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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.
| 951. |
A `DC` ammeter cannot measure alternating current becauseA. `AC` changes its directionB. `DC` instruments will measure the average valueC. `AC` can damage the `DC` instrumentD. `AC` produces more heat |
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Answer» Correct Answer - B Average value in AC comes out to be zero |
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| 952. |
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 theta`C. `E_("rms")//I_("rms")sin theta`D. zero |
| Answer» Correct Answer - B | |
| 953. |
Energy dissipates in LCR circuit in:A. L onlyB. C onlyC. R onlyD. all of these |
| Answer» Correct Answer - C | |
| 954. |
For an LCR series circuit with an aac source of angular frequency `omega`.A. circular will be capacitive if `omegagt(1)/(sqrt(LC))`B. circular will be inductive if `omega=(1)/(sqrt(LC))`C. Power factor of circuit will be unity if capacitive reactance equals inductive reactanceD. circular will be leading voltage if `omegagt(1)/(sqrt(LC))` |
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Answer» Correct Answer - C The circuit will have inductive nature if `omega gt (1)/(sqrt(LC))(omega L gt(1)/(sqrt(LC)))` Hence, (a) is false. Also if circuit has inductive nature, The current will lag behind voltage, Hence (d) is also false. If `omega =(1)/(sqrt(LC)) (omegaL=(1)/(omega C))`, the circuit will have resistance nature, Hence (b) is false. Power factor, `cos (phi) = (R )/(sqrt(R^(2)+ (omegaL=(1)/(omega C))^(2))=1` If `omega L =(1)/(omega C). Hence (c) is true. |
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| 955. |
Which of the following components of a LCR circuit, with ac supply, dissipates energyA. LCRB. RC. CD. All of these |
| Answer» Correct Answer - b | |
| 956. |
Assertion : In series `L-C-R, AC` circuit, current and voltage are in same phase at resonance. Reason : In series `L-C-R, AC` circuit, resonant frequency does not depend on the value of resistance. Hence, current at resonance does not depend on resistance.A. If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.B. If both Assertion and Reason are true but Reason is not the correct explanation of AssertionC. If Assertion is true, but the Reason is false.D. If Assertion is false but the Reason is true. |
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Answer» Correct Answer - C At resonasnce `X_L=X_C` `implies Z=R` Hence, `I=V/Z=V/R` So, current at resonance depends on `R`. |
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| 957. |
In a series LCR circuit `R= 200(Omega)` and the voltage and the frequency of the main supply is 220V and 50 Hz respectively. On taking out the capacitance from the circuit the current lags behind the voltage by `30(@)`. On taking out the inductor from the circuit the current leads the voltage by `30(@)`. The power dissipated in the LCR circuit isA. `305 W`B. `210 W`C. zeroD. `242 W` |
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Answer» Correct Answer - 4 The given circuit is under resonace as `X_(L) = X_(C )` Hence, power dissipated in the circuit is `P = (V^(2))/(R ) = 242 W` |
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| 958. |
An inductor L and a capacitor C are connected in the circuit as shown in the figure. The frequency of the power supply is equal to the resonant frequency of the circuit. Which ammeter will read zero ampere A. `A_(1)`B. `A_(2)`C. `A_(3)`D. None of these |
| Answer» Correct Answer - c | |
| 959. |
In a series LCR circuit `R= 200(Omega)` and the voltage and the frequency of the main supply is 220V and 50 Hz respectively. On taking out the capacitance from the circuit the current lags behind the voltage by `30(@)`. On taking out the inductor from the circuit the current leads the voltage by `30(@)`. The power dissipated in the LCR circuit isA. 305WB. 210WC. 0WD. 242W |
| Answer» Correct Answer - D | |
| 960. |
The current in series `LCR` circuit will be the maximum when `omega` isA. As large as possibleB. Equal o natural frequency of LCR systemC. `sqrt(LC)`D. `sqrt(1//LC)` |
| Answer» Correct Answer - d | |
