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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.
| 851. |
A `0.7` henry inductor is connected across a `120 V-60 Hz AC` source. The current in the inductor will be very nearlyA. `4.55 amp`B. `0.355 amp`C. `0.455 amp`D. `3.55 amp` |
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Answer» Correct Answer - C `i=V/(X_(L))=120/(2xx3.14xx60xx0.7)=0.455A` |
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| 852. |
An iron cored coil is connected in series with an electric bulb, with an AC source, as shoen in figure. As the iron piece is taken out of the coil, how will tthe brightness of bulb challenge ? |
| Answer» As the iron rod is taken out of the coil, the self-inductance of the coil decreases, the impedance of circuit decreases, hence current increases. So, power consumed (and hence brightness of bulb) by bulb increases. | |
| 853. |
An inductor of `1` henry is connected across a `220 v, 50 Hz` supply. The peak value of the current is approximately. |
| Answer» `i_(0) = (E_(0))/(X_(L)) = (sqrt(2)E_("rms"))/(omega L) = (sqrt(2)E_("rms"))/(2 pi f L) = (sqrt(2)(220))/(2 pi xx 50 xx 1) = 0.99 A` | |
| 854. |
A coil having an inductance of `1//pi` henry is connected in series with a resistance of `300 Omega`. If 20 volt from a 200 cycle source are impressed across the combination, the value of the phase angle between the voltage and the current is :A. 5/4.B. 4/5.C. 3/4.D. 4/3. |
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Answer» Correct Answer - D tan`phi` = `underset(L)(X)`/R ( In case of L-R circuit) L`omega`/R = L`2pi`f/R = `2 xx 200 / 300 ` = 4/3 |
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| 855. |
Find the maximum value of current when inductance of two henry is connected to 150 volt, 50 cycle supply. |
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Answer» Here, `L = 2H, E_("rms") = 150 V, f - 50 Hz` `X_(L) = L omega = L xx 2 pi f = 2 xx 2 xx 3.14 xx 50 = 628` ohm `RHS` value of current of thorugh the inductor `I_("rms") = (E_("rms))/(X_(L)) = (150)/(628) = 0.24 A` Maximum value (or peak value) of current is given by `I_("rms") = (I_(0))/(sqrt(2))` or `I_(0) = sqrt(2) I_("rms) 1.414 xx 0.24 = 0.339 A` |
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| 856. |
An alternating voltage `E=200sqrt(2) sin (100 t)V` is connected to a `1 (mu)F` capaacitor through an ac ammeter (it reads rms value). What will be the reading of he ammeter? |
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Answer» Correct Answer - 20mA `X_(C ) =(1)/(omega C) =(1)/(100xx10^(-6)) =(10^4) (Omwega)` As ac instruments read rms value, The reading of ammeter will be, `I_(rms)=(E_(rms))/(X_C) =(E_0)/(sqrt(2)X_(C ))` `{as E_(rms)=(E_0)/(sqrt(2))}` i.e. `I_(rms)=(200 sqrt(2))/(sqrt(2)xx10^(4)) = 0.02A = 20mA`. |
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| 857. |
The average value of current `i=I_(m) sin omega t from t=(pi)/(2 omega )` to `t=(3 pi)/(2 omega)` si how many times of `(I_m)`? |
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Answer» `ltIgt =(underset(0)overset((2pi)/(omega))int sin omegatdt)/((2pi)/(omega))=((I_(m))/(omega)(1-cos omega(2pi)/(omega)))/((pi)/(omega))=(2I_(m))/(pi) ` (ii) `ltIgt =(underset((pi)/(2omega))overset((2pi)/(omega))int I_(m) sin omegatdt)/((pi)/(omega))=0` |
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| 858. |
An alternating voltage `E=200sqrt(2) sin (100 t)V` is connected to a `1 muF` capacitor through an ac ammeter (it reads rms value). What will be the reading of he ammeter?A. `10 mA`B. `40 mA`C. `80 mA`D. `20 mA` |
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Answer» Correct Answer - 4 `I_("rms") = (E_("rms"))/(X_(C )) = (E_(0) omega C)/(sqrt(2))` |
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| 859. |
An alternating voltage `E=200sqrt(2) sin (100 t)V` is connected to a `1 muF` capacitor through an ac ammeter (it reads rms value). What will be the reading of he ammeter? |
