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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.
| 901. |
The frequency of an alternating voltage is `50` cycles/sec and its amplitude is `120V`. Then the r.m.s value of voltage isA. `101.3 V`B. `84.8 V`C. `70.7 V`D. `56.5 V` |
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Answer» Correct Answer - B `V_(rms)=(V_(0))/(sqrt(2))=120/1.414=84.8 V` |
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| 902. |
In the given `LR` circuit the source has angular frequency `omega` The power factore of the circuit is .A. `L//R`B. `R//omegaL`C. `(R)/(sqrt(R^(2) + omega^(2) L^(2)))`D. `R + omegaL` |
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Answer» Correct Answer - C `cos phi =(R )/(Z) = R/(sqrt(R^(2) + X_(L)^(2))) = R /(sqrt(R^(2) + omega^(2) L^(2)))` . |
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| 903. |
In an `AC` circuit, the power factorA. unity when the circuit contains only an inductanceB. unity when the circuit contains only a resistanceC. zero when the circuit contains only a resistanceD. unity when the circuit contains only a capacitance |
| Answer» Correct Answer - B | |
| 904. |
The frequency of an alternating voltage is `50` cycles/sec and its amplitude is `120V`. Then the r.m.s value of voltage isA. 101.3 VB. 84.8 VC. 70.7 VD. 56.5 V |
| Answer» Correct Answer - b | |
| 905. |
An ac source is connected across a resistance of `10 Omega` The power dissipated in the resistor is `100W` The rms valuse of the current and voltge are .A. `sqrt10 A ,sqrt(1000)V`B. `2 sqrt(10) A ,2 sqrt(1000)V`C. `2 sqrt(10) A, 2sqrt(1000)V`D. `sqrt(10) A , 2sqrt(1000) V` |
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Answer» Correct Answer - A `P = i^(2)_(rms) R implies 100 i^(2) _(rms) xx 10` `i_(rms) = sqrt10A` `V_(rms) = i_(rms) R =10 sqrt10 = sqrt(1000)V` . |
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| 906. |
An `AC` supply gives `30 V_(rms)`which passes through a `10 Omega` resistance. The power dissipated in it isA. `90sqrt(2) W`B. `90 W`C. `45 sqrt(2) W`D. `45 W` |
| Answer» Correct Answer - b | |
| 907. |
An alternating emf is applied across a parallel combination of a resistance R, capacitance C and an inductance L. If `underset(R )(I)`, `underset(L)(I)` and `underset(C )(I)` are the currents through R,L and C respectively, the phase relationship among `underset(R )(I)`, underset(L)(I)` and `underset(C )(I)` and source emf E, is given byA. B. C. D. |
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Answer» Correct Answer - C Current in inductance `underset(L)(I)` lags behind , cuurent in resistance `underset(g)(I)` is in phase with voltage , while current in capacitance `undetset(C )(I)` leads by a phase of `pi`/2 |
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| 908. |
An AC supply gives 30 `underset(rms)(V)` which passes `10 Omega` resistance. The power dissipated in it isA. 90`sqrt 2` WB. 90 WC. 45`sqrt 2` WD. 45 W |
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Answer» Correct Answer - B Power dissipated in circuit P = ``/R = `(30)^(2)`/10 = 90 W |
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| 909. |
An alternating potential V = `underset(o)(V) sin omega`t is applied across a circuit. As a result the current. I = `underset(o)(I)sin(omegat - pi/2)` flows in it. The power consecutive in the circuit per cycle isA. zeroB. 0.5`V_(0)` and `I_(0)`C. 0.707`V_(0)` and `I_(0)`D. 1.414`V_(0)` and `I_(0)` |
| Answer» Correct Answer - A | |
| 910. |
In the given `AC`, circuit, which of the following in incorrect: A. Voltage across resistance is lagging by `90^(@)` than the voltage across capacitorB. Voltage across capacitor is lagging by `180^(@)` than voltage across inductorC. Voltage across inductor is leading by `90^(@)` than voltage across resistanceD. resistance of the cicuit is equal to impedance reactance of circuit |