| 961. |
A coil has `L=0.04H` and `R=12Omega`. When it is connected to `220V, 50Hz` supply the current flowing through the coil, in ampere isA. 10.7B. 11.7C. 14.7D. 12.7 |
| Answer» Correct Answer - d | |
| 962. |
An electron lamp is conected to `220V, 50Hz` supply. Then the peak value of voltage isA. `210V`B. `211V`C. `310V`D. `320V` |
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Answer» Correct Answer - C `V_(0)=V_(rms)xxsqrt(2)=220xxsqrt(2)=310` |
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| 963. |
Voltage and current for a circuit with two elements in series are expressed as `V(t)=170sin(6280t+pi/3)volt` `i(t)=8.5sin(6280t+pi/2)amp` (a) Plot the two waveforms. (b) Determine the frequency in Hz. (c) Determine the power factor starting its nature. (d) What are the values of the elements?A. `R = 27.32 Omega, C = 25.92 mF`B. `R = 17.32 Omega, C = 15.92 mF`C. `R = 7.32 Omega, C = 5.92 mF`D. `R = 10.32 Omega, C = 5.92 mF` |
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Answer» Correct Answer - B `tan phi = (1)/(omega CR)` and `phi = (pi)/(6)` |
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| 964. |
An AC generator produced an output voltage `E=170sin 377t` volts , where `t` is in seconds. The frequency of `AC` voltage isA. `50Hz`B. `110Hz`C. `60Hz`D. `230Hz` |
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Answer» Correct Answer - C `2piv=377 implies V=60.33Hz` |
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| 965. |
When `100 V` DC is applied across a solenoid, a current of `1.0 A` flows in it. When `100 V` AC is applied across the same coil. The current drops to `0.5A`. If the frequency of the ac source is `50 Hz`, the impedance and inductance of the solenoid areA. 200Omega and 0.55HB. 100Omega and 0.86HC. 200Omega and 1.0HD. 100Omega and 0.93H |
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Answer» Correct Answer - A |
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| 966. |
When `100 V` DC is applied across a solenoid, a current of `1.0 A` flows in it. When `100 V` AC is applied across the same coil. The current drops to `0.5A`. If the frequency of the ac source is `50 Hz`, the impedance and inductance of the solenoid areA. `200 Omega` and 0.55 HB. `100 Omega` and 0.86 HC. `200 Omega` and 1.0 HD. 1100`Omega` and 0.93 H |
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Answer» Correct Answer - A Impedance = `V_AC/I_AC = 100/0.5 = 200 Omega = Z` `R = V_DC/I_DC = 100 / 1 = 100 Omega` `X_L = sqrt Z^(2) - R^(2) = 100sqrt 3Omega = (2pifL)` `L = 100sqrt 3/ 2pifL = 100 xx 1.732 / 2 xx 3.14 xx 50` = 0.55 H |
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| 967. |
When `100 V` DC is applied across a solenoid, a current of `1.0 A` flows in it. When `100 V` AC is applied across the same coil. The current drops to `0.5A`. If the frequency of the ac source is `50 Hz`, the impedance and inductance of the solenoid areA. 200 ohm and 0.55 henryB. 100 ohm and 0.86 henryC. 100 ohm and 1.0 henryD. 100 ohm and 0.93 henry. |
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Answer» Correct Answer - A A Solenoid consists of inductance and resistance. When `100 V` dc is applied, `omega = 0 implies Z = R` `Z = (V_("rms"))/(l_("rms")) implies R = (100)/(1) = 100 Omega` When `100 V, 50 Hz` ac is applied, `Z = (V_("rms"))/(l_("rms")) = (100)/(0.5) = 200 Omega` `Z^(2) = R^(2) + X_(L)^(2) implies 200^(2) = (100)^(2) + X_(C )^(2)` `implies X_(L) = - 100 sqrt(3) implies 2 pi f L = 100 sqrt(3)` `implies L = (100 sqrt(3))/(2 xx pi xx 50) = 0.55 H` |
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| 968. |
A transformer steps an `A.C` voltage from `230 V` ot `2300 V`. If the number of turns in the secondary coil is 1000, the number of turns in the primary coil will beA. 100B. `10,000`C. 500D. 1000 |
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Answer» Correct Answer - 1 `(n_(s))/(n_(p)) = (V_(s))/(V_(p))` |
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| 969. |
L,C,R represents the inductance, capacitance and reactance respectively. Which of the following combinations have the same dimensions as that of frequency?A. `1/(RC)`B. `R/L`C. `1/sqrt(LC)`D. `C/L` |
