impedence |
Answer» Correct Answer - B Let `I_(1)` and `I_(2)` be currents through `B_(1)` and `B_(2)` then `I_(1)xxsqrt(R^(2)xxX_(L)^(2))=220` Let `I_(1)` and `I_(2)` be currents through `B_(1)` and `B_(2)` then `I_(1)xxsqrt(R^(2)xxX_(L)^(2))=220` `(I_(1))/(I_(2))=(sqrt(R^(2)xxX_(C)^(2)))/(sqrt(R^(2)xxX_(L)^(2)))=(sqrt(R^(2)+(1/(c omega))^(2)))/(sqrt(R^(2)+L^(2)omega^(2)))` `=(sqrt(R^(2_)+(1/(500xx10^(-6)xx2pixx50))^(2)))/(sqrt(R^(2)+(10xx10^(-3)xx2pixx50)^(2)))` `=sqrt(R^(2)+40)//sqrt(R^(2)+9.87), I_(2)gtI_(1)` Bulb `B_(2)` will be brighter. As frequency increases `X_(C)` decreases `X_(L)` increases. `I_(2)` becomes less `I_(1)` increases. So, brightness of `B_(1)` will increases and that of `B_(2)` decreases. |
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1277. |
In the circuit given below, what will be the reading of the voltmeter? A. `300 V`B. `900 V`C. `200 V`D. `400 V` |
Answer» Correct Answer - C `V^(2)=V_(R)^(2)+(V_(L)-V_(C))^(2)` Since `V_(L)=V_(C)` hence `V=V_(R)=200V` |
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1278. |
In a region of uniform magnetic inductance `B=10^(-2)`tesla. A circular coil of radius `30 cm` and resistance `pi^(2) ohm` is rotated about an axis which is perpendicular to the direction of `B` and which forms a dimater of the coil. If the coil rotates at `200` r.p.m the amplitude of the alternatic current induced in the coil isA. `4pi^(2)mA`B. `30 mA`C. `6 mA`D. `200 mA` |
Answer» Correct Answer - C Amplication of `ac=i_(0)=(V_(0))/R=(omegaNBA)/R=((2piv)NB(pir^(2)))/R` `implies i_(0)=(2pixx200/60xx1xx10^(-2)xxpixx(0.3)^(2))/(pi^(2))=6mA` |
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1279. |
In a region of uniform magnetic induction `B=10^(2)` tesla, a circular coil of radius `30 cm` and resistance `pi^(2)` ohm is rotated about an axis which is perpendicular to the directon of `B` and which form a diameter of the coil. If the coil rotates at `"200rpm"` the amplitude of the alternating current induced in the coil is |
Answer» Correct Answer - 6 `i_(0)=(NBAomega)/(R )` |
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1280. |
A power transformer is used to step up an alternating e.m.f. of `220 V` to `11kv` to trannsmit `4.4 kW` of power. If the primary coil has `1000` turns, what is the current rating of the secondary?(Assume `100%` efficiency for the transformer)A. 4AB. 0.4AC. 0.04AD. 0.2A |
Answer» Correct Answer - C `I_(s)=Ps//Vs Rightarrow "Is"=(4.4xx10^(3))/(11xx10^(3))=0.4A` |
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1281. |
An inductor coil is connected to a 12V battery and drawing a current 24 A. This coil is connected to capacitor and an AC source of rms voltage rating 24 V in the series connection. The rms current through the circuit would found to beA. 48 AB. 36 AC. 0D. 24 A |
Answer» Correct Answer - A The resistance of inductor(coil), `R=12/24 = 0.5Omega` If phase difference between current and emf is zero, then reactance of circuit will must be zero. The impedance of the circuit equals to resistance i.e, `Z=0.5Omega` Now, rms current, `I_(rms) =V_(rms)/Z = 24/0.5 =48 A` |
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1282. |
