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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.
| 701. |
Find capacitive reactance of a capacitor connected to a dc source. |
| Answer» Correct Answer - Infinite | |
| 702. |
Assertion: Current versus time graph is as shown in figure, rms value of current is 4A. Reason: For a constant current, rms current is equal to that constant values. Reason: For a constant current, rms current is equal to that constant value. 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 - A | |
| 703. |
Assertion: In an AC, only capacitor circuit has instaneous power equal to zero at any instant of time. Reason: Phase difference current function and voltage function is `90^(@)`.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. |
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Answer» Correct Answer - D d) If `V=V_(0)sinomegat`, then current function will lead the voltage function by `90^(@)`. `I=I_(0)cosomegat` `therefore` Instaneous power, `P=VI=V_(0)I_(o)sinomegatcosomegat` or `PneO` at all time. |
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| 704. |
Find inductive reactance of the inductor when it is connected in D.C. circuit. |
| Answer» Correct Answer - Zero | |
| 705. |
Assertion: If an inductor coil is connected to DC source, the current supplied by it is `I_(1)`. If the same coil is connected with an AC source of same voltage. Then current is `I_(2)`, then `I_(2) lt I_(1)`. Reason: In AC circuit, inductor coil offers more resistance.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. |
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Answer» Correct Answer - A `Z_(DC) = r` `Z_(AC) = sqrt(r^(2)+X_(L)^(2))` Here, r= internal resistance of the coil. Also, `I alpha I/Z` `therefore I_(1) gt I_(2)` |
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| 706. |
Assertion: Inductive reactance of an inductor in DC circuit is zero. Reason: Angular frequency of DC circuit is zero.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. |
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Answer» Correct Answer - A a) `X_(L) = omegaL` In DC circuit, `omega=0` `therefore X_(L) =0` |
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| 707. |
In an alternating current circuit consisting of elements in series, the current increases on increasing the frequency of supply. Which of the following elements are likely to consitute the circuit ?A. only resistorB. Resistor and an inductorC. Resistor and capacitorD. Only a capacitor |
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Answer» Correct Answer - C::D According to the question, the current increases on increasing the frequency of supply. Hence, the reactance of the circuit must be decreases as increasing frequency. For a capacitive circuit. `X_(C) = 1/(omegaC) = 1/(2pi//C)` Clearly when frequency increases, `X_(C)` decreases. For R-C circuit, `X=sqrt(R^(2)+(1/omegaC)^(2))` When frequency increases X decreases. |
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| 708. |
Assertion: If the frequency of alternating current in an `AC` circuit consisting of an inductance coil is increased then current gets decreased. Reason: The current is inversely proportional to frequency of alternating current.A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the a ssertion.C. If assertion is true but reason is false.D. If the assertion and reason both are false. |
| Answer» Correct Answer - a | |
| 709. |
The potential difference `V` across and the current `I` flowing through an instrument in an `AC` circuit are given by: `V=5 cos omega t` volt `I=2 sin omega t` Amp.A. zeroB. 5wattC. 10wattD. 2.5watt |
| Answer» Correct Answer - A | |
| 710. |
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. 10 WC. 5 WD. 2.5 W |
| Answer» Correct Answer - a | |
| 711. |
The potential difference `V` across and the current `I` flowing through an instrument in an `AC` circuit are given by: `V=5 cos omega t` volt `I=2 sin omega t` Amp.A. zeroB. `5` wattC. `10` wattD. `2.5` watt |
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Answer» Correct Answer - A `P_(av)=v_(rms)i_(rms)cos phi` `phi=90^(@) P_(av)=0` |
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| 712. |
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 |
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Answer» Correct Answer - C `tan varphi=(X_(L))/R=(2pivL)/R` `implies tan 30^(@)=(2pixx50xxL)/(pisqrt(3)) implies L=0.01 H`. |
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| 713. |
In an `LCR` circuit the energy is dissipated inA. `L` onlyB. `C` onlyC. `R` onlyD. all of these |
| Answer» Correct Answer - C | |
| 714. |
How does an inductor behave in a dc circuit? |