| Answer» Correct Answer - A::B::C | |
| 970. |
The current through an ac (series) circuit increases as the source frequency is increased. Which of the followings are the most suitable combinations of the circuit?A. only resistorB. resistor and an inductorC. resistor and a capacitorD. only capacitor |
| Answer» Correct Answer - C::D | |
| 971. |
A `1.5 mu F` capacitor is charged of 60 V. The charging battery is then disconnected and a 15 mH coil is connected in series with the capacitor so that LC oscillations occur. Assuming that the circuit contains no resistance, the maximum current in this coil shall be close toA. `1.4 A`B. ` 1.2 A`C. `0.8 A`D. `0.6 A` |
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Answer» Correct Answer - D Here `q_(0)=CV=1.5xx10^(-6)xx60=9xx10^(-5)C` Oscillating charge in LC circuit, `q=q_(0)sin (omega t+(pi)/(2))` `therefore` Oscillating current, `I=(dq)/(dt)=q_(0)omega cos(omega t+(pi)/(2))` `thereofre` Maximum current, `I_(0)=q_(0)omega=(q_(0))/(sqrt(LC))" "(because omega = (1)/(sqrt(LC)))` `=(9xx10^(-5))/(sqrt(15xx10^(-3)xx1.5xx10^(-6)))=0.6 A` |
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| 972. |
A complex current wave is given by `i=95+5sin100omegatA`. Its given value over one time period is given asA. 10AB. 5AC. `sqrt 50 ` AD. 0 |
| Answer» Correct Answer - B | |
| 973. |
The peak value of an alternating emf E given by E = `underset(o)(E) ` cos `omega`t is 10 V and frequency is 50 Hz . At time t = (1/600) s, the instantaneous value of emf isA. 10 VB. 5`sqrt 3` VC. 5 VD. 1 V |
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Answer» Correct Answer - B E = 10 cos(`2pi`ft) = 10 cos (`2pi xx 50 xx 1/600`) = 10 cos`pi`/6 = `5sqrt 3`V |
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| 974. |
If the rms current in a 50 Hz ac circuit is 5 A, the value of the current `1//300` second after its value becomes zero isA. `5 sqrt 2`AB. 5`sqrt 3/2 `AC. 5/6 AD. 5/`sqrt 2` A |
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Answer» Correct Answer - B Given, v = 50 Hz, `underset(rms)(I)` = 5 A `t = 1/300` s We have to find I(t) `underset(o)(I)` = Peak value = `sqrt 2underset(rms)(I)` = `sqrt 2 xx 5` = `5sqrt 2` A I = `underset(o)(I) sin omegat = 5sqrt 2sin 2pivt = 5sqrt 2sin 2pivt = 5sqrt 2sin 2pi xx 50 xx 1/300` `5sqrt 2 sin pi/3` = `5sqrt 2 xx sqrt 3 / 2` = `5sqrt 3/2` A |
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| 975. |
If the rms current in a 50 Hz ac circuit is 5 A, the value of the current `1//300` second after its value becomes zero isA. `5sqrt(2)A`B. `5sqrt((3)/(2))A`C. `(5)/(6)A`D. `(5)/(sqrt(2))A` |
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Answer» Correct Answer - B Here, `I_("rms")=5A, upsilon = 50 Hz, t=(1)/(300)s` `I_(0) = sqrt(2)I_("rms")=5sqrt(2)A`, `I=I_(0)sin omega t=I_(0)sin 2pi upsilon t = 5sqrt(2) sin (2pixx50xx(1)/(300))` `=5sqrt(2)"sin" (pi)/(3)=5sqrt(2)(sqrt(3))/(2)=5 sqrt((3)/(2))A` |
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| 976. |
If the rms current in a 50 Hz ac circuit is 5 A, the value of the current `1//300` second after its value becomes zero isA. `5sqrt(2)`AB. `5sqrt(3//2)` AC. `5//6` AD. `5//sqrt(2)` A |
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Answer» Correct Answer - B Given, `v=50Hz, I_(rms)=5A` `t=1/300s` We have to find I(t) `I_(0)= "Peak value"=sqrt(2), I_(rms)=sqrt(2) xx5` `=5sqrt(2)` A `I=I_(0)sinomegat=5sqrt(2)sin2pivt=5sqrt(2)sin 2pi xx 50 xx 1/300` `=5sqrt(2)sin pi/3=5sqrt(2) xx sqrt(3)/2 = 5sqrt(3//2)` A |
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| 977. |
If the rms current in a 50 Hz ac circuit is 5 A, the value of the current `1//300` second after its value becomes zero isA. `5 sqrt(2) A`B. `5 sqrt((3)/(2))A`C. `(5)/(6) A`D. `(5)/(sqrt(2))A` |
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Answer» Correct Answer - 2 Here, `I_("rms") = 5 A, upsilon = 50 Hz, t = (1)/(300) s` `I_(0) = sqrt(2) I_("rms") = 5 sqrt(2) A`. From `I = I_(0) sin omega t = I_(0) sin 2 pi upsilon t` `I = 5 sqrt(2) sin (2 pi xx 50 xx (1)/(300))` `= 5 sqrt(2) "sin" (pi)/(3) = 5 sqrt(2) (sqrt(3))/(2) = 5 sqrt((3)/(2)) A` |
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| 978. |