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Answer» Comparing `E = 200 sqrt(2) sin (100 t)` with `E = E_(0) sin omega t, E_(0) = 200 sqrt(2) V` and `omega = 100 ("rad"//s)` `X_(C ) = (1)/(oemga C) = (1)/(100 xx 10^(-6)) = 10^(4) Omega` `I_("rms") = (E_("rms"))/(Z) = (E_(0))/(sqrt(2)X_(C )) = (200 sqrt(2))/(sqrt(2) xx 10^(4)) = 20 mA` |
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| 860. |
An alternating voltage `E=200sqrt(2) sin (100 t)V` is connected to a `1 (mu)F` capaacitor through an ac ammeter (it reads rms value). What will be the reading of he ammeter?A. `10 mA`B. `5 mA`C. `5 sqrt(2)mA`D. `10 sqrt(2)mA` |
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Answer» Correct Answer - B `X_(C) = 1/(omega C) =(1)/(100 xx 10 ^(-6)) = 10^(4) Omega` so, ` I_(rms) = (E_(rms))/(Z) = (50 sqrt(2))/(sqrt(2) xx 10^(4)) = 5 mA`. |
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| 861. |
When a voltage `v_(s)=200sqrt2 sin (omega t=15^(@))` is applied to an AC circuit the current in the circuit is found to be `i=2som (omegat+pi//4)` then average power consumed in the circuit isA. 200 wattB. `400sqrt2 watt`C. `100sqrt6 watt`D. `200sqrt2watt` |
| Answer» `P_("av")=V_("rms") I_("rms") cos theta=(200sqrt2)/(sqrt2).(2)/(sqrt2)cos (30^(@))=100sqrt6 "watt"` | |
| 862. |
A capacitor acts as an infinite resistance forA. `DC`B. `AC`C. `DC` as well as `AC`D. neither `AC` nor `DC` |
| Answer» `X_(C)=1/(omegaC)`for `DC_(omega)=0`.so, `x_(C)=oo` | |
| 863. |
Find the effective value of current. `i=2 sin 100 (pi)t + 2 cos (100 pi t + 30^(@))`. |
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Answer» The equation can be written as `I = 2 sin 1000 pit +2 sin(100 pit+90^(@))`. so phase difference `phi=120^(@)` `I_(m))_(res)=sqrt(A_(1)^(2)+A_(2)^(2)+2A_(1)A_(2)costheta)` `=sqrt(4+4+2xx2xx2(-1/2))=2`so effective value or `rms` value =`2//sqrt2=sqrt2A` |
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| 864. |
An AC source producing `emf` `epsilon = epsilon_0 [cos(100 pi s^(-1)) t + cos (500 pi s^(-1))t]` is connected in series with a capacitor and a resistor. The steady-state current in the circuit is found to be `I = i_1 cos [(100 pi s^(-1) t + varphi_1] + i_2 cos [(500 pi s^(-1))t +phi_2]`.A. `i_(1) ge i_(2)`B. `i_(1) = i_(2)`C. `i_(1) le i_(2)`D. the information is insufficient to find the relation between `i_(1)` and `i_(2)`. |
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Answer» Correct Answer - C `i_(1) = (in_(0))/(sqrt(R^(2) + ((1)/(100 pi C))^(2))) , i_(2) = (in_(0))/(sqrt(R^(2) + ((1)/(500 pi C))^(2)))` Clearly `i_(1) lt i_(2)` |
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| 865. |
Choose the currect option .A. In an ac circuit having resistance only voltage and current are in same plane .B. In a ac circuit having inductance only voltage leads the current by `pi//2`C. In a ac circuit having capacitance only current leads the voltage by `pi//2`D. All |
| Answer» Correct Answer - D | |
| 866. |
An inductor-coil having some resistance is connected to an AC source. Which of the following quantities have zero average value over a cycle?A. induced emf in the inductorB. currentC. joule heatD. magnetic energy stored in the indcutor |
| Answer» Correct Answer - A::B::C::D | |
| 867. |
The peak voltage in a `220 V AC` source isA. `220V`B. about 160VC. about 310VD. 440 V |
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Answer» Correct Answer - C `V_(0) =sqrt2 V_(rms) = -220sqrt2 = 310 V` . |
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| 868. |
An AC source producing `emf` `epsilon = epsilon_0 [cos(100 pi s^(-1)) t + cos (500 pi s^(-1))t]` is connected in series with a capacitor and a resistor. The steady-state current in the circuit is found to be `I = i_1 cos [(100 pi s^(-1) t + varphi_1] + i_2 cos [(500 pi s^(-1))t +phi_2]`.A. `i_(1)gti_(2)`B. `i_(1) =i_(2)`C. `i_(1)lt i_(2)`D. none |
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Answer» Correct Answer - C Impedance of `RC` circuit `Z = sqrt(R^(2) + ((1)/(omegaC))^(2)` `i_(1) = (E_(0))/(sqrt(R^(2) + ((1)/(100piC))^(2))) i_(2) = (E_(0))/(sqrt(R^(2) + ((1)/(500 pi C))^(2)))` `i_(1) lt i_(2)` . |
