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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.
| 251. |
The graphs given below depict the dependence of two reactive impedences `X_(1)` and `X_(2)` on the frequency of the alternating e.m.f. applied individually to them. We can then say that A. `X_(1)` is an inductor and `X_(2)` is a capacitorB. `X_(1)` is an resistor and `X_(2)` is a capacitorC. `X_(1)` is an capacitor and `X_(2)` is a inductorD. `X_(1)` is an inductor and `X_(2)` is a resistor |
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Answer» Correct Answer - C |
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| 252. |
In an `LCR` series ac circuit the voltage across `L,C` and `R` are `V_(1), V_(2)` and `V_(3)` respectively The voltage of the source is .A. equal to `V_(1)+V_(2)+V_(3)`B. equal to `V_(1)-V_(2)+V_(3)`C. more than `V_(1)+V_(2)+V_(3)`D. none of these is ture |
| Answer» Correct Answer - D | |
| 253. |
In an `LCR` series ac circuit the voltage across `L,C` and `R` are `V_(1), V_(2)` and `V_(3)` respectively The voltage of the source is .A. `V_(1) + V_(2) + V_(3)`B. `sqrt(V_(1)^(2) + (V_(2) + V_(3))^(2))`C. `V_(1) - V_(2) - V_(3)`D. `sqrt(V_(1)^(2) - (V_(2) - V_(3))^(2))` |
| Answer» Correct Answer - 4 | |
| 254. |
An alternating voltage `E=200sqrt(2)sin(100t)` is connected to a `1` microfarad capacitor through an AC ammeter. The reading of the ammeter shall beA. `10mA`B. `20mA`C. `40mA`D. `80mA` |
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Answer» Correct Answer - B Reducing of ammeter `=i_(rms)=(V_(rms))/(X_(C))=(V_(0)omegaC)/(sqrt(2))` `=(200sqrt(2)xx100xx(1xx10^(-6)))/(sqrt(2))=2xx10^(-2)A=20mA` |
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| 255. |
In an `LCR` series ac circuit the voltage across `L,C` and `R` are `V_(1), V_(2)` and `V_(3)` respectively The voltage of the source is .A. `sqrt((V_(1) -V_(2))^(2) + V_(3)^(2))`B. `sqrt(V_(1)^(2) + (V_(2) -V_(3)^(2)`C. `sqrt(V_(2)^(2) + (V_(1) -V_(3))^(2))`D. `V_(1) + V_(2) +V_(3)` |
| Answer» Correct Answer - A | |
| 256. |
Two sinusoidal voltage of the same frequency are shown in the diagram. What is the frequency, and the phase relationship between the voltage? Frequency in `Hz` phase lead of `N` over `M` in radiusA. `{:(,"Frequency in Hz","Phase lead of N over M in radian"),((a),0.4,-pi//4):}`B. `{:(,"Frequency in Hz","Phase lead of N over M in radian"), ((b),2.5,-pi//2):}`C. `{:(,"Frequency in Hz","Phase lead of N over M in radian"), ((c ),2.5,+pi//2):}`D. `{:(,"Frequency in Hz","Phase lead of N over M in radian"),((d),2.5,-pi//4):}` |
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Answer» Correct Answer - B |
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| 257. |
Two sinusoidal voltage of the same frequency are shown in the diagram. What is the frequency, and the phase relationship between the voltage? Frequency in `Hz` phase lead of `N` over `M` in radiusA. Frequency/Hz =0.4, phase lead of N over M in rad `s^(-1)`= `-(pi)/(4)`B. Frequency/Hz =2.5, phase lead of N over M in rad `s^(-1)`= `-(pi)/(2)`C. Frequency/Hz =2.5, phase lead of N over M in rad `s^(-1)`= `2(pi)/(2)`D. Frequency/Hz =2.5, phase lead of N over M in rad `s^(-1)`= `-(pi)/(4)` |
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Answer» Correct Answer - B the period of sinusoidal voltage , T=0.4s. `f=1/T=1/(0.4)=2.5Hz` N lags M by 0.1 s which is equivalent to `((0.1)/(0.4)) 2 pi or (pi)/(2) rad`. Thus, the lead of N over M is `-(pi)/(2) rad`. |
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| 258. |