To reduce the resonant frequency in an LCR series circuit with a generator,A. the frequency of the generator should be reducedB. another capacitor should be connected is parallel with the first capacitorC. the iron core of the inductor should be removedD. the dielectric in the capacitor should be removed |
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Answer» Correct Answer - B `f=1/(2pisqrt(LC))` `therefore` To reduce f either L or C or both has to be increased. |
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| 979. |
Which of the following combinations should be selected for better turning of an LCR circuit used for communication ?A. R = 20 `Omega` , L = 1.5h, C = `35mu`FB. R = `25Omega` , L = 2.5 H , C = `45mu`FC. R = `15Omega` , L = 3.5 H , C = `30mu`FD. R = `25Omega`, L = 3.5 H, C = `45mu`F |
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Answer» Correct Answer - C (c) Quality factor (Q) of an L-C-R circuit is given by, `Q = 1/R sqrt (L/C)` Where R is resistance, L is inductance and C is capacitance of the circuit . To make Q high , R should be low, L should be high and C should be low . these conditions are best satisfied by the values given in option (c) Note : we shoiuld be careful while writing formula for quality factor, because we are considering series L-C-R circuit . |
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| 980. |
The internal resistance and internal reactance of an alternating current generator are `R_g` and `X_g` respectively. Power from this source is supplied to a load consisting of resisting `R_g` and reactance `X_L`. For maximum power to be delivered from the generator to the load. The value of `X_L` is equal toA. zeroB. `X_g`C. `-X_g`D. `R_g` |
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Answer» Correct Answer - C For maximum power, total reactance is zero. |
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| 981. |
An alternating current generator has an internal resistance `R_(g)` and an internal reactance `X_(g)`. It is used to supply power to a passive load consisting of a resistance `R_(g)` and a rectance `X_(L)`. For maximum power to be delivered from the generator to the load, the value of `X_(L)` is equal toA. zeroB. `X_(g)`C. `- X_(g)`D. `R_(g)` |
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Answer» Correct Answer - 3 For maximum power to be delivered from the generator to the load, the total reactance must vanish. i.e., `X_(L) + X_(g) = 0` or `X_(L) = - X_(g)` |
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| 982. |
An alternating current generator has an internal resistance `underset(g)(R )` and an internal reactance `underset(g)(X )` . It is used to supply power to a passive load consisting of a resistance `underset(g)(R )` and a reactance `underset(L)(X )`. For maximum power to be delivered from the generator to the load , the value of `underset(L)(X )` is equal toA. zeroB. `underset(g)(X )`C.D. `underset(g)(R )` |
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Answer» Correct Answer - C For delivering maximum power from the generator to the load, total internal reactance must be equal to conjugate of total external reactance. Hence, `underset(int)(X )` = `` |
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| 983. |
In an `AC` circuit, the instantaneous values of e.m.f and current are `e=200sin314t` volt and `i=sin(314t+(pi)/3)` ampere. The average power consumed in watt isA. 200B. 100C. 0D. 50 |
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Answer» Correct Answer - 4 `P_("avg") = I_("rms") E_("rms") cos phi = (1)/(sqrt(2)) xx (200)/(sqrt(2)) cos 60^(@)` `50 W` |
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| 984. |
Two identical incandescent light bulbs are connected as shown in figure. When the circuit is an AC voltage source of frequency f, which of the following observations will be correct. A. Both bulbs will glow alternativelyB. Both bulbs will glow with same brightness provided f = 1/2`pi` `sqrt 1/LC`C. Bulb `underset(1)(b)` will light up initially and goes OFF, bulb `underset(2)(b)` will be ON constantlyD. Bulb `underset(1)(b)` will blink and bulb `underset(2)(b)` will be ON constantly |