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Answer» An inductor, having a pure resistance R and a self-inductance L, has an impedance , `Z=sqrt(R^2+omega^2L^2)`. Here, `omega=2pif` where f=frequency of the electric source. Naturally, f=0 for dc circuits, so, `omega=0` and Z=R This means that the inductor behaves as a pure resistance in dc-circuits, whereas its inductance L plays no role. |
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| 715. |
In an alternating current circuit consisting of elements in series, the current increases on increasing the frequency of supply. Which of the following elements are likely to consitute the circuit ?A. Only resistorB. Resistor and inductorC. Resistor and capacitorD. Only inductor |
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Answer» Correct Answer - C Reactance of a capacitor, `X_(C )=(1)/(omegaC)=(1)/(2pi upsilon C)` As frequency increases, `X_(C )` decreases and therefore current increases. As R does not vary with frequency, therefore, likely elements constituting the circuit may be capacitor and resistor. |
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| 716. |
A `30 mu F` capacitor is connected to a 150 V, 60 Hz ac supply. The rms value of current in the circuit isA. 17 AB. `1.7 A`C. `1.7 mA`D. `2.7 A` |
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Answer» Correct Answer - B Here, `C = 30xx10^(-6)F, V_("rms")=150V, upsilon = 60 Hz` Capacitive reactance `X_(C )=(1)/(omega C)=(1)/(2pi upsilon C)=88.46 Omega` `I_("rms")=(V_("rms"))/(X_(C ))=(150)/(88.46)=1.7 A` |
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| 717. |
`60muF` capacitor to a 110 V, 60 Hz ac supply. Determine the rms value of the current in the circuit. What is the net power absorbed by the circuit over a complete cycle? |
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Answer» `I_(rms)=V_(rms)/X_C` where, `X_C=1/(2pifC)=1/(2pitimes60times60times10^-6)Omega` and `V_(rms)=110V` `thereforeI_(rms)=110(2pitimes60times60times10^-6)A=2.49A` The net power absorbed is zero as in case of a ideal capacitor there is no power loss. |
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| 718. |
The rms value of current in an ac circuit is 25 A, then peak current isA. `35.36 mA`B. `35.36 A`C. `3.536 A`D. `49.38 A` |
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Answer» Correct Answer - B Here, `I_("rms")=25A` `therefore I_(m) = sqrt(2) I_("rms")=sqrt(2)xx25=35.36 A`. |
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| 719. |
In the question number 30, the net power absorbed by the circuit in one complete cycle isA. 5 WB. 10 WC. 15 WD. zero |
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Answer» Correct Answer - D As `P = V_("rms")I_("rms")cos phi` In a pure capacitance circuit, the phase difference between alternating voltage and current is `pi//2` . Hence `P = V_("rms")I_("rms")cos 90^(@) = 0`. |
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| 720. |
In the question number 3, the net power consumed over a full cycle isA. 586 WB. 242 WC. `48.4 W`D. 484 W |
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Answer» Correct Answer - D The given circuit is pure resistive circuit. Hence voltage and current both are in phase i.e., `phi = 0` `therefore = P = V_("rms")I_("rms")cos phi=220xx2.2xx1=484 W` |
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| 721. |
Assertion: In one complete cycle, power is consumed only across a resistance in series L-C-R circuit. Reason: Average power consumed across an inductor or a capacitor is zero.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 | |
| 722. |
In the figure shown an ideal alternative current (A,C) source of 10 Volt is connected. Find half of the total average power (in watts) given by the source to the circuit. |
| Answer» Correct Answer - 9 | |
| 723. |
Assertion: At resonance, power factor of series L-C-R circuit is zero. Reason: At resonance, current function and voltage functions are in same phase.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. |
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Answer» Correct Answer - D At resonance, `phi = 0` `therefore cosphi =1` or power factor =1 |
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| 724. |
Why the current does not rise immediately in a circuit containing inductanceA. because of induced emfB. because of high voltage dropC. both 1 and 2D. because of joule heating |
| Answer» Correct Answer - 3 | |
| 725. |
In an `AC` circuit containing only capacitance the currentA. leads the voltafe by ` 180^(@) `B. remians in phase with the volatageC. leads the voltage by ` 90^(@) `D. lags the voltage by ` 90^(@) ` |
| Answer» Correct Answer - C | |
| 726. |
In an `AC` circuit containing only capacitance the currentA. leads the voltage by `180^(@)`B. lags the voltage by `90^(@)`C. leads the voltage by `90^(@)`D. remains in phase with the voltage |
