InterviewSolution
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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.
| 1. |
If a circuit made up of a resistance 1 `Omega` and inductance 0.01 H, an alternating emf of 200 voit at 50 Hz is connected, then find the phase difference between the current and the emf in the circuit.A. `tan^(-1)(pi)`B. `tan^(-1)((pi)/(2))`C. `tan^(-1)((pi)/(4))`D. `tan^(-1)((pi)/(3))` |
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Answer» (a)`"tan"phi=((X_(L))/(R))` and `X_(L)=OmegaL=(2pifL)=(2pi)(50)(0.01)=piomega` also `R=1Omegaimplies phi="tan"^(-1)(pi)` |
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| 2. |
An Ac source of volatage V=100 sin `100pit` is connected to a resistor of ressistance 20` Omega`The rsm value of current through resistor isA. `10 A`B. `10/sqrt 2 A `C. `5/sqrt 2 A`D. non of these |
| Answer» (c) `I_(0) = (V_(0))/(R) = (100)/(20) = 5 A` therefore `I_(rms) = (I_(0))/(sqrt 2) = (5)/(sqrt 2) A` | |
| 3. |
The time taken by the current to rise to 0.63 of its maximum value in a DC circuit containing inductance (L) and resistance ® depends onA. L onlyB. R onlyC. `L/R`D. LR |
| Answer» (c) It depends upon its time constant, `tau =(L)/(R)`. | |
| 4. |
The reactance of an inductor connected with DC voltage isA. zeroB. `infty`C. 1`Omega`D. None of these |
| Answer» (a)`X_(L)=omegaL=0xxL=0` | |
| 5. |
An Ac voltage is represented by e=220 sin `(100pi)` t volt and is applied over a resistance of 110 ohm. Calculate the heat produced in 7 min.A. `11xx10^(3)` calB. `22xx10^(3)` calC. `33xx10^(3)` calD. `25xx10^(3)` cal |
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Answer» (b)`H=l_(v)^(2)(RT)/J cal` `=((sqrt2)^(2)xx110xx7xx60)/4.2 ( :.l_(rms)=(l_(0))/sqrt2=(220//110)/sqrt2=sqrt2)` `=22xx10^(3) cal` |
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| 6. |
A current l=3+8 sin 100t is passing through a resistor of resistance 10 `Omega`.The effective value of current isA. `5 A`B. `10 A`C. `4sqrt2 A`D. 3sqrt2 A |
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Answer» `(a) l_(effective)^(2)=l_(rms) (overset(T)underset(0)int l^(2) dt)/(overset(T)underset(0)int dt)` After solving,` l_(effective)^(2)=l_(rms)=5A` |
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| 7. |
A metallic square loop ABCD is moving in its own plane with velocity v in a uniform magnetic field perpendicular to its plane as shown in the figure. An electric field is induced A. in AD,but not in BCB. in BC,but not in ADC. neither in AD nor in BCD. in both AD and BC |
| Answer» (d) In both AD and BC, because the electrons and positive ions will be separated by magnetic force in these two arms. | |
| 8. |
An Ac source of volatage V=100 sin `100pit` is connected to a resistor of ressistance 20` Omega`The rsm value of current through resistor is ,total heat generated in one cycle isA. `sqrt2 J`B. 5 JC. `4sqrt2 J`D. zero |
| Answer» (b) `DeltaH = I_(rms)^(2) RT = ((5)/(sqrt 2))^(2) xx 20 xx (1)/(50) = 5J` | |
| 9. |
An Ac source of volatage V=100 sin `100pit` is connected to a resistor of ressistance 20` Omega`The rsm value of current through resistor is total charge transferred through resistor in long time isA. zeroB. `(2l_(0))/(pi)`C. `(l_(0))/(25pi)`D. non of these |
| Answer» (a) `q = int_(0)^(t = 1//50s) I "dt" = int_(0)^(t = 1//50s) 100sin 100 pit = 0` | |
| 10. |
An Ac source of volatage V=100 sin `100pit` is connected to a resistor of ressistance 20` Omega`The rsm value of current through resistor is ,total charge transferred in 1/100 second isA. `1/(10 pi) C`B. `1/(5 pi) C`C. zeroD. non of these |
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Answer» (a) When `t le (T)/(2), i.e. t le (1)/(100)s` Then, `q = I_(av)t = (2I_(0))/(pi)t = (10)/(pi) xx (1)/(100) = (1)/(10pi)` |
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| 11. |
An Ac source of volatage V=100 sin `100pit` is connected to a resistor of ressistance 20` Omega`The rsm value of current through resistor is the averge value for half cycle isA. `10/pi A`B. `5/pi A`C. zeroD. non of these |
| Answer» (a) `I_(av) = (2I_(0))/(pi) = (2 xx 5)/(pi) = (10)/(pi)A` | |
| 12. |
Caluculate the peak and rms value of current in AC circuit. The current is represented by the eqution `i=5"sin"(300t-(pi)/4),where t is in second and I in ampere.A. 5 A,3.535 AB. 5 A,5.53 AC. 3 A,3.53 AD. 6.25 A, 5.33 A |