Two sinusoidal voltage of the same frequency are shown in the diagram. What is the frequency, and the phase relationship between the voltage? Frequency in `Hz` phase lead of `N` over `M` in radiusA. `{:("Frequency m Hz","Phase lead of N over M in radians"),(0.4,-pi//4):}`B. `{:("Frequency m Hz","Phase lead of N over M in radians"),(2.5,-pi//2):}`C. `{:("Frequency m Hz","Phase lead of N over M in radians"),(2.5,+pi//2):}`D. `{:("Frequency m Hz","Phase lead of N over M in radians"),(2.5,-pi//4):}` |
| Answer» Correct Answer - b | |
| 259. |
When an ac source of emf`e=E_(0) sin (100 t)` is connected across a circuit, the phase difference between emf e and currnet I in the circuit is observed to be `(pi)//(4)` as shown in fig. If the circuit consists possibly only of R-C or R-C of L-R series, find the relationship find the relationship between the two elements. A. `R = 1 kOmega, C = 10mF`B. `R = 1 kOmega, C = 1mF`C. `R = 1 kOmega, L = 10mF`D. `R = 1 kOmega, L = 1H` |
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Answer» Correct Answer - A `e = E_(0) sin (100 t) =E_(0) sin omega t` `omega = 100 rad//sec` `phi = (pi)/(4)` is possible in `RC` not in `LC` `tan phi = X_(C)/(R) implies (tan) ((pi)/(4)) = (1 //omegaC)/(R)` `RC = (1)/(omega) = (1)/(100)` . |
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| 260. |
In a series `LCR` circuit the frequency of a `10 V`, `AC` voltage soure is adjusted in such a fashion that the reactance of the inductor meausers `15 Omega` and that of the capacitor `11 Omega`. If `R = 3 Omega`, the potentail difference across the series combination of `L` and `C`will be:A. `8 V`B. `10 V`C. `22 V`D. `52 V` |
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Answer» Correct Answer - A `i = (10)/(sqrt(3^(2) + (15 - 11)^(2))) = 2 A` `:. V_(L) - V_(C ) = i (X_(L) - X_(C )) = 2 (15 - 11) = 8 V` |
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| 261. |
When an ac source of emf`e=E_(0) sin (100 t)` is connected across a circuit, the phase difference between emf e and currnet I in the circuit is observed to be `(pi)//(4)` as shown in fig. If the circuit consists possibly only of R-C or R-C of L-R series, find the relationship find the relationship between the two elements. A. `R=1 kOmega, C=10 muF`B. `R=1 kOmega, C=1 mu F`C. `R=1 kOmega, L=10 H`D. `R=1 kOmega, L=1H` |
| Answer» Correct Answer - a | |
| 262. |
When an ac source of emf`e=E_(0) sin (100 t)` is connected across a circuit, the phase difference between emf e and currnet I in the circuit is observed to be `(pi)//(4)` as shown in fig. If the circuit consists possibly only of R-C or R-C of L-R series, find the relationship find the relationship between the two elements. A. `R= 1kOmega`, C=`10 muF`B. R=`1kOmega`, `C=1muF`C. `R=1kOmega`, L=10HD. `R=1Omega` L=1H |
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Answer» Correct Answer - C c) EmF and AC source `E=E_(0)sin(100t)` We have, `omega=100` and `phi =pi/4=45^(@)` `tanphi = (omegaL)/R` `rArr I=(100 xx L)/(R)` If L=10H, then R=`kOmega` |
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| 263. |
An LCR series circuit consists of a resistance of a `10 Omega` a capacitance of reactance `60 Omega` and an inductor coil The circuit is found to resonate when put across a `300 V ,100 ` Hz supply The inductance of the coil is `(taken pi =3)` .A. 0.1 HB. 0.01 HC. 0.2 HD. 0.02 H |
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Answer» Correct Answer - A Angular velocity, `omega_o = 2pinu= 2pi xx 100` `omega_o = 2 xx 3 xx 100` = `600 rad s^(-1)` [since`pi` = 3] Further , `omega_o = 1 / sqrt LC` …..(i) Also, `X_C = 1 / COmega_o = 60 Omega` `implies C = 1 / omega_(o) xx 60 = 1 / 600 xx 60impliesC = 1 / 36 xx 10^(3)` F so, put values in Eq(i), we get `600 = 1 / sqrt(L(1/36 xx 10^(3)))` `implies 36 x 10^(4) = 36 xx 10^(3) / L implies 36 xx 10^(3) / 36 xx 10^(4) = 1/10` = 0.1 H |