| Answer» Correct Answer - 3 | |
| 727. |
A capacitor of capacity `C` is connected in `A.C` circuit. If the applied emf is `V = V_(0) sin omega t`, then the current isA. `I = (V_(0))/(L omega) sin omega t`B. `I = (V_(0))/(omega C) sin (omega t + (pi)/(2))`C. `I = V_(0) C omega sin omega t`D. `t = V_(0) C omega sin (omega t + (pi)/(2))` |
| Answer» Correct Answer - 4 | |
| 728. |
A bulb is connected first with `DC` and the then `AC` of same voltage then it will shine brightly withA. `AC`B. `DC`C. Equally with bothD. Brightness will be in ratio `1//4` |
| Answer» Correct Answer - 3 | |
| 729. |
The r.m.s current in an `AC` circuit is `2A`. If the wattless current be `sqrt(3)A`, what is the power factor?A. `1/2`B. `1/3`C. `1/sqrt(3)`D. `1/sqrt(2)` |
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Answer» Correct Answer - A We know that, `I_(WL) = I_(rms)sinphi` `rArr sqrt(3) = 2sinphi` or `sinphi= sqrt(3)/2 rArr phi=60^(@)` So, power factor = `cosphi=cos60^(@)=1/2` |
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| 730. |
A bulb is connected first with `DC` and the then `AC` of same voltage then it will shine brightly withA. `AC`B. `DC`C. Brightness will be in ratio `1//1.4`D. Equally with both |
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Answer» Correct Answer - D Brightness `prop P_(consumed) prop 1/R` for Bulb, `R_(ac)=R_(dc)`, so brightness will be equal in both the cases. |
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| 731. |
If power factor is `1//2` in a series `RL,` circuit `R=100 Omega. AC` mains is used then `L` isA. `(sqrt(3))/(pi)` henryB. `pi` henryC. `(pi)/(sqrt(3))` HenryD. None of these |
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Answer» Correct Answer - A `cos phi=1/2 implies phi=60^(@) tan 60^(@)=(omegaL)/R implies L=(sqrt(3))/(pi)H` |
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| 732. |
In an LCR circuit, the capacitance is made one-fourth, when in resonance. Then what should be the change in inductance, so that the circuit remains in resonance?A. `4` timesB. `1//4` timesC. `8` timesD. `2` times |
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Answer» Correct Answer - A At resonance `omegaL=1/(omegaC) rArr l prop 1/C`. |
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| 733. |
A resistor `R`, an inductor `L`, a capacitor `C` and voltmeters `V_(1),V_(2)` and `V_(3)` are connected to an oscillator in the circuit as shown in the adjoining diagram.When the frequency of the oscillator is increased, upto resonance frequency, the voltmeter reading (at resonance frequency) is zero in the case of: A. voltmeter `V_(1)`B. voltmeter `V_(2)`C. voltmeter `V_(3)`D. all of three voltmeters |
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Answer» Correct Answer - B At resonance voltages across `C` and `L` are in opposite phase so net voltage will be zero. So` V_(2)=0`. |
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| 734. |
In figure `i_(1) = 10 e^(-2t) A, i_(2) = 4 A, v_(C ) = 3 e^(-2t) V` The potential difference across `AB (V_(AB))` is:A. `8 e^(-2t) V`B. `(1)/(2) e^(-3t) V`C. `17 e^(-2t)`D. `16 e^(-2t) V` |
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Answer» Correct Answer - C `V_(A) = - i_(1)R_(1) + V_(C ) = V_(B), V_(AB) = V_(A) - V_(B) = i_(1) R_(1) - V_(C )` Substitution the value we have, `V_(AB) = (10 e^(2t)) (2) - 3e^(-2t) , V_(AB) = 17 e^(-2t) V` Thus, `V_(AB)` decreases exponentially from `17 V` to 0. |
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| 735. |
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 - A From `KVL` we have `V_(A) - i_(1)R_(1) + i_(2)R_(2) = V_(C )` , `V_(A) - V_(C ) = i_(1) R_(1) - i_(2) R_(2)` Substituting the values, we have `V_(AC) = (10 e^(-2t)) (2) - (4) (3)` `V_(AC) = (20 e^(-2t) - 12) V` At `t = 0, V_(AC) = 8V`, At `t = oo, V_(AC) = 12 V` Therefore, `V_(AC)` decreases exponentially from `8V` to - `- 12V` |
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| 736. |
The input signal given to a `CE` amplifier having a voltage gain of `150` is `V_(i)=2cos(15t+(pi)/3)`. The corresponding output signal will beA. `300 cos (15t+(4pi)/3)`B. `300 cos (15t+(pi)/3)`C. `75 cos (15t+(2pi)/3)`D. `2 cos (15t+(5pi)/3)` |
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Answer» Correct Answer - A Input signal `V_(i n)=2 cos(15t+(pi)/3)` Voltage gain`=150` `CE` amplifier gives phase difference of `pi` between input and output signals `A_(v)=(V_(0))/(V_(i n))` so `V_(0)=A_(v)V_(i n)` so `V_(0)=150xx2cos (15 t+(pi)/3+pi)` `V_(0)=300 cos(15t+(4pi)/3)` |
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| 737. |
The frequency of ac mains in India isA. 30 cpsB. 50 cpsC. 60 cpsD. 120 cps |
| Answer» Correct Answer - B | |
| 738. |