A series LCR circuit with `L= 0.12H, C=480 nF,` and R=23 Omega` is connected to a 230-V variable frequency supply. (a) What is the source frequency for which current amplitude is maximum? Find this maximum value. (b) What is the source frequency for which average power absorbed by the circuit is maximum? Obtain the value of maximum power. (c ) For which frequencies of the source is the power transferred to the circuit half the power at resonant frequency? (d) What is the Q-factor of the circuit? |
Answer» (a) Current amplitude is maximum at resonace and source frequency at resonance is given by `f_(0)=(1)/(2 pi sqrt(LC))=(1)/(2 pi sqrt(0.12xx480xx10^(-9))) = 663.14 Hz` and `omega_(0) 2pi f_(0)=4166.66 rad s^(-1)` The maximum current amplitude is given by `I_(0(max))=(E_0)/(R ) =(230 sqrt(2))/(23)= 14.14A` (b) Average power absorbaed by the circuit is maximum at resonance for wich frequency is calculate above as 663.14 Hz. Maximum power absorbed is given by `P_(max)=(E_(v)^(2))/(R ) =((230)^(2))/(23) =2300W` (c ) `P=E_(v) I_(v) cos phi =E_(v) I_(v) (R )/(Z) =(E_(v)^(2))/(Z^2) R and P_(max)=(E_(v)^(2))/(R )` Given `P=P_(max)//2` `implies (E_(v)^(2)R)/(Z^2)=1/2 (E_(v)^(2))/(R ) implies Z^(2)=2 R^(2)` `implies R^(2) + (X_(C )-X_(L))^(2)=2R^(2)` `implies X_(C )-X_(L) = +_ R` Solve to get `omega_(1)=(-RC+sqrt(R^(2)C^(2)+4LC))/(2LC)=4263.6 rad s^(-1)` (d) Q-factor is given by `Q=1/R sqrt((L)/(C )) =1/23 sqrt(0.12)(480xx10^(-9))=21.74`. |
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1283. |
A series LCR circuit with `L = 0.12 H, C = 480 nF, R = 23 Omega` is connected to a 230 V variable frequency supply. What is the source frequency for which current amplitude is maximum. Obtain this maximum vlaue. |
Answer» Inductance `L = 0.12 H`, Capacitance `C = 480 nF = 480xx10^(-9)F` Resistance `R = 23 Omega` The rms value of voltage Vrms = 230 V Current is maimum at resonance. At resonance impedance `Z = R = 23 Omega` The rms value of current `I_("rms")=(V_("rms"))/(Z)=(230)/(23)=10A`. The maximum value of current `I_(0)=sqrt(2)` `I_("rms")=1.414xx10=14.14A`. At natural frequency, the current amplitude is maximum. `omega = (1)/(sqrt(LC))(1)/(2pi sqrt(0.12xx480xx10^(-9)))=4166.6` = 4167 rad/s. |
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1284. |
A transformer with 20 turns in its secondary coil is used to step down the input 220 V ac emf. The output of the transformer is fed to a rectifier circuit which convents the ac input into dc output. The input and output of rectifier are as shown in Figure (there is no change in peak voltage). The rectifier output has an average emf of 8.98 volt. Calculate the number of turns in the primary coil of the transformer. |
Answer» Correct Answer - 440 |
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1285. |
An inductor-coil , a capacitor and an AC source of rms voltage `24 V` are connected in series. When the frequency of the source is varied, a maximum rms current of `6.0 A` is observed. If this inductor coil is connected to a battery of `emf 12 V` and internal resistance `4.0 Omega`, what will be the current? |
Answer» Correct Answer - `1.5 A` `R=24/6=4 Omega , I=12/4+4=1.5` Amp. |
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1286. |
A coil a capacitor and an `AC` source of rms voltage `24 V` are connected in series. By varying the frequency of the source, a maximum rms current of 6 A is observed. If coil is connected is at DC batteryof emf 12 volt and internal resistance `4Omega`, then current through it in steady state isA. 2.4 AB. 1.8 AC. 1.5 AD. 1.2 A |