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Answer» Correct Answer - A Since the circuit is at resonance so current in the circuit is in the phase with applied voltage. Voltage across inductro leads the current by `pi//2` and across a capacitor lag by `pi//2`. So the voltage across resistance is lagging by `90^(@)` than the voltage across capacitor. |
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| 911. |
The phase difference between the voltage and the current in an AC circuit is `pi//4`. If the frequency is `50Hz` then this phase difference will be equivalent to a time ofA. `0.02s`B. `0.25s`C. `2.5 ms`D. `25 ms` |
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Answer» Correct Answer - C Time difference `=T/(2pi)xxvarphi=((1//50))/(2pi)xx(pi)/4` `=1/400s=2.5 ms` |
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| 912. |
The phase difference between the voltage and the current in an AC circuit is `pi//4`. If the frequency is `50Hz` then this phase difference will be equivalent to a time ofA. 0.02 sB. 0.25 sC. 2.5 msD. 25 ms |
| Answer» Correct Answer - c | |
| 913. |
In previous question the current in the circuit at resonance is .A. `10A`B. `15A`C. `30A`D. `60A` |
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Answer» Correct Answer - C `I = (V)/(Z) = (V)/ ( R) =(300)/(10) =30 A` . |
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| 914. |
If the power factor is `1//2` in a series `RL` circuit with `R = 100 Omega`. If `AC` mains, `50 Hz` is used then `L` isA. `(sqrt(3))/(pi)` henryB. `pi` henryC. `sqrt(3)` henryD. `sqrt(3) pi` henry |
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Answer» Correct Answer - A `Z = sqrt(R^(2) + omega^(2) L^(2))` `cos phi = (R )/(2) implies (1)/(2) = (100)/(2) implies Z = 200 Omega` `:. R^(2) + omega^(2) L^(2) = Z^(2) implies omega^(2) L^(2) = Z^(2) - R^(2)` `implies omega^(2) L^(2) = 4 xx 10^(4) - (100)^(2) = 3 xx 10^(4) implies omega L = 100 sqrt(3)` `implies 2 xx pi xx 50 xx L = 100 sqrt(3) implies L = (100 sqrt(3))/(2 pi xx 50) = (sqrt(3))/(pi)` |
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| 915. |
What will be the self-inductance of a coil, to be connected in a series with a resistance of `pi sqrt (3) Omega` such that the phase difference between the e.m.f. and the current at `50 Hz` frequency is `30^@`?A. 0.5 HenryB. 0.03 HenryC. 0.05 HenryD. 0.01 Henry |
| Answer» Correct Answer - d | |
| 916. |
In an`LCR` circuit .A. current always lags behind voltage ifB. current and voltage are always in phaseC. current in the voltage if `omega gt (1)/sqrt(LC)`D. current lags behind the voltage `if omega lt (1)/(sqrt(LC)` |
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Answer» Correct Answer - C If `Omega lt omega_(r), X_(C) gt X_(L)` current leads `omega = omega_(r) X_(C) = X_(L)` voltage and current in same phase `omega gt omega_(r), X_(L) gt X_(C)` voltage leads . |
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| 917. |
If power factor is `1//2` in a series `RL,` circuit `R=100 Omega. AC` mains is used then `L` isA. `sqrt(3)/pi` HeneryB. `pi` HenryC. `pi/sqrt(3)` HenryD. None of these |
| Answer» Correct Answer - a | |
| 918. |
A resistor `R`, an inductor `L` and a capacitor `C` are connected in series to a source of frequency `n`. If the resonant frequency is `n_(r)`, then the current lags behind voltage whenA. `omega lt omega_(0)`B. `omega gtomega_(0)`C. `omega = omega_(0)`D. `omega = omega_(0)` |
| Answer» Correct Answer - B | |
| 919. |
A resistor `R`,an inductor `L`, and a capacitor `C` are connected in series to an oscillator of freqency `upsilon`.if the resonant frequency is `upsilon_(r)`, then the current lags behind voltage, when:A. v=0B. `v lt v_(r)`C. `v=v_(r)`D. `v lt u_(r)` |
| Answer» Correct Answer - D | |
| 920. |
A resistor `R`, an inductor `L` and a capacitor `C` are connected in series to a source of frequency `n`. If the resonant frequency is `n_(r)`, then the current lags behind voltage whenA. `n=0`B. `n lt n_(r)`C. `n= n_(r)`D. `n gt n_(r)` |
| Answer» Correct Answer - d | |
| 921. |