| Answer» Correct Answer - (a) | |
| 13. |
An Ac source of volatage V=100 sin `100pit` is connected to a resistor of ressistance 20` Omega`The rsm value of current through resistor is average value of current for half cycle isA. zeroB. `5/sqrt 2 A`C. `10 A`D. non of these |
| Answer» (a) `I = (V)/(R) = (100)/(20) sin100pit = 5sin100pit` therefore `I = (5int_(0)^(T) sin100pit "dt")/(int_(0)^(T) "dt") = 0` | |
| 14. |
An Ac source of volatage V=100 sin `100pit` is connected to a resistor of ressistance 20` Omega`The rsm value of current through resistor is , power factor isA. 1B. 0C. `1/2`D. non of these |
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Answer» (a) `"cos " phi = (R)/(Z) = (R)/(sqrt(R+ (X_(L)-X_(C))^(2))) = (R)/(sqrt(R^(2)+ 0)) = 1` `rArr cos phi = 1` |
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| 15. |
The average value for half cycle in a 200 V AC source isA. 180 VB. 200 VC. 220 VD. none |
| Answer» (a) `V_(rms) = (V_(0))/(sqrt(2))` therefore `V_(0) = sqrt2 V_(rms)` therefore `V_(av) = (2V_(0))/ (pi) = (2)/(pi) sqrt 2V_(rms) = (2sqrt2 xx 200)/(pi) = 180 V` | |
| 16. |
An AC source is connected in parallel with an L-C-R circuit as shown.Let `l_(s), l_(L), l_(C) " and " l_(R)` denote the currents through and `V_(s), V_(L), V_(C) " and "V_(R)`the voltage across the corresponding componts.Then,A. `l_(S)=l_(L)+L_(C)+l_(R)`B. `V_(S)=V_(L)+V_(C)+V_(R)`C. `(l_(L), l_(C), l_(R))ltl_(S)`D. `l_(L), l_(C)` may be greater than `l_(S)` |
| Answer» (d) In parallel resonant circuit, current through L and C may be greater than the source current , i.e. `I_(L), I_(C)` may be greater than `L_(s)` | |
| 17. |
An Ac circuit with f=1000 Hz consists of a cail of 200 mH and negligible resistance. Calculate the voltage across the coil, if the effective current of 5 mA is flowing.A. `7.64` V (rms)B. `7.452` V (rms)C. `6.28` V (rms)D. `74.62` V (rms) |
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Answer» (c) Inductive reactance , `X_(L)=Lomega=2pifL` also , `(E)/(I)=Lomega=X_(L)` or `E=IX_(L) implies E=(5)/(1000)xx1256` `E=6.28` volt (rms) |
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| 18. |
Find the energy of photon of electromagnetic radiation of wavelength 200 Å.A. `1.76xx10^(-18)J`B. `0.99xx10^(-18)J`C. `0.54xx10^(-18)J`D. `0.63xx10^(-18)J` |
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Answer» (b) `E=(hc)/(lambda)=(6.62xx 10^(-34)xx3xx 10^(8))/(2000xx10^(-10))=0.0099xx 10^(-16)` `E=0.99xx 10^(-18) J` |
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| 19. |
Calculate the wavelength of a radio wave of frequency of 1 MHz.A. 400 mB. 300 mC. 350 mD. 200 m |
| Answer» Correct Answer - (b) | |
| 20. |
An object is placed at some distance from a radio station. If the interval between transmission and reception of pulses is `2.66xx10^(-2)`S, then find the distance.A. 4000 kmB. 2000 kmC. 3000 kmD. 2500 km |
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Answer» (a) `2x=ct` `x=(ct)/(2)=(3xx10^(8)xx2.66xx10^(-12))/(2)implies x ~=4000 Km` |
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| 21. |
A radio wave of intensity / isn reflcted by a surface. Find intersity (l), if pressure exerted on the surface is `2xx10^(-8)N//m^(2)`.A. `3N//m^(2)`B. `4N//m^(2)`C. `6N//m^(2)`D. `7N//m^(2)` |
| Answer» Correct Answer - (a) | |
| 22. |
A TvV tower has a height of 100 m. Find the area covered by the TV broadast, if radius of the earth is 6400 km.A. `380xx10^(7)m^(2)`B. `402xx10^(7)m^(2)`C. `595xx10^(7)m^(2)`D. `440xx10^(7)m^(2)` |
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Answer» (b) Area `=pir^(2)=pi=((2hR)^(1//2))^(2)` Area `=2piRh=2xxpixx6400xx 10^(3)xx100=402 xx 10^(7)m^(2)` |
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| 23. |
Find the speed of light in air, if an electromagnetic wave is travelling in air whose dielectric constant is `K=1006.`A. `3xx10^(8)" "m//s`B. `3.88xx10^(8)" "m//s`C. `2.5xx10^(8)" "m//s`D. `4.6xx10^(8)" "m//s` |
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Answer» (a) `v=(1)/(sqrt(mu_(0)kepsi_(0)))implies (c)/(sqrt(k))` `v=(c)/(sqrt(k))=(3xx10^(8))/(sqrt(1.006))implies v~=3xx10^(8)m//s` |
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| 24. |