Answer» Correct Answer - C Let R be the resistace of coil + capacitor . Then, R = 24/6 = 4`Omega` In this second case , `implies` `underset(rms)(I)` = 12 / 4 + 4 = 12/8 = 1.5 A |
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1287. |
A pure resistive circuit element X when connected to an ac supply of peak voltage 400 V gives a peak current of 5 A which is in phase with the voltage. A second circuit element Y, when connected to the same ac supply also gives the same value of peak current but the current lags behind by `90^(@)`. If the series combination of X and Y is connected to the same suply, what will be the rms value of current?A. `10/sqrt 2` AB. `5/sqrt 2` AC. 5/2 AD. 5A |
Answer» Correct Answer - C R = `underset(L)(X)` = 200 / 5 = `40 Omega` `therefore` Z = `` = `40sqrt 2 Omega` `underset(rms)(I)` = `underset(rms)(V)`/Z = `200/sqrt 2` / 40`sqrt 2` = 5/2 A |
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1288. |
In an LCR circuit voltages across R,L and C are 10V,10V and 20V respectively. Voltage between the two end points of the whole combination isA. 30VB. `10sqrt3V`C. 20VD. `10sqrt2V` |
Answer» Correct Answer - D | |
1289. |
In an `LCR` series circuit the rms voltages across `R,L` and `C` are foundd to be `10 V, 10 V` and `20 V` respectively. The rms voltage across the entire combination isA. `30 V`B. `1 V`C. `20 V`D. `10 sqrt(2) V` |
Answer» Correct Answer - 4 `V = sqrt(V_(R )^(2) + (V_(L) - V_(C ))^(2))` |
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1290. |
An electric bulb is designed to consume `55 W` when operated at 110 volts. It is connected to a `220 V, 50 Hz` line through a choke coil in series. What should be the inductance of the coil for which the bulb gets correct voltage? |
Answer» Correct Answer - `(2.2sqrt3)/pi=1.3 H H=(7sqrt3)/10 H` | |
1291. |
A series LCR circuit containing a resistance of `120Omega` has angular resonance frequency `4times10^5rad//s` .At resonance, the voltages across resistance and inductancr are 60V and 40V, respectively. Find the values of L and C. At what frequency , does not current in the circuit lag behind the voltage by `45^@`? |
Answer» At resonance, `X_L=X_C` `thereforeI=V_R/R` `because`voltage acorrs the resistance, `V_R=60V`) =0.5A Now, voltage across the indicator, `V_L=IX_L=IomegaL` or,`L=V_L/(Iomega)=40/(0.5times4times10^5)` [`because` angular frequency, `omega=4times10^5rad//s`] `=2times10^-4H` We know that at resonance, `X_L=X_C`or,`omegaL=1/(omegaC)` or,`C=1/(omega^2L)=1/((4times10^5)^2times0.2times10^-3)=3.125times10^-8F` Let the angular frequency be `omega_1` when the current lags behind the voltage by `45^@`. `thereforetan45^@=(omega_1L-1/(omega_1C))/R` `[C=3.125times10^-8F=1/32times10^6F]` or,`1times120=omega_1times2times10^-4-1/(omega_1(1/32)times10^-6)` or,`omega_1^2-6times10^5omega_1-16times10^10=0` The physically meaningful solution of the above equation is, `omega_1=(6times10^5times10times10^5)/2=8times10^5rad//s` |
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1292. |
In series LCR circuit, the voltages across R, L and C are shown in the figure. The voltage of applied source A. 110 VB. 10 voltC. 50 voltD. 70 volt |
Answer» Correct Answer - C `v=sqrt(V_(R)^(2)+(V_(L)-V_(C))^(2))` `v=sqrt(40^(2)+(50-20)^(2))=50` |
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1293. |