220 volt a.c. is more dangerous than 220 volt d.c why?A. the AC attractsB. the DC repelsC. the body offers less resistance to ACD. peak voltage for AC is much larger than 220 V |
| Answer» Correct Answer - D | |
| 739. |
Alternating current is transmitted to distant places atA. high voltage and low currentB. high voltage and high currentC. low voltage and low currentD. low voltage and high current |
| Answer» Correct Answer - 1 | |
| 740. |
An AC voltage is given by `E=E_(0) sin(2pit)/(T)`. Then the mean value of voltage calculated over time interval of T/2 secondsA. is always zeroB. is never zeroC. `"is"(2E_(0)//pi)"always"`D. may be zero |
| Answer» Correct Answer - D | |
| 741. |
Alternating current is transmitted to distant places atA. at high voltage and low currentB. at high voltage and high currentC. at low voltage and low currentD. at low voltage and high current |
| Answer» Correct Answer - A | |
| 742. |
An AC voltage is given by E = `underset(o)(E) ` sin 2` pi `t / T Then , the mean value of volatage calculated over any time interval of T / 2A. is always zeroB. is never zeroC. is always (2`underset(o)(E) ` / `pi `)D. may be zero |
| Answer» Correct Answer - C | |
| 743. |
The instantaneous current in an AC circuit is l = `sqrt 2`sin(50 t + `pi` / 4 ) . The rms value of current isA. `sqrt 2`AB. 50 AC. 90 AD. 1A |
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Answer» Correct Answer - D `underset(rms)(I)` = `underset(o)(I)`/ `sqrt 2` = `sqrt 2`/`sqrt 2` = 1A |
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| 744. |
In an ac circuit , V and I are given by V=100sin(100t)V,and I=100sin `(100t+pi/3)` a respectively. The power dissipated in the circuit isA. `10^4`WB. 10 WC. 2500 WD. 5 W |
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Answer» Correct Answer - C From the given equation for V and I we get, `V_0=100V,_0`and`theta=pi/3` power`P=(V_0I_0)/2costheta=(100times100)/2cos""pi/3=2500W` |
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| 745. |
The relation between an ac voltage source and time in SI units is `V=120 sin(100 pi t)cos (100 pi t) V`. The value of peak voltage and frequency will be respectivelyA. 120 V and 100 HzB. `(120)/(sqrt(2)) V` and 100 HC. 60 and 200 HzD. 60 V and 100 Hz |
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Answer» Correct Answer - D `V = 120 sin (100pi t) cos (100 pi t)V` `=60 sin (200pi t)V (because sin theta 2 sin theta cos theta)` Compare it withstandard equation, `V = V_(0) sin omega t` We get, `V_(0)=60 V` and `omega = 200pi` or `2pi upsilon=200pi` or `upsilon = 100 Hz` |
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| 746. |
220 V, 50 Hz , AC is applied to a resistor . The instantaneous value of voltage isA. 220`sqrt 2`sin100`pi`tB. 220sin100`pi`tC. 220`sqrt 2`sin50`pi`tD. 220sin50`pi`t |
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Answer» Correct Answer - A Here , `underset(rms)(V)` = 220V, `nu` = 50 Hz Peak value of voltage , `underset(o)(V)` = `sqrt 2``underset(rms)(V)` = 220`sqrt 2`V The instantaneous value of voltage is V = `underset(o)(V)`sin`2pi``nu`t = 220`sqrt 2`sin`2pi` `xx` 50t = `220sqrt 2` sin 100`pi`t |
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| 747. |
The natural frequency of an LC oscillations is `f_0`. Show that , during periodic oscillations the maximum current `I_0` flowing through the oscillator is related to the maximum charge `Q_0`, on the capacitor as `I_0=2pif_0Q_0` . |
| Answer» Use the relations , `f_0=1/(2pisqrt(LC))` and `1/2Q_0^2/C=1/2LI_0^2` | |
| 748. |
STATEMENT-1 : A transformer is used for impedance impedance matching so as to deliver maximum current to a circuit. and STATEMENT-2 : The resistance R of external circuit connected across secondary, as seen by the generator in primary is `R_(eq)=((N_(P))/(N_(S)))^(2)R` . By adjusting . `(N_(P))/(N_(S))`, impedance of generator can be matched to external circuit.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-1C. Statement-1 is True, Statement-2 is FalseD. Statement-1 is False, Statement-2 is True |
| Answer» Correct Answer - A | |
| 749. |
The phase relationship between current and voltage in a pure resistive circuit is best represented byA. B. C. D. |
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Answer» Correct Answer - C In the pure resistive circuit current and voltage both are in phase. Hence graph (c ) is correct. |
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| 750. |
A light bulb is rated at 100 W for a 220 V ac supply . The resistance of the bulb isA. `284 Omega`B. `384 Omega`C. `484 Omega`D. `584 Omega` |
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Answer» Correct Answer - C Here, `P = 100 W, V_("rms")=220 V` Resistance of the bulb is `R=(V_("rms")^(2))/(P)=((220)^(2))/(100)=484 Omega` |
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