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| 869. |
What reading would you expact of a square-wave current, suitching rapodly between +0.5 A and -0.5 A, when passed through an ac ammeter? |
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Answer» Correct Answer - B `(0.5)^(2)R((T)/(2))+(0.5)^(2)R((T)/(2))=I_(rms)^(2) RT` or `I_(rms)=1/2 A = 0.5 A`. |
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| 870. |
A constant current of `2.8 A` exists in a resistor. The rms current isA. `2.8A`B. about 2AC. `1.4A`D. none |
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Answer» Correct Answer - A `i_(rms) = i_(dc) = 2.8A` . |
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| 871. |
The magnetic field energy in an inductor changes from maximum value to minimum value in `5.0 ms` when connected to an AC source. The frequency of the source isA. `20HZ`B. `50HZ`C. `200HZ`D. `500HZ` |
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Answer» Correct Answer - B `U = (1)/(2) Li^(2)` U is maximum when I is maximu `= i_(0)` U is minimum when `I =0` Time taken by current to change its value from `I = i_(0) to I = 0 is T//4` `(T)/(4) = 5 xx 10^(-3) implies T = 0.02 sec` `f = (1)/(T) = (1)/(0.02) =50` Hz . |
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| 872. |
The peak voltage in a `220 V AC` source isA. `200 V`B. about `160 V`C. about `310 V`D. `440 V` |
| Answer» `V_(0)=sqrt2 V_(rms)=sqrt2xx220=330 V` | |
| 873. |
What is the r.m.s. value of an alternating current which when passed through a resistor produces heat which is thrice of that produced by a direct current of `2` amperes in the same resistor?A. `6A`B. `3A`C. `2A`D. `2sqrt3A` |
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Answer» Correct Answer - D `i_(rms)^(2) R = 3 i_(dc)^(2) R` `i_(rms) = i_(dc) sqrt3 = 2 sqrt3 A` . |
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| 874. |
An AC source is rated `220 V, 50 Hz`. The average voltage is calculated in a time interval of `0.01 s`. ItA. must be zeroB. may be zeroC. is never zeroD. is `(220//sqrt2V` |
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Answer» Correct Answer - B may be zero |
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| 875. |
An AC source is rated `220 V, 50 Hz`. The average voltage is calculated in a time interval of `0.01 s`. ItA. must be zeroB. may be zeroC. is never zeroD. is `200//sqrt2V` |
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Answer» Correct Answer - B `f = 50 Hz, T = (1)/(f) = (1)/(50) =0.02 sec` if `t =0 to t = (T)/(2) bar(V) = 0` `if t = (T)/(4) to t = (3T)/(4) bar V =0` . |
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| 876. |
When a voltage `v_(S)=200sqrt2 sin (omega t+15^(@))` is applied to an `AC` circuit the current in the circuit is found to be `i=2 sin (omegat+pi//4)` then average power concumed in the circuit is (A)`200` watt , (B)`400sqrt2` watt , (C )`100sqrt6` watt , (D)`200sqrt2` watt |
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Answer» `P_(av)=v_(rms)I_(rms)cos phi` `=(200sqrt2)/sqrt2 2/sqrt2.cos(30^(@))=100sqrt6` watt |
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| 877. |
The heat produced in a given resistor in a given time by the sinusoidal current `I_(0)` sin `omegat` will be the same as that by a steady current of magnitude .A. `I_(0)/(sqrt2`B. `I_(0)`C. `I_(0)sqrt2`D. `I_(0)/2` |
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Answer» Correct Answer - A Heat produced by rms value of ac is equal to heat produced by dc `i_(rms) = i_(dc)` `I_(0)/(sqrt2) = i_(dc)` . |
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| 878. |
An AC source is rated `220 V, 50 Hz`. The average voltage is calculated in a time interval of `0.01 s`. ItA. must be zeroB. may be zeroC. is never zeroD. is `(220//sqrt2)V` |
| Answer» Correct Answer - May be zero | |
| 879. |