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| 264. |
When an AC source of `emf e=E_(0) sin(100t)` is connected across a circuit i in the circuit, the phase difference between the emf e and the current i in the circuit is observed to be `(pi//4)`, as shown in the diagram. If the circuit consists possibly only of R-C or R-L or L-C in series, find the relationship between the two elements A. `R=1 k Omega, C=10 muF`B. `R=1 k Omega, C=1 muF`C. `R=1 k Omega, L=10 H`D. `R=1 k Omega, L=1 H` |
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Answer» Correct Answer - A As the current `i` leads the voltage by `pi/4`, it is `RC` circuit, hence `tan varphi=(X_(C))/Rimplies tan (pi/4)=1/(omegaCR)` `implies omegaCR=1` as `omega=100 rad//sec` `implies CR=1/100sec^(-1)`. From all the given option only option (a) is correct . |
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| 265. |
In a series LCR circuit, the total reactance is `4 Omega` and resistance is ` 3 Omega` . Its power factor isA. `4/3`B. `3/4`C. `4/5`D. `3/5` |
| Answer» Correct Answer - D | |
| 266. |
Power factor is maximum in an `LCR` circuit whenA. `X_(L)=X_(C)`B. `R=0`C. `X_(L)=0`D. `X_(C)=0` |
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Answer» Correct Answer - A In `LCR` circuit, in the conditions of resonance `X_(L)=X_(C)`, i.e., circuit behaves as resistive circuit. In resistive circuit power factor is maximum. |
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| 267. |
A short-circuited coil is placed in a time-varying magnetic field. Electrical power is dissipated due to the current induced in the coil. If the number of turns were to be quadrupled and the wire radius halved, the electrical power dissipated would beA. halvedB. the sameC. doubledD. quadrupled |
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Answer» Correct Answer - B Power `P=(e^(2))/R`, hence `e=-((dphi)/(dt))` where `phi=NBA` `:. E=-NA((dB)/(dt))` Also `R prop1/(r^(2))` Where `R` =resistance, `r`=radius, `l`=length `:. P prop(N^(2)r^(2))/l implies (P_(1))/(P_(2))=1` |
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| 268. |
A capacitor of capacitance `2 mu F` is connected in the tank circuit of an oscillator oscillating with a frequency of 1 kHz. If the current flowing in the circuit is `2 mA`, the voltage across the capacitor will be |
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Answer» `V_(C ) = IX_(C ) = I xx (1)/(omega C) = (I)/(2 pi fC)` `= (2 xx 10^(-3))/(2 pi xx 10^(3) xx 2 xx 10^(-6)) = 0.16 V` |
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| 269. |
A capacitor of capacitance `2 mu F` is connected in the tank circuit of an oscillator oscillating with a frequency of 1 kHz. If the current flowing in the circuit is `2 mA`, the voltage across the capacitor will beA. `0.32 V`B. `0.16 V`C. `79.5 V`D. `159 V` |
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Answer» Correct Answer - B Reactance fo capacitor or capacitive reactance is given by `X_(C)=1/(omegaC)=1/(2pifC)` where `f` is frequency of the alternating current Given, `f=1xx10^(3)Hz, C=2xx10^(6)F` `:. X_(C)=1/(2xx3.14xx10^(3)xx2xx10^(-6))` `=(10^(3))/(4xx3.14)` Hence, voltage across the capacitor is `V=iX_(C)=2xx10^(-30xx(10^(3))/(4xx3.14)=0.16 V` |
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| 270. |
A lamp consumes only `50%` of peak power in an `a.c.` circuit. What is the phase difference between the applied voltage and the circuit currentA. `(pi)/6`B. `(pi)/3`C. `(pi)/4`D. `(pi)/2` |
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Answer» Correct Answer - B `P=1/2V_(0)i_(0)cos phi implies P=P_(Peak) cos phi` `implies 1/2 (P_(Peak))=P_(Peak) cos phi` `implies cos phi=1/2 implies phi=(pi)/3` |
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| 271. |
