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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.
| 101. |
The reactance of a coil when used in the domestic AC power supply `(220V,50cycl es)` is `50ohm`. The inductance of the coil is nearlyA. `2.2` henryB. `0.22` henryC. `1.6` henryD. `0.16` henry |
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Answer» Correct Answer - D `X_(L) =2pivLimplies L=(X_(L))/(2piv)=50/(2xx3.14xx50)=0.6H` |
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| 102. |
The reactance of a coil when used in the domestic AC power supply `(220V,50cycl es)` is `100ohm`. The selfinductance of the coil is nearlyA. 3.2 henryB. 0.32 heneryC. 2.2 henryD. 0.22 henry |
| Answer» Correct Answer - b | |
| 103. |
In a `LCR` circuit having `L=8.0` henry, `C=0.5 mu F` and `R=100` ohm in series. The resonance frequency in per second isA. `600` radianB. `600 Hz`C. `500` radianD. `500 Hz` |
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Answer» Correct Answer - C Resonance frequency in radian/second is `omega=1/(sqrt(LC))=1/(sqrt(8xx0.5xx10^(-6)))=500 rad//sec` |
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| 104. |
A circuit has a coil of resistance 60 ohm and inductance 3 henry. It is connected in series with a capacitor of `4 mu F` and `A.C` supply voltage of `200 V` and 50 `"cycle"//"sec"`A. the impedance of the coil is `943 Omega`B. the impedance of the coil is `843 Omega`C. the p.d. across the inductor coil is `1110V`D. the p.d. across the capacitor is `924 V` |
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Answer» Correct Answer - A,C,D `Z = sqrt(R^(2) + (X_(L) - X_(C ))^(2)) , I_("rms") = (V)/(2)` Impedance of coil `= Z_(1) = sqrt(R^(2) + X_(L)^(2))` Potential drop across the coil `= I_("rms") Z_(1) = I_("rms") sqrt(R^(2) + X_(L)^(2))` Potential drop across capacitor `= V_(c ) = I_("rms") X_(C )` |
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| 105. |
In a `LCR` circuit having `L=8.0` henry, `C=0.5 mu F` and `R=100` ohm in series. The resonance frequency in per second isA. 600 radianB. 600 HzC. 500 radianD. 500 Hz |
| Answer» Correct Answer - c | |
| 106. |
An inductive circuit a resistance of `10ohm` and an inductance of `2.0` henry. If an AC voltage of `120`volt and frequency of `60Hz` is applied to this circuit, the current in the circuit would be nearlyA. 0.32 ampB. 0.16 ampC. 0.48 ampD. 0.80 amp |
| Answer» Correct Answer - b | |
| 107. |
An inductive circuit a resistance of `10ohm` and an inductance of `2.0` henry. If an AC voltage of `120`volt and frequency of `60Hz` is applied to this circuit, the current in the circuit would be nearlyA. 0.32 ampB. 0.016 ampC. 0.48 ampD. 0.80 amp |
| Answer» Correct Answer - b | |
| 108. |
An inductive circuit a resistance of `10ohm` and an inductance of `2.0` henry. If an AC voltage of `120`volt and frequency of `60Hz` is applied to this circuit, the current in the circuit would be nearlyA. `0.32 amp`B. `0.16 amp`C. `0.48 amp`D. `0.80 amp` |
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Answer» Correct Answer - B `Z=sqrt(R^(2)+X_(L)^(2))=sqrt(10^(2)+(2pixx60xx2)^(2))=0.016A` `:. i=120/753.7=0.159A` |
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| 109. |
A rectangle loop with a sliding connector of length `l=1.0 m` is situated in a uniform magnetic field `B=2T` perpendicular to the plane of loop. Resistance of connector is `r=2Omega`. Two resistance of `6Omega` and `3Omega` are connected as shown in figure. the external force required to keep the connector moving with a constant velocity `v=2m//s` is A. 6NB. 4NC. 2ND. 1N |
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Answer» Correct Answer - C |
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| 110. |
An equilateral triangular loop `ADC` having some resistance is pulled with a constant velocity `v` out of a uniform magnetic field directed inot the paper. At time `t = 0`, side `DC` of the loop at is at edge of the magnetic field. The induced current `(i)` versus time `(t)` graph will be asA. B. C. D. |
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Answer» Correct Answer - B |
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| 111. |
If the instantaneous current in a circuit is given by `I = 20 cos (omega t + phi) A`, the rms value of the current isA. `2A`B. `sqrt(2) A`C. `2 sqrt(2) A`D. zero |
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Answer» Correct Answer - 2 `I_("rms") = (I_(0))/(sqrt(2))` |
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| 112. |
If the instantaneous current in a circuit is given by `i=2cos (omegat-phi)`ampere, the r.m.s. value of the current isA. `2` ampereB. `sqrt(2)` ampereC. `2sqrt(2)` ampereD. zero ampere |
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Answer» Correct Answer - B `I_(r.m.s.)=1/(sqrt(2))2/(sqrt(2))=sqrt(2)A` |
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| 113. |
The rms and the average value of the voltage wave shown in figure are A. `sqrt((32)/(3)) V , 1V`B. `sqrt((11)/(3)) V , 1V`C. `sqrt((11)/(3)) V , 3V`D. `sqrt((32)/(3)) V , 3V` |
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Answer» Correct Answer - A `i_("ave") = (int idt)/(int dt)` , `i_("rms") = sqrt((int i^(2) dt)/(int dt))` |