When a voltage `V_(s)=200 sqrt(2) sin (100 t) V` is applied to an ac circuit the current in the circuit is found to be `i=2 sin [omega t + (pi//4)]A`. Find the average power consumed in the circuit. |
| Answer» `P=V_("RMS") I_("RMS")cos phi=(200sqrt2)/(sqrt2)xx(2)/(sqrt2)xx"cos" (pi)/(4)=200W` | |
| 880. |
An alternating current having peak value `14 A` is used to heat a metal wire. To produce the same heating effect, a constant current `i` can be used where `i` isA. `14A`B. about `20A`C. `7A`D. about `10A` |
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Answer» Correct Answer - D `I = I _(0)/(sqrt2) = (14)/(1.4) =10 A` |
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| 881. |
Find the effective value of current `i=2+4 cos 100 pi t`. |
| Answer» `I_(rms)=[int_(0)^(T)((2+4cos 100pit)^(2)dt)/T]^(1//2)=2sqrt3` | |
| 882. |
An alternating current having peak value `14 A` is used to heat a metal wire. To produce the same heating effect, a constant current `i` can be used where `i` isA. `14 A`B. about `20 A`C. `7 A`D. about `10 A` |
| Answer» `I_(RMS)=I_(0)/sqrt2=14/sqrt2=10` | |
| 883. |
An inductor (L=2000mH) is connected to an AC source of peak current. What is the instantaneous voltage of the source when the current is at its peak value? |
| Answer» Because phase difference between voltage and current is `pi//2` for pure inductor. So, Ans is zero | |
| 884. |
An alternating current having peak value `14 A` is used to heat a metal wire. To produce the same heating effect, a constant current `i` can be used where `i` isA. 14AB. about 20AC. 7AD. about 10A |
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Answer» Correct Answer - D `I_("RMS")=(I_(0))/(sqrt2)=(14)/(sqrt2)=10` |
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| 885. |
The peak value of an alternating current is `5 A` and its frequency is `60 Hz`. Find its rms value. How long will the current take to reach the peak value starting from zero? |
| Answer» `I_(rms)=I_(0)/sqrt2=5/sqrt2A,t=T/4=1/240s` | |
| 886. |
In the question number 11, the peak voltage of the source isA. 305 VB. 310 VC. 311 VD. 315 V |
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Answer» Correct Answer - C `V_(m)=sqrt(2)V_("rms")=220 sqrt(2) V~~ 311 V`. |
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| 887. |
The peak value of an alternating current is 5A and its frequency is 60Hz. Find its rms value. How long will the cukrgrenQt current IS 5 A and its frequency is 60 Hz. Find its runs value. |
| Answer» `I_("rms")=(I_(0))/(sqrt2)=(5)/(sqrt2)A, t=(T)/(4)=(1)/(240)s` | |
| 888. |
An ac source is of `(200)/(sqrt(2))` V, 50 Hz. The value of voltage after `(1)/(600)s` from the start isA. 200 VB. `(200)/(sqrt(2))V`C. 100 VD. 50 V |
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Answer» Correct Answer - C `V_("rms")=(200)/(sqrt(2))V` `V_(0)=sqrt(2)xx(200)/(sqrt(2))=200 V` `V=V_(0)sin 2pi upsilon t=200 sin (2pixx50xx(1)/(600))` `=200 "sin"(pi)/(6)=200xx(1)/(2)=100 V` |
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| 889. |
An ac source of voltage `V=V_(m)sin omega t` is connected across the resistance R as shown in figure. The phase relation between current and voltage for this circuit is A. both are in phaseB. both are out of phase by `90^(@)`C. both are out of phase by `120^(@)`D. both are out of phase by `180^(@)` |
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Answer» Correct Answer - A The given circuit is a pure resistive circuit. In this circuit the voltage and current both are in phase. |
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| 890. |
The rms value and frequency of an ac current are 5A and 50 Hz respectively. The value of the current after `1/300s` from the time when its value becomes zero isA. `5sqrt2A`B. `5sqrt(3/2)A`C. `5/6A`D. `5/sqrt2A` |
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Answer» Correct Answer - B `I=I_0sinomegat=5sqrt2(100pitimes1/300)` `5sqrt2timessqrt3/2=5sqrt(3/2)A` |
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| 891. |