In a circuit, the current lags behind the voltage by a phase difference of `pi//2`, the circuit will contain which of the following ?A. Only `R`B. Only `L`C. `R` and `C`D. only `C` |
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Answer» Correct Answer - B When a circuit inductance only, then the current lags behind the voltage by the phase difference of `pi//2` or `90^(@)`. While in a purely capacitance circuit, the current leads the voltage by a phase angle of `pi//2` or `90^(@)`. In a purely resistive circuit current is in phase with the applied voltage. |
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| 272. |
A transformer is used to light a `100 W` and `110 V` lamp from a `220V` mains. If the main current is `0.5 A`, the Efficiency of the transformer is approximately:A. `30 %`B. `50 %`C. `90 %`D. `10 %` |
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Answer» Correct Answer - C The efficiency of transformer `=("Energy obtai n ed form the secondary coil")/("Energy given t o the primary coil")` or `eta=("Output power")/("Input power")` or `eta=(V_(s)I_(s))/(V_(p)I_(p))` Given, `V_(s)I_(s)=100 W, V_(p)=200V, I_(p)=0.5 A` Hence, `eta=100/(220xx0.5) 0.90=90%` |
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| 273. |
Two concentric and coplanar coils have radii a and `b ( gt gt a)` as shows in Fig. Resistance of the inner coil is `R`. Current in the outer coil is increased from `0` to `i`, then the total charge circulating the inner coil is A. `(mu_(0)ia^(2))/(2Rb)`B. `(mu_(0)ib)/(2R)`C. `(mu_(0)i)/(2a) (pib^(2))/(R)`D. `(mu_(0)iB)/(2piR)` |
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Answer» Correct Answer - A |
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| 274. |
In a step - down transformer having primary to secondary turn ratio `20:1` the input voltage applied is `250V` and output current is `8A` Assuming `100%` efficiency calculate the (a) voltage across secondary coil (b) current in primary coil (c) power output . |
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Answer» `N_(s)/(N_(P)) = (1)/(20) V_(p) = 250 V, i_(s) = 8A` `(V_(s))/(V_(p)) = N_(s)/(N_(p)) implies V_(s) = (N_(s))/(N_(p)) xx V_(p) = (1)/(20) xx 250 = 12.5V` `(i_(p))/(i_(s)) = N_(s)/N_(p) implies i_(p) = (N_(s))/(N_(p) )i_(s) = (1)/(20) xx 8= 0.4A` Power output `= V_(s) i_(s) i_(s) = 12.5 xx 8 =100 W` . |
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| 275. |
A power transmission line feeds input power at 2400 V to a step down transformer with its primary windings having 4000 turns. What should be the number of turns in the secondary windings in order to get output power at 240 V?A. 400B. 420C. 424D. 436 |
| Answer» Correct Answer - A | |
| 276. |
In a step-down transformer having primary to secondary turn ratio 10:1, the input voltage applied is 250V and outout current is 10 A. Assuming 100% efficiency, calculate the (i) voltage across secondary coil (ii) current in primary coil (iii) power output |
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Answer» Given, `N_(S)/N_(P) = 1/10,V_(P)=250V, I=8A` i) `V_(S)/V_(P)=N_(S)/N_(P) rArr V_(S) = N_(S)/N_(P) xx V_(P) = 1/10 xx 250 = 25V` ii) `I_(P)/I_(S) = N_(S)/N_(P)rArr I_(P) = N_(S)/N_(P)I_(S) = 1/10 xx 10 = 1A` iii) Power output=`V_(S)I_(S) = 25 xx 10 = 250W` |
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| 277. |
(i) The primary of a transformer has 400 turns while the secondary has 2000 turns. If the power output from the secondary at 1100 V is 12.1 kW, calculate the primary voltage. (ii) If the resistance of the primary is `0.2 Omega` and that of the secondary is `2.0 Omega` and the efficiency of the transformer is `90 %`, calculate the heat losses in the primary and the secondary coils. |