A resistor `R`,an inductor `L`, and a capacitor `C` are connected in series to an oscillator of freqency `upsilon`.if the resonant frequency is `upsilon_(r)`, then the current lags behind voltage, when:A. `upsilon=0`B. `upsilon lt upsilon_(r)`C. `upsilon = upsilon_(r)`D. `upsilon gt upsilon_(r)` |
| Answer» Correct Answer - D | |
| 922. |
An LC circuit has L =20 mH , `C=50muF` and initial charge 10 mC the resistance being negligible. At what time is the total energy shared equally between the inductor and the capacitor? |
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Answer» `E=q^2/(2C)=1/(2C)q_0^2cos^2omegat` when `omegat=45^@,cosomegat=1/sqrt2` `E=1/(2C).q_0^2. 1/2=1/2(q_0^2/(2C))=1/2times` total energy So, when `omegat=45^@`or`t=T/8,(3T)/8,(5T)/8,....` the energy is shared equally between the capacitor and inductor. |
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| 923. |
An LC circuit has L =20 mH , `C=50muF` and initial charge 10 mC the resistance being negligible. What is the natural frequency of the circuit? |
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Answer» `omega=1/sqrt(LC)` `therefore` Reasonance frequency, `omega=1/sqrt(20times10^-3times50times10^-6)=10^3rad.s^-1` `f=omega/(2pi)=10^3/(2times3.14)=159Hz` |
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| 924. |
An LC circuit has L =20 mH , `C=50muF` and initial charge 10 mC the resistance being negligible. After what time interval from the moment the circuit is switched on the energy stored is (i) completely electrical i.e., stored only in the capacitor and (ii) completely magnetic ,i.e., stored only in the indicator? |
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Answer» `q=q_0cosomegat` (i) when t=0, `T/2,T,(3T)/2,.....` then `q=pmq_0` i.e,. The energy is completely electrical. (ii) when `t=T/4,(3T)/4,(5T)/4,.......` then q=0 i.e., the energy is purely magnetic. |
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| 925. |
A resistor `R`, an inductor `L` and a capacitor `C` are connected in series to an oscillator of frequency `n`. If the resonant frequency is `n_r,` then the current lags behind voltage, whenA. `n=0`B. `n lt n_(r)`C. `n=n_(r)`D. `n gt n_(r)` |
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Answer» Correct Answer - D The current will lag behind the voltage when reactance of inductance is more than the reactance of condenser. Thus, `omegaLgt1/(omegaC)` or `omegagt1/(sqrt(LC))` or `ngt1/(2pisqrt(LC))` or `ngtn_(r)`, where `n_(r)`=resonant frequency. |
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| 926. |
An LC circuit has L =20 mH , `C=50muF` and initial charge 10 mC the resistance being negligible. If a resistor is inserted in the circuit, how much energy is dissipated as heat? |
| Answer» When a resistor is connected in the circuit, all the energy stored in the circuit (i.e., 1J) will be displaced as heat energy since the LC oscillation will be damped ans stop ultimately. | |
| 927. |
A `10` ohm resistance, `5 mH` coil and `10mu F` capacitor are joined in series. When a suitable frequency alternating current source is joined to this combination, the circuit resonates. If the resistance is halved, the resonance frequencyA. Is halvedB. Is doubledC. Remain unchangedD. In quadrupled |
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Answer» Correct Answer - C Resonant frequency `=1/(2pisqrt(LC))` does not depend on resistance. |
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| 928. |
A `10` ohm resistance, `5 mH` coil and `10mu F` capacitor are joined in series. When a suitable frequency alternating current source is joined to this combination, the circuit resonates. If the resistance is halved, the resonance frequencyA. Is halvedB. Is doubledC. Remains unchangedD. In quadrupled |
| Answer» Correct Answer - c | |
| 929. |
A coil of inductance 0.50 H and resistance `100Omega` is connected to a 240 V, 50Hz ac supply. what is the maximum current in the coil? |