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| 114. |
Select the correct options among the following In an `R-C` circuit (a) instantaneous `A.C` is given by `I = I_(0) sin (wt + phi)` (b) the alternating current in the circuit leads the emf by a phase angle `phi`. (c) Its impedance is `sqrt(R^(2) + (omega C)^(2))` (d) It capacitive reactance is `omega c`A. `a,b` are trueB. `b,c,d` are trueC. `c,d` are trueD. `a,c` are ture |
| Answer» Correct Answer - 1 | |
| 115. |
If `i_(1) = i_(0) sin (omega t), i_(2) = i_(0_(2)) sin (omega t + phi)`, then `i_(3) =` A. `sqrt(i_(0_(1))^(2) + i_(0_(2))^(2)) sin {phi + omega t}`B. `(i_(0_(1)) + i_(0_(2))) sin ((phi)/(2) + omega t)`C. `(sqrt(i_(0_(1))^(2) + i_(0_(2))^(2) + 2i_(0_(1)) i_(0_(2)) cos phi)) sin [phi + omega t]`D. `(sqrt(i_(0_(1))^(2) + i_(0_(2))^(2) + 2i_(0_(1)) i_(0_(2)) cos phi)) sin [alpha + omega t]` where `alpha=Tan^(-1) [(i_(0_(2)) sin phi)/(i_(0_(1))+i_(0_(2)) cos phi)]` |
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Answer» Correct Answer - D Apply principle of superposition |
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| 116. |
The average and effective values for the waveshaphe shown in figure are: A. `(2)/(pi) V_(m)` and `(V_(m))/(2)`B. `(V_(m))/(pi)` and `(V_(m))/(sqrt(2))`C. `(2)/(pi) V_(m)` and `(V_(m))/(sqrt(2))`D. `(V_(m))/(pi sqrt(2))` and `(V_(m))/(sqrt(2))` |
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Answer» Correct Answer - C `V_("ave") = (V_(m) int_(0)^(pi) sin theta d theta)/(pi) = (2 V_(m))/(pi)` `V_("eff") = sqrt((V_(m)^(2) int_(0)^(pi) sin^(2) theta d theta)/(pi)) = (V_(m))/(sqrt(2))` |
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| 117. |
An ideal transformer is used to decrease an alternating voltage from 880 V and 220 V. If the number of turns of its primary coil is 4000, then what is that in the secondary coil?A. 16000B. 4000C. 2000D. 1000 |
| Answer» Correct Answer - D | |
| 118. |
r.m.s. value of current `i=3+4 sin (omega t+pi//3)` is:A. `5 A`B. `sqrt17 A`C. `5/sqrt2 A`D. `7/sqrt2 A` |
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Answer» Correct Answer - B `I_(rms)=[(underset(0)overset(T)inti^(2)dt)/T]^(1/2)=[underset(0)overset(T)int[(3+4sin(omegat+pi//3))]^(2)/Tdt]^(1//2)=sqrt17` |
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| 119. |
In a transformer ratio of secondary turns `(N_(2))` and primary turns `(N_(1))`,i.e. `N_(2)/N_(1)=4`.if the voltage applied in primary is `200 V,50 Hz`, find (a)voltage inducted in secondary (b)If current in primary is `1A`, find the current in secondary if the transformer is (i)Ideal and (ii)`80%` efficient and there is no flux leakage. |
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Answer» Correct Answer - (a)`800 V` , (b)(i)`0.25 A` ,(ii)`0.2 A`. (a)`V_(2)/V_(1)=N_(2)/N_(1)=4 , rArrV_(2)=200xx4=800` (b)(i)`I_(2)/I_(1)=N_(2)/N_(1)=1/4 rArr i_(2)=1/4=0.25 A` (ii)`I_(1)V_(1)xx0.80=V_(2)I_(2) rArr I_(2)=0.8xx0.25=0.2 A` (a)`V_(2)/V_(1)=N_(2)/N_(1)=4 , rArrV_(2)=200xx4=800` (b)(i)`I_(2)/I_(1)=N_(2)/N_(1)=1/4 rArr i_(2)=1/4=0.25 A` (ii)`I_(1)V_(1)xx0.80=V_(2)I_(2) rArr I_(2)=0.8xx0.25=0.2 A` |
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| 120. |
in a transformer the number of rurns in the primary and secondary coils are `1000` and `3000` respectively If the primary is connected across `80V` `AC` the potential difference across each turn of the secondary will be .A. `240V`B. `0.24V`C. `0.8V`D. `0.08V` |
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Answer» Correct Answer - D `V_(s)/(V_(p)) = N_(s)/N_(p)` `V_s = N_s/N_(p) V_(p) = (3000)/(1000)xx80 = 240 V` p.d across each turn ` = (240)/(3000) = 0.08V` . |
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| 121. |
A transformer has 50 turns in the primary and 100 in the secondary. If the primary is connected to a `220 V DC` supply, what will be the voltage across the secondary? |
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Answer» Correct Answer - zero Transformer does not work on `D.C.` |
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| 122. |
An ideal transformer is used to step up an alternating emf of `220V` to `4.4kV` to transmit `6.6kW` of power The current rating of the secondary isA. `30A`B. `3A`C. `1.5A`D. `1A` |
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Answer» Correct Answer - C `P_(0) =V_(s) i_(s)` `6.6 =4.4 xx i_(s) implies i_(s) =1.5A` . |
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| 123. |
Assertion : A transformer cannot work on dc supply. Reason : dc changes neither in magnitude nor in direction.A. If both assertion ans reason are true ans reaason 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 istrue but reason is false.D. If both assertion and reason are false. |
| Answer» Correct Answer - A | |
| 124. |
Consider the following two statements `A` and `B` and identify the correct answer. (A) In a transformer a large alternating current at low voltage can be transformed into a small alternating current at high voltage (B) Energy in current carrying coil is stored in the form of magnetic field.A. `A` is true but `B` is falseB. Both `A` and `B` is trueC. `A` is false but `B`d is trueD. Both `A` and `B` are false |
| Answer» Correct