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Answer» i) Given, `N_(1) = 400, `N_(2)`=2000, `V_(2)=1100V As `V_(1)=V_(2).N_(1)/N_(2)= 1100 xx 400/2000 = 220V` ii) Resistance of primary, `R_(1) = 0.2Omega` Resistance of secondary, `R_(2) = 2.0 Omega` Output power =`V_(2)I_(2)`=12.1kW = 12100 W `therefore` Current in the secondary, `I_(2)=P_(2)/V_(2) = 12100/1100 = 11A` As, efficiency = ("output power")/("input power") `rArr 90/100 = (12100W)/("input power")` Input power, `P_(1) = (12100 x 100)/90 = 13.44 xx 10^(3)`W Also, input power, `P_(1) = V_(1)I_(1)` `therefore` Current in the primary, `I_(1) = P_(1)/V_(1) = (13.44 xx 10^(2))/(220) = 61.1`A Power loss in the primary, `=I_(1)^(2)R_(1) = (61.1)^(2)xx 0.2 = 746.61`W Power loss in the secondary, `=I_(2)^(2)R_(2) = (11)^(2) xx 20 = 242W` |
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| 278. |
A 10 kW transformer has 20 turns in primary and 100 turns in secondary circuit. A.C. voltage `E_(1) = 600 sin 314 t` is applied to the primary. Find max. value of flux and max. value of secondary voltage. |
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Answer» i) Flus linked with each turn of primary, `phi=Bacosomegat=phi_(0)cosomegat` Here, `phi_(0)` BA=maximum value of flux linked with each turn `therefore V_(1)=-N_(1)(dphi)/(dt) = -N_(1)d/(dt)(phi_(0)cosomegat) = omegaN_(1)phi_(0)sinomegat` Peak value of `V_(1) = V_(0) = omegaN_(1)phi_(0)` or `phi_(0) = V_(0)/(omegaN_(1))` Given, `V_(1) = 600 sin 314t=V_(0)Sinomegat` `therefore V_(0)=600V, omega=314 rad s^(-1)` Here, `phi^(0) = 600/(314xx20)=0.0955Wb` ii) `V_(2)^(0)/V_(1)^(0) = N_(2)/N_(1)` `therefore` Maximum value of secondary voltage is, `V_(2)^(2) = N_(2)/N_(1)V_(1)^(0)= 100/20 xx 600 = 3000V` |
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| 279. |
A step up transformer operates on a `230 V` line and a load current of 2 ampere. The ratio of the primary and secondary windings is `1 : 25`. What is the current in the primary?A. `15A`B. `50A`C. `25A`D. `12.5A` |
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Answer» Correct Answer - B |
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| 280. |
A transformer is used to light a 140 W, 24 V lamp from 240 V AC mains. The current in mains cable is 0.7 A, find the efficiency of transformer.A. `63.8%`B. `74%`C. `83.3%`D. `48%` |
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Answer» Correct Answer - C Output power = 140 W Input power `= 240xx0.7=168 W` Efficiency `=("output power")/("input power")xx100=(140)/(168)xx100=83.3%` |
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| 281. |
The primary coil of an ideal transformer has 100 turns and the secondary coil has 600 turns. ThenA. The power in the primary circuit is less than that in the secondary circuitB. The current in two circuits are sameC. The voltage in the two circuits are sameD. The primary current is six times the secondary current |
| Answer» Correct Answer - D | |
| 282. |
A 60 W load is connected to the secondary of a transformer whose primary draws line voltage. If a current of 0.54 A flows in the load, what is the current in the primary coil? Comment on the types of transformer being used. |
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Answer» Correct Answer - A Given, `P_(s)=60 W, I_(s) = 0.54 A` Current in the primary `I_(p)=?` Taking line voltage as 220V. We can write Since, `P_(L) = 60 W, I_(L) = 0.54A` `rArr = V_(L)= 60/0.54=110V` Voltage in the secondary `(E_(s))` is less than voltage in the primary `(E_(p))`. Hence, the transfomer is step down transformer. Since, the transformation ratio `r=V_(s)/V_(p)= I_(p)/I_(s)` Substituting the values, `(110V)/(220V) = I_(p)/(0.54A)` On Solving `I_(p) =0.27A` |
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| 283. |
A power transformer (step up) with an `1:8` turns ratio has `60 Hz,120 V` across the primary, the load in the secondary is `10^(4) Omega`.The current in the secondary isA. 1.2AB. 0.96AC. 12mAD. 96mA |
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Answer» Correct Answer - D |
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| 284. |
A transformer has 500 primary turns and 10 secondary turns If `V_(p)` is 120 V (rms) and a resistive load of `15 Omega` is connected across the secondary , the currents in the primary and secondary coil of transformer are -A. 3.2 mA, 0.16 AB. 0.16 A, 3.2 mAC. 3.2 mA, 3.2 mAD. 0.16A, 0.16A |
| Answer» Correct Answer - A | |
| 285. |
in a step-up transformer, the turn ratio is `1:2` leclanche cell (e.m.f. 1.5V) is connected across the primary. The voltage devloped in the secondary would beA. 3 VB. `1.5 V`C. `0.75 V`D. zero |
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Answer» Correct Answer - A `(V_(s))/(V_(p))=(N_(s))/(N_(p))=(V_(s))/(1.5)=(2)/(1)=V_(s)=3 V` |
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| 286. |
A 60 W load is connected to the secondary of a transformer whose primary draws line voltage. If a current of 0.54 A flows in the load, what is the current in the primary coil? Comment on the types of transformer being used.A. `0.27 mA`B. `2.7 A`C. `0.27 A`D. `10 A` |
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Answer» Correct Answer - C Here, `P_(s)=60 W, I_(s)=0.54 A, V_(p)=220 V` `therefore V_(s)=(P_(s))/(I_(s))=(60)/(0.54)=111 V` As`(V_(s))/(V_(p))=(I_(p))/(I_(s))`, for anideal transformer `therefore I_(p)=(V_(s))/(V_(p))xxI_(s)=(111)/(220)xx0.54=0.27 A` |
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| 287. |
In the question number 96, the number of turns in the secondary isA. 20B. 80C. 120D. 160 |
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Answer» Correct Answer - C As, `(V_(s))/(V_(p))=(N_(s))/(N_(p))rArr N_(s) = (V_(s))/(V_(p))xxN_(p)` `=(220)/(11000)xx6000=120`. |
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| 288. |
A step down transformer converts transmission line voltage from 11000 V to 220 V. The primary of the transformer has 6000 turns and efficiency of the transformer is 60%. If the output power is 9 kW, then the input power will beA. 11 kWB. 12 kWC. 14 kWD. 15 kW |
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Answer» Correct Answer - D Here, `V_(p)=11000 V, V_(s)=220 V` `N_(p) = 6000 eta, 60% , P_(o)=9 kW=9xx10^(3) W` Efficiency, `eta = ("Output power")/("Input poers")=(P_(o))/(P_(i))` `therefore P_(i)=(P_(o))/(eta)=(9xx10^(3))/(60//100)=1.5xx10^(4)=15 kW` |
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| 289. |
A circuit containing a 80 mH inductor and a `60muF` 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» `I_(max)=V_(max)/sqrt(R^2+(X_L-X_C)^2)` `X_L=omegaL=100pitimes80times10^-3=25.12Omega` `X_C=1/(omegaC)=1/(100pitimes60times10^-6)=53.03Omega` `thereforeI_(max)=(230sqrt2)/(sqrt((25.12-53.03)^2))=11.6A` `I_(rms)=I_(max)/sqrt2=11.6/sqrt2=8.20A` |
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| 290. |
A circuit containing a 80 mH inductor and a `60muF` capacitor in series is connected to a 230 V , 50 Hz supply. The resistance of the circuit is negligible. Obtain rms values of potential drop across the inductor and capacitor. |
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Answer» `V_L=I_(rms).omegaL=8.20times25.12approx206V` `V_C=I_(rms).1/(omegaC)=8.20times53.03approx435V` |
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| 291. |