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Answer» Given, Inductance L = 0.50 H Resistance `R = 100 Omega` `V_("rms")=240V, f=50Hz` Impedance of circuit `Z=sqrt(R^(2)+X_(L)^(2))=sqrt(R^(2)+(2pi fL)^(2))` `= sqrt((100)^(2)+(2xx3.14xx50xx0.50)^(2))` `= 186.14 Omega` the rms value of current `I_("rms")=(V_("rms"))/(z)=(240)/(186.14)1.29 A` the maximum value of current in the circuit `I_(0)=sqrt(2)I_("rms")=1.414xx1.29=1.824A` |
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| 930. |
An alternating current source of frequency `100 Hz` is joined to a combination of a resistance, a capacitance and a coil in series. The potential difference across the coil, the resistance and the capacitor is `46, 8 and 40` volt respectively. The electromotive force of alternating current source in volt isA. 94B. 14C. 10D. 76 |
| Answer» Correct Answer - c | |
| 931. |
A coil of inductance 0.50 H and resistance `100Omega` is connected to a 240 V, 50Hz ac supply. IF the circuit is connected to a high frequency supply (240V,10 kHZ), what wull be the answer to (a) and (b) . From the answer explain the statement that at a very high frequency the presence of a inductor in the circuit nearly amounts to an open circuit. |
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Answer» `omega=2pif=2pitimes10^4rad.s^-1` `I_(max)=V_(max)/sqrt(R^2+omega^2L^2)=(240sqrt2)/(sqrt(100^2+4pi^2. 10^8 .5^2))=0.011A` `thereforephi=tan^-1{X_L/R}=tan^-1((2pifL)/R)` `=tan^-1 ""(2pitimes10^4times0.5)/100` `tan^-1(100pi)approxpi/2` `therefore` Time interval=`phi.T/360=phi/(360f)` `=90/(360times10times10^3)=0.25times10^-4s` `I_(max)` is very small. So it can be concluded that at high frequencies an inductance behaves as an open circuit. |
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| 932. |
If resistance of `100Omega` , inductance of `0.5` henry and capacitor of `10xx10^(-6)F` are connected in series through `50Hz` AC supply, then impedence isA. 1.876B. 18.76C. 189.72D. 101.3 |
| Answer» Correct Answer - c | |
| 933. |
The power factor of a good choke coil isA. Nearly zeroB. Exactly zeroC. Nearly oneD. Exactly one |
| Answer» Correct Answer - a | |
| 934. |
The value of current at half power points is -A. `sqrt(2)I_(m)`B. `I_(m)//sqrt(2)`C. `I_(m)//2`D. `2I_(m)` |
| Answer» Correct Answer - B | |
| 935. |
The power factor of a good choke coil isA. zeroB. oneC. 0.5D. none of the above |
| Answer» Correct Answer - A | |
| 936. |
A `50 Hz AC` source of `20 V` is connected across `R ` and `C` as shown in figureure. The voltage across `R` is `12 V`. The voltage across `C` isA. `8V`B. `16V`C. `10V`D. not possible to determine unless value of `R` and `C` are given |
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Answer» Correct Answer - B `V_C=sqrt(V^2-V_R^2)=sqrt((20)^2-(12)^2)` `= 16V`. |
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| 937. |
A `50 Hz AC` source of `20 V` is connected across `R ` and `C` as shown in figureure. The voltage across `R` is `12 V`. The voltage across `C` isA. 8 VB. 16 VC. 10 VD. not possible to determine unless value of R and C are given |
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Answer» Correct Answer - B `V_(C) = sqrt(V^(2)-V_(R)^(2)) =16 V` |
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| 938. |
The power factor of a good choke coil isA. Nearly zeroB. Exectly zeroC. Nearly oneD. Exactly one |
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Answer» Correct Answer - A `cos varphi=R/Z`. In choke coil `varphi~~90^(@)` so `cos varphi=0` |
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| 939. |
For wattless power is an `AC` circuit, the phase angle between the current and voltagge isA. `90^(@)`B. `45^(@)`C. `180^(@)`D. `60^(@)` |
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Answer» Correct Answer - A If the current is wattless then power is zero. Hence phase difference `varphi~~90^(@)` |
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| 940. |
A generator at a utility company produces 100 A of current at 4000 V. The voltage is stepped up to 240000 V by a transformer before it is sent on a high voltage transmission line. The current in transmission line isA. 3.67 AB. 2.67 AC. 1.67 AD. 2.40 A |