In a `LCR` circuit the `P.D` between the terminals of the inductance is `60V`, between the terminals of the capacitor is `30V` and that between the terminals of resistance is `40V`. The supply voltage will be equal to......A. 50 VB. 70 VC. 130 VD. 10 V |
Answer» Correct Answer - a | |
1294. |
In a `LCR` circuit the `P.D` between the terminals of the inductance is `60V`, between the terminals of the capacitor is `30V` and that between the terminals of resistance is `40V`. The supply voltage will be equal to......A. `50V`B. `70V`C. `130V`D. `10V` |
Answer» Correct Answer - A `V=sqrt(V_(R)^(2)+(V_(L)-V_(C))^(2))=sqrt((40)^(2)+(60-30)^(2))=50 V` |
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1295. |
A series `LCR` circuit containing a resistance of `120 Omega` has angular resonance frequency `4 xx 10^(5) "rads"^(-1)`. At resonance the voltages across resistance and inductance are `60 V` and `40 V` respectively. (a) The value of `L` and `C` are `0.2` mH, `1/32 muF` (b) If angular frequency is changed to `8xx10^(5) "rad"//s`, the current lags the voltage by `45^(@)` (c) If angular frequency is charged to `6xx10^(5) "rad"//s`, the current lags the voltage by `45^(@)`A. (a),(c ) are correctB. (a),(b) are correctC. (a),(b) (c ) are correctD. (a),(b),(c ) are wrong |
Answer» Correct Answer - B `i = (V)/(R ), V_(L)` and `iomega_(0) L` `V_(c ) = (i)/(omega_(0)C)` and `Tan 45^(@) = ((omega L + (1)/(omega C)))/(R )` |
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1296. |
An AC circuit consists of a resistance and a choke coil in series . The resistance is of 220 `Omega` and choke coils is of 0.7 H . The power abosorbed from 220 V and 50 Hz , source connected with the circuit , isA. 55 WB. 110 WC. 220 WD. 440 W |
Answer» Correct Answer - B `X_L = (2pifL) = 2pi xx 50 xx 0.7 = 220Omega` `therefore Z = sqrt (R^(2) + X_L^(2) = 220sqrt 2 Omega` `I_rms = V_rms/Z = 220 / 220sqrt 2 = 1/sqrt 2 A` `P = I_rms^(2).R = 1/2 xx 220` = 110 W |
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1297. |
Current in resistance is 1 A , then A. `underset(s)(V)` = 5 VB. impedance of network is `5 Omega`C. power factor of given circuit is (0.6) lagging (current is lagging )D. All the above |
Answer» Correct Answer - D `V_S^(2)` = `(3)^(2)` + `(8 - 4)^(2)` , `V_S` = 5 v Now, Z = `V_S`/t = 5/t = 5 `Omega` Also, `V_R` = IR or R = 3/1 = `3 Omega` So, PF = R/Z = 0.6 as `V_L gt V_C` `implies` I lags V, so this a lagging nature. |
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1298. |
In an `LCR` series circuit the voltages across `R,L` and `C` at resonance are `40 V and `60V` respectively the applied voltage is .A. `60V`B. `40V`C. `160V`D. `sqrt((40)^(2) + (120)^(2) V)` |
Answer» Correct Answer - B | |
1299. |
In figure which voltmeter reads zero when `omega` is equal to the resonant frequency of series `LCR` circuit .A. `V_(1)`B. `V_(2)`C. `V_(3)`D. none |
Answer» Correct Answer - B `V_(L) =V_(C)` and hence `V_(2) =0` . |
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1300. |
In the series L-C-R circuit , the voltmeter and ammeter readings are respectively A. V = 200 V , I = 4 AB. V = 150 , I = 2 AC. V = 100 , I = 5 AD. V = 100 V, I = 2 A |
Answer» Correct Answer - D `V_L` = `V_C` `implies` `X_L` = `X_C` or circuit shown is in resonance . `therefore` `V_applied` = `V_R` = 100 V I = `V_R` . R = 2A |
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