For a transformer, the turns ratio is 3 and its efficiency is 0.75. The current flowing in the primary coil is 2A and the voltage applied to it is 100 V. Then the voltage and the current flowing in the secondary coild are..............respectively.A. 150 V, 1.5 AB. 300 V, 0.5 AC. 300 V, 1.5 AD. 150 V, 0.5 A |
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Answer» Correct Answer - B `N_(S)/N_(P) = 3` `V_(S)/V_(P) = I_(P)/I_(S) = N_(S)/N_(P)` `V_(s) = V_(P).N_(S)/N_(P) = 100 xx 3 = 300 V` `rArr I_(S) = I_(P).N_(P)/N_(S) = 2 xx 1/3 ~~ 0.5 A` |
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| 292. |
A circuit containing a 80 mH inductor and a `60muF` 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 capacitor? |
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Answer» Average power transferred to the capacitor, `P_C=VIcosphi=VIcos"" pi/2=0` |
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| 293. |
A circuit containing a 80 mH inductor and a `60muF` capacitor in series is connected to a 230 V , 50 Hz supply. The resistance of the circuit is negligible. What is the total average power absorbed by the circuit? |
| Answer» Average power absorbed by the circuit=0 | |
| 294. |
A 0.21 H inductor and a `12 Omega ` resistance are connected in series to a `220 V, 50 Hz ac source. The current in the circuit isA. `(220)/(sqrt(4400))A`B. `(22)/(3sqrt(5)) A`C. `(220)/(sqrt(4500))A`D. `(22)/(5sqrt(3)) A` |
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Answer» Correct Answer - B here, `X_(L)=omega L =2 pi L = 2pi xx 50 xx 0.21 = 21 pi Omega` so, `Z=sqrt(R^(2)+X_(L)^(2))=sqrt(12^(2)+(21 pi)^(2))` `=sqrt(12^(2)+(21 xx 21//7)^(2))=sqrt(4500))=30 sqrt(5) Omega` so, (a) `I=(V)/(Z) =(220)/(3 sqrt(5))A` and (b) `phi=tan^(_1)((X_L)/(R )) = tan^(-1) ((21 pi)/(12)) = tan^(-1)((7 pi)/(4))` i.e. the current will lag the applied voltage. |
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| 295. |
A coil of self-inductance `L` is connected in series with a bulb `B` and an `AC` source. Brightness of the bulb decreases whenA. frequency of the AC source is decreasedB. number of turns in hte coil is reducedC. a capacitance of reactance `X_(C) = X_(L)` is included in the same circuit.D. an iron rod is inserted in the coil |
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Answer» Correct Answer - D `Z= sqrt(R^(2)+X_(L)^(2)) = sqrt(R^(2)+(2pifvL)^(2))` As, `I=V/Z, P=I^(2)R` i.e, `V uarr,Luarr= Zuarr,Idarr` and `Pdarr` |
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| 296. |
A circuit containing an `80mH` inductor and a `60(mu)F` capacitor in sereis is connected to a `230V-50Hz` Supply. The resistance in the circuit is negligible . (a) Obtain the current amplitude and rms currents. (b) Obtain the rms values of voltage across inductor and capacitor. (c ) What is the average power transferred to the inductor and to hte capacitor? (d) What is the total power absorbed by the circuit? |
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Answer» (a) Given `L=80MH, C=60(mu)F, E_(v)=230V, f=50Hz` Since the circuit has no resistance, so impedance of the circuit is equal to its reactance. `Z=X=X_(C)_X_(L)=(1)/(100 pi C)-100pi L=28 (Omega)` Current amplitude is `I_(0)=(E_0)/(Z)=(230 sqrt(2))/(28)=11.6A` (b) The rms volatge across capacitor is `V_(Lv)=I_(v)X_(L)=(111.6//sqrt(2))100 (pi)L=206 V` (c ) Average power transferred to inductor and capacitor will be zero because whatever energy is used in building up of charges on capacitor (current in case of inductor ) is returned in discharging the capacitor (decay of current in case of inductor). (d) As there is no resistance in the circuit, power factor of hte circuit is `cos(phi) =R//Z=0`. Hence, no power is absorbed by the circuit. |
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| 297. |
A circuit containing an `80mH` inductor and a `60muF` capacitor in series is connected to a `230V-50Hz` Supply. The resistance in the circuit is negligible . (a) Obtain the current amplitude and rms currents. (b) Obtain the rms values of voltage across inductor and capacitor. (c ) What is the average power transferred to the inductor and to the capacitor? (d) What is the total power absorbed by the circuit? |
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Answer» Inductance, L=80mH=`80xx10^(-3)H` Capacitan ce, `C=60muF=60xx10^(-6)F` Supply voltage, V=230V Frequency, V= 50Hz Angular frequency , `oemga=2nv=100`n rad/s peak voltage, `v_(0)=Vsqrt(2)=23-sqrt(2)V`. (a) Maximum current is given as: `I_(0)=(V_(0))/((omegaL-( 1)/(omegaC)))` `=(230sqrt(3))/((100pixx80xx10^(-3)-(1)/(100pixx6010^(-6)))` `=(230sqrt(2))/((8pi-(1000)/(6pi)))=-11.63A` the negative sign appears because `omegaLlt(1)/(omegaC)`. Amplitude of maximum current. `|I_(0)|=11.63A` Hence, rms value of current. `I=(I_(0))/sqrt(2)=(-11.63)/sqrt(3)=-8.22A` (b) Potential difference across the inductor, `V_(L)=IxxomegaL` `=8.22xx100nxx80xx10^(-8)` `=206.61V` Potential difference across the capacitor, `V_(C )=Ixx(1)/(omegaC)` `=8.22xx(1)/(100pixx60xx10^(-6))=436.3V` (c) Average power consumed by the inductor is zero as actual voltage leads the current `(pi)/(2)` (d) Average power consumed by the capacitor is zero as voltage lags current by `(pi)/(2)`. (e) The total power absorbed (averaged over one cycle ) is zero. |
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| 298. |
A 0.21-H inductor and a `88-Omega` resistor are connected in series to a 220-V, 50-Hz AC source. The current in the circuit and the phase angle between the current and the source voltage are respectively. `(Use pi= 22//7)`A. `2A, tan^(-1)3//4`B. `14.4A, tan^(-1)7//8`C. `14.4A, tan^(-1)8//7`D. `3.28 A, tan^(-1)2//11` |
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Answer» Correct Answer - A `I_(rms)=V_(rms)/Z=V_(rms)/sqrt(R^(2)+(omegaL)^(2))=2A` `tanphi=(omegaL)/R=66/88=3/4` |
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| 299. |
A 0.01 H inductor and `sqrt(3)pi Omega` resistance are connected in series with a 220 V, 50 Hz AC source. The phase difference between the current and emf isA. `pi/2rad`B. `pi/6` radC. `pi/3` radD. *- |
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Answer» Correct Answer - B `phi = tan^(-1)((X_(L))/R) = tan^(-1)((2pivL)/R)` `=(tan^(-1))((2pi xx 50xx0.01))/(sqrt(3pi))=tan^(-1)(1/(sqrt(3))) = pi/6rad` |
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| 300. |
In an RC series circuit(in fig), the rms voltage of source is 200V and its frequency is 50 Hz, if `R=100 Omega and C=(100)/(po) (mu)F`, find (a) inpedance of the circuit. (b) power factor angle, (c ) power factor, (d) current, (e) maximum current (f) voltage across R, (g) voltage across C, (h) maximum voltage across R, (i) maximum voltage across C, (j) ltPgt, (k) ` gtP_(C )lt`. `(m) i(t)`, `V_(R )(t),and V_(C)(t)`. |
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Answer» `X_(C)=(10^(6))/((100)/(pi)(2pi50))=100Omega` (i) `Z=sqrt(R^(2)+X_(c^(2))=sqrt(100^(2)+(100)^(2)=100sqrt2Omega` (ii) `tan phi =(X_(c))/(R )=1 therefore phi-=45^(@)` (iii) `"Power factor"=cos phi=(1)/(sqrt2)` `(iv)"Current" I_("rms")=(V_(rms"))/(Z)=(200)/(100sqrt2)=sqrt2A` `(v)"Maximum current"=I_("rms")sqrt2=2A` `(vi)"voltage across"R=V_(R,"rms")=I_("rms")R=sqrt2xx100"volt"` `(viii)"max voltage across R "=sqrt2 V_(R,"rms")X_(C)=sqrt2xx100"volt"` `(ix)"max voltage acorss C"=sqrt2V_(C,"rms")=200"volt"` `(x) lt P gt =V_("rms")I_("rms")cos phi=200xxsqrt2xx (1)/(sqrt2)=200"volt"` `("xi") lt P_(R)=I_("rms")cos phi=200W` `("xii") lt P_(C) =0` |
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