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Answer» Correct Answer - A This is a parallel circuit, for oscillation, the energy in I. and C will be alternately maximum. |
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| 985. |
In a series `LCR` circuit, resistance `R=10Omega` and the impedence `Z=20Omega` the phase difference between the current and the voltage isA. `30^(@)`B. `45^(@)`C. `60^(@)`D. `90^(@)` |
| Answer» Correct Answer - c | |
| 986. |
In the figures shown current in the long straight wire varies as `I=I_(0)((t_(0)-t)/(t_(0))` where `I_(0)` is the initial current The charge flown through resistance in the time `t_(0)` is A. `(mu_(0))/(2pi)(lbl_(0))/(R)`B. `(mu_(0))/(2pi)(ll_(0))/(R)In((b)/(a))`C. `(mu_(0))/(2pi)(ll_(0))/(R)In((a+b)/(a))`D. `(mu_(0))/(2pi)(al_(0))/(R)In((b)/(l))` |
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Answer» Correct Answer - B |
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| 987. |
In the circuit shown, the capacitor initially charged with a 12V batteryy, when switch `S_(1)` is open and switch `S_(2)` is closed. The maximum value of current in the circuit when `S_(2)` is opened and `S_(2)` is closed is A. `10^(-6)A`B. `7.2muA`C. `720muA`D. `360muA` |
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Answer» Correct Answer - C |
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| 988. |
The primary winding of transformer has `500` turns whereas its secondary has `5000` turns. The primary is connected to an ac supply of `20V,50Hz`. The secondary will have an output ofA. `2V,5Hz`B. `200V,500Hz`C. `2V,50Hz`D. `200V,50Hz` |
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Answer» Correct Answer - D |
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| 989. |
For an RLC circuit driven with voltage of amplitude `v_m` and frequency `omega_0=1/sqrt(LC)` the current exibits resonance. The quality factor, Q is given byA. `R/((omega_0C))`B. `(CR)/(omega_0)`C. `(omega_0L)/R`D. `(omega_0R)/L` |
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Answer» Correct Answer - C Q factor of RLC circuit= `(omega_0)/(Deltaomega)=omega_0/(R/L)=(omega_0L)/R` |
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| 990. |
The output of a step-down transformer is measured to be `24 V` when connected to a 12 watt light bulb. The value of the peak current isA. `1//sqrt(2)A`B. `sqrt(2)A`C. 2AD. `2sqrt(2)`A |
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Answer» Correct Answer - A secondary voltage `V_(S) = 24V` Power associated with secondary `P_(S) =12 W` `I_(s)=P_(S)/V_(s)=12/24` `=1/2A=0.5A` Peak value of the current in the secondary `I_(0) = I_(S) sqrt(2)` `=(0.5)(1.414)=0.707=1/sqrt(2)A` |
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| 991. |
When a voltage measuring device is connected to a.c. mains the meter shows the steady input voltage of `220 V`. This meansA. input voltage cannot be AC voltage, but a DC voltageB. maximum input voltage is 220 VC. the meter reads not v but `ltv^(2)gt` and is calibrated to read `sqrt(ltv^(2)gt)`D. the pointer of the meter is stuck by some mechanical defect. |
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Answer» Correct Answer - C The voltmeter connected to AC mains reads mean value `v^(2)` and is calibrated in such a way that it gives value of `v^(2)` which is multiplied by from factor to given rms value. |
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| 992. |
The r.m.s. voltage of the wave form shown is A. 10 VB. 7 VC. 6.37 VD. none of the above |
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Answer» Correct Answer - A `V_(rms) = sqrt(V_(av)) = sqrt(1/Tint_(0)^(T)V^(2)dt)) = sqrt(1/Tint_(0)^(T)10^(2)dt)=10V` |
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| 993. |
An altenating voltage of 220volts r.m.s at a frequency of 40 cycles/sec is supplied to a circuit containing a pure inductance of 0.01H and a pure resitance of 6ohm in series. Calculate (i) the current, (ii) potential difference across the resistance, (iii) potential difference across the inductance, (iv) the time lag, (v) power factor. |