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Answer» Correct Answer - C For step-up transforer, `V_(s) gtV_(P)` and `I_(S) lt I_(P)` For an ideal transformer, `V_(S)I_(S) = V_(P)I_(P)` `therefore 240000I_(S) = 100 xx 4000` or `I_(S) = 1.67 A` |
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| 941. |
Assertion: An AC can be transmitted over long distances without much power loss. Reason: An AC can be stepped up or down with the help of a transformer.A. If both Assertion and Reason are true and Reason 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 is true but Reason is false.D. If Assertion is false but Reason is true. |
| Answer» Correct Answer - B | |
| 942. |
Angular frequency `omega` in an AC, L-C-R series circuit is gradually increased. Then, match the following two columns. |
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Answer» Correct Answer - (a-q), (B-p); (C-r); (D-s) `X_(C) = 1/(omegaC) propto 1/(omega), X_(L) = omegaL propto L` R is not a function of `omega` Z is minimum at resonance frequency. |
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| 943. |
In the following circuit the current is in phase with the applied voltage. Therefore, the current in the circuit and the frequency of the source voltage respectively, are A. `(V)/(R) and (1)/(2pisqrtLC)`B. `"zero and" (1)/(sqrtLC)`C. `sqrt((C)/(L))v, and (2)/(pisqrtLC)`D. `sqrt((C)/(LR^(2)))and (2)/(sqrtLC)` |
| Answer» Correct Answer - A | |
| 944. |
Match the following two columns for L-C-R series AC circuit. |
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Answer» Correct Answer - (A-q); (B-p);(C-q); (D-s) Power factor, `cosphi = R/Z` At resonance frequency, `X_(L)=X_(C)` At frequency gt resonance frequency, `X_(L) gt X_(C)` as `X_(L) propto omega` |
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| 945. |
In an AC, series L-C-R circuit, `R=X_(L)=X_(C)` and applied AC, voltage is V. Then match the following two columns. |
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Answer» Correct Answer - (A-q); (B-q); (C-r); (D-p) Z=R [As `X_(L)=X_(C))]` `therefore I=V/Z=V/R` Now, `V_(R) = IR=V` `V_(C) = IX_(C)=V=V_(L)` `V_(RL) = sqrt(V^(2)+V^(2)) = sqrt(2)V` `V_(LC) = V-V=0` |
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| 946. |
The average power delivered to a series AC circuit is given by (symbols have their usual meaning):A. `E_("rms")I_("rms")cos phi`B. `(I_("rms"))^(2)R`C. `(E_("max").^(2)R)/(2(|z|)^(2))`D. `(I_("max").^(2)|z|cos phi)/(2)` |
| Answer» Correct Answer - A::B::C::D | |
| 947. |
In a series L-C-R, AC circuit assuming that symbols have their usual meanings match the following two columns. |
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Answer» Correct Answer - (A-q); (B-s); (C-s); (D-p) `I=V/sqrt(R^(2)+(X_(L)-X_(C))^(2))` |
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| 948. |
The average power delivered to a series AC circuit is given by (symbols have their usual meaning):A. `E_(rms)I_(rms)cosphi`B. `(I_(rms))^(2)R`C. `(E_(max)^(2)R)/(2(|z|)^(2))`D. `(I_(max)^(2)|z|cos phi)/2` |
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Answer» Correct Answer - A,B,C,D `P_(av)=(A)=(B)=( C)=(D)`. |
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| 949. |
The average power delivered to a series AC circuit is given by (symbols have their usual meaning):A. `E_(rms)I_(rms)`B. `E_(rms)I_(rms) cos phi`C. `E_(rms)I_(rms) sin phi`D. zero |
| Answer» Correct Answer - B | |
| 950. |
The potential differences `V` and the current `i` flowing through an instrument in an `AC` circuit of frequency `f` are given by `V=5 cos omega t` and `I=2 sin omega t` amperes (where `omega=2 pi f`). The power dissipated in the instrument isA. zeroB. `10W`C. `5W`D. `2.5 W` |
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Answer» Correct Answer - A `V=5 cos omega t=5sin(omega t+(pi)/2)` and `i=2 sin omegat` Power `=V_(rms)xxi_(rms)xxcos varphi=0` (Since `varphi=(pi)/2`, therefore `cos varphi=cos((pi)/2)=0`) |
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