The line the draws power supply to your house from street hasA. `220sqrt(2)V` average voltage.B. 220 V average voltage.C. Voltage and current out of phase by `pi//2`.D. Voltage and current possibly differing in phase`phi`such that `|phi|lt (pi)/(2)`. |
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Answer» Correct Answer - D As the line has some resistance `(R ne 0)`, voltage and current differ in phase `phi` such that `|phi| lt (pi)/(2)`. |
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| 892. |
The line that draws power supply to your house hasA. zero average currentB. 220 V average voltageC. voltage and current out of phase by `90^@`D. voltage and current possibly differing in phase `phi` such that `|phi|ltpi/2` |
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Answer» Correct Answer - A::D Since the line drawc ac, average current is zero, Again since the line has some resistance (`Rne0`), there is some phase difference between the voltage and current. |
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| 893. |
When an ac voltage of 220 V is applied to a capacitor CA. the maximum voltage between plated is 220VB. the current is in phase with the applied voltageC. the charge on the plates is in phase with the applied voltageD. power delivered to the capacitor is zero |
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Answer» Correct Answer - C::D `P=V_(rms)I_(rms)cosphibecauseP=0[thereforephi=90^@]` |
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| 894. |
When `100`volts `DC` is applied across a coil, a current of `1amp` flows through it. When `100` volt `AC` at `50` cycles `s^(-1)` is applied to the same coil, only `0.5` ampere current follows.A. `100 Omega`B. `200 Omega`C. `300 Omega`D. `400 Omega` |
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Answer» Correct Answer - B When DC is supplied `R=V/i=100/1=100Omega` When AC is supplied `Z=V/i=100/0.5=200Omega` |
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| 895. |
An LCR series ac circuit is at resonance with 10 V each across L, C and R. If the resistance is halved, the respective voltage across L, C and R areA. 10 V, 10 V and 5 VB. 10 V, 10 V and 10 VC. 20 V, 20 V and 5 VD. 20 V, 20 V and 10 V |
| Answer» Correct Answer - D | |
| 896. |
Voltage and current is an AC circuit are given by `V=sin(100pit-(pi)/6)` and `I=4sin(100pit+(pi)/6)`A. Voltage leads the current by `30^(@)`B. Current leads the current by `30^(@)`C. Current leads the current by `60^(@)`D. Voltage leads the current by `60^(@)` |
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Answer» Correct Answer - C Phase difference `Deltavarphi=varphi_(2)-varphi_(1)=(pi)/6-((-pi)/6)=(pi)/3` |
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| 897. |
At resonant frequency the current amplitude in series LCR circuit isA. maximumB. minimumC. zeroD. infinity |
| Answer» Correct Answer - A | |
| 898. |
When `100`volts `DC` is applied across a coil, a current of `1amp` flows through it. When `100` volt `AC` at `50` cycles `s^(-1)` is applied to the same coil, only `0.5` ampere current follows.then the impedence of the coil isA. `100 Omega`B. `200 Omega`C. `300 Omega`D. `400 Omega` |
| Answer» Correct Answer - b | |
| 899. |
A resistance of `20ohms` is connected to a source of an alternating potential `V=220sin(100pit)`. The time taken by the current to change from its peak value to r.m.s. value isA. `0.2 sec`B. `0.25 sec`C. `25xx10^(-3) sec`D. `2.5xx10^(-3) sec` |
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Answer» Correct Answer - D Peak value to r.m.s value means, current becomes `1/(sqrt(2))` times. So form `i=i_(0)sin100pit implies 1/(sqrt(2))xxi_(0)=i_(0)sin100pit` `implies sin(pi/4)=sin100pit implies t=1/400sec=2.5xx10^(-3)sec` |
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| 900. |
In a series resonant circuit, the `AC` voltage across resistance `R`, inductance `L`, and capacitance `C` are `5V,10V` and `10V` respectively. The `AC` voltage applied to the circuit will beA. 20 VB. 10 VC. 5 VD. 25 V |
| Answer» Correct Answer - c | |