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Answer» `(i) z=sqrt((omegaL)^(2)+R^(2))=sqrt((2pixx40xx0.01^(2))^(2)+6^(2))=sqrt((42.4))` `I_(rms)=(220)/(z)=33.86"amp"` `(ii) V_("rms")=I_("rms")xxR=202.98"volts"` `(iii) mLxxI_("rms")=96.86"volts"` `(iv) t=T(phi)/(2pi)=0.01579"sec"` `(v)=cos phi=(R)/(Z)=0.92` |
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| 994. |
In an a.c. circuit, the instantaneous e.m.f. and current are given by e = 100 sin 30 t `i = 20 sin( 30t-(pi)/(4))` In one cycle of a.c., the average power consumed by the circuit and the wattless current are, respectively :A. `(50)/(sqrt2),0`B. 50,0C. 50,10D. `(1000)/(sqrt2),10` |
| Answer» Correct Answer - C | |
| 995. |
Using an ac voltmeter, the potential difference in the electrical line in a house is read to be 234 V. If the line freqency is known to be 50 cycles per second, the equation for the line voltage isA. 165 sin `(200pit)`B. 234 sin `(100pit)`C. 331 sin `(100pit)`D. 440 sin `(200pit)` |
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Answer» Correct Answer - C Peak voltage, `V_(0) = sqrt(2)V_(rms) = 331V` `omega=2pif=100pi` `rArr V_(L) = 331sin(100pit)` Where, `V_(L)` is line voltage. |
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| 996. |
Using an ac voltmeter, the potential difference in the electrical line in a house is read to be 234 V. If the line freqency is known to be 50 cycles per second, the equation for the line voltage isA. (A) `V=165 sin (100 pi t)`B. (B) `V=331 sin (100 pi t)`C. (C) `V=220 sin (100 pi t)`D. (D) `V=440 sin (100 pi t)` |
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Answer» Correct Answer - B `E=(E_0) sin omega t ` Voltage read is rms value. `:. (E_0)= sqrt(2) xx 234 V =331 V` amd ` omega t = 2 pi f t = 2 pi zz 50 xx t = 100 pi t ` Thus, the equation of the line voltage is given by `E=331 sin (100 pi t)`. |
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| 997. |
An alternating voltage of `220` volts `r.m.s.` at a freqency of `40` cycles/sec is supplied to a circuit containing a pure inductance of `0.01 H` and a pure resistance of `6 ohms` in series.Calculate (i)the current , (ii)potential diffence across the resistance, (iii)potential difference across the inductance, (iv)the time lag, (v)power factor. |
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Answer» (i)`z=sqrt((omegaL)^(2)+R^(2))=sqrt((2pixx40xx0.01^(2))^(2)+6^(2))=sqrt((42.4))` `I_(rms)=220/z=33.83 amp` (ii)`V_(rms)=I_(rms)xxR=202.98` volts , (iii)`omegaLxxI_(rms)=96.83` volts (iv)`t=T pi/(2pi)=0.01579 sec` , (v)`cos phi=R/Z=0.92` |
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| 998. |
An inductor has a resistance `R` inductance `L`. It is connected to an `A.C` source of e.m.f. `E_(V)` and angular frequency `omega` then the current `I_(v)` in the circuit isA. `(E_(V))/(omega L)`B. `(E_(V))/(R )`C. `(E_(V))/(sqrt(R^(2) + omega^(2) L^(2)))`D. `sqrt(((E_(V))/(R ))^(2) + ((E_(V))/(omega L))^(2))` |
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Answer» Correct Answer - 3 `i= (E_(0))/(sqrt(R^(2) + X_(L)^(2))), X_(L) = L omega` |
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| 999. |
A radio tuner has a frequency range from `500 kHz` to `5 MHz`. If its `LC` circuit has an effective inductance of `200 mu H`, what is the range of it varialbe capacitors? (Take `pi^(2) = 10 )`.A. 2.5 pF to 250 pFB. 5.0 pF to 500 pFC. 7.5 pF to 750 pFD. 10 pF to 1000 pF |
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Answer» Correct Answer - A Resonant frequency `= (1)/(2 pi sqrt(LC))` |
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| 1000. |
Using an ac voltmeter, the potential difference in the electrical line in a house is read to be 234 V. If the line freqency is known to be 50 cycles per second, the equation for the line voltage isA. `V = 165 sin (100 pi t)`B. `V = 331 sin (100 pi t)`C. `V = 220 sin (100 pi t)`D. `V = 440 sin (100 pi t)` |
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Answer» Correct Answer - 2 `E = E_(0) sin omega t`, voltage read is r.m.s value `E_(0) = sqrt(2) xx 234 V = 331` volt and `omega t = 2 pi nt = 2 pi xx 50 xx t = 100 pi t` Thus, the eqn of line voltage is given by `V = 331 sin (100 pi t)` |
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