An electromagnetic wave with pointing vector 5 `W//m^(2)`is absorbed by a surface of same area. If the force on the surface is `10^(-7)`N, then area isA. `6m^(2)`B. `3m^(2)`C. `60m^(2)`D. `4m^(2)` |
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Answer» (a) We know `,P=(I)/(c)=(5)/(3xx10^(8))` and `F=PxxA implies A=(F)/(P)` `A=(10^(-7))/(5)xx3 xx10^(8)implies A=6 m^(2)` |
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| 25. |
The electric in an electromagetic wave is given by E=(100 N/C)`sinomega(t-(X)/(C ))`. If the energy contained in a cylinder of coss-section `10 cm^(2)` and length 50 cm along the X-axis is `4.4xx10^(-8)J//m^(3)`, then find intensity of the wave.A. `12.4 W//m^(3)`B. `13.2 W//m^(3)`C. `15.7 W//m^(3)`D. `11.9W//m^(3)` |
| Answer» (b) `l=(1)/(2)epsiE_(0)^(2)c=(4.4xx10^(-8))xx(3xx10^(8))=132 W//m^(2)` | |
| 26. |
If at a certain instant, the magnetic induction of the electromagnetic wave in vacuum is `6.7xx10^(-12)`T, then the magnitude of of electric field intensity will beA. `2xx10^(-3)" "N//C`B. `3xx10^(-3)" "N//C`C. `4xx10^(-3)" "N//C`D. `1xx10^(-3)" "N//C` |
| Answer» Correct Answer - (a) | |
| 27. |
If a dry cell of emf=1.5 V is connected across the primary of a step-up transformer of turn ratio 3:5, then calculate the voltage developed across the secondary.A. 30 VB. 5 VC. zeroD. None of these |
| Answer» (c) Transformer is an AC device , it does not work on DC, so output voltage is zero. | |
| 28. |
The number of turns in primary and secondary coils of a transformer is 50 and 200, respectively. If the current in the primary coil is 4 A, then current in the secondary coil isA. `1 A `B. 2AC. 4AD. 5A |
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Answer» (a) In a transformer, `therefore (N_(P))/(N_(s))=(I_(s))/(I_(p))=(50)/(200)=(I_(s))/(4) implies I_(s) =1A` |
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| 29. |
Figure represents two bullbs `B_(1)" and " B_(2)`resister R and inductor L. When the switch S in turned off,then A. both `B_(1)" and "B_(2)` die out promptlyB. both `B_(1) " and " B_(2)` die out with some selayC. `B_(1)` dies out promptly but `B_(2)` with some selayD. B_(2) dies out promptly but `B_(1)` with some delay |
| Answer» (c) Bulb `B_(I)` dies cut out promptly, but self-induced emf across L, during decay. | |
| 30. |
The transformation ratio in the step -up transformer isA. 1B. greater than oneC. less than oneD. the ratio greater or less than sepenods on the other factors |
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Answer» (b) Transformation ratio `, k=(N_(s))/(N_(p))=(V_(s))/(V_(p))` for step-up transformer, `N_(s) gt N_(p),` i.e., `V_(s) gt V_(p)` hence,` k gt 1` |
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| 31. |
The squre root of the prioduct of indutance and capacitance has the dimension ofA. lengthB. massC. timeD. no dimersion |
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Answer» (c) We know `f=(1)/(2pisqrt(LC))` or `sqrt(LC)=(1)/(2pif)`=time Thus, `sqrt(LC)` has the dimension of time. |
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| 32. |
Calculate the maximum current in the circuit, if a capacitor of capacitance `1muF` is charged to a potential of 2 V and is connceted in parallel to an inductor of inductance `10^(-3)H`.A. `sqrt4000" "mA`B. `sqrt2000" "mA`C. `sqrt1000" "mA`D. `sqrt5000" "mA` |
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Answer» (a) charge on capacitor `q_(0)=CV" "q_(0)=2xx10^(-6)C` Here, `q=q_(0)sinomegat` For maximum current` =I_(0)=(dq)/(dt)=omegaq_(0)` `omega=(1)/(sqrtLC)=(10^(9))^(1//2),I_(0)=(10^(9))^(1//2)(2xx10^(-6))` `=sqrt10xx10^(4)xx2xx10^(-6)=2sqrt10xx10^(-2)A=sqrt4000" "mA` |
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| 33. |
In a circuit consisting of inductor (L), capacitor (C) and resistor (R) are in series, if `omegaLlt(1)/(omegaC)`, then the emfA. leads the currentB. lags behind the currentC. is in phase with currentD. is zero |
| Answer» (a) When `omegaLlt(1)/(omegaC),` emf leads the current by `(pi)/(2)`. | |
| 34. |
What is the reactance of a capacitor connected to a constant `DC` source?A. zeroB. `infty`C. 1`Omega`D. None of these |
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Answer» (b) For DC source ,frequency is zero `:.X_(c)=(1)/(omegaC)=1/0xxC=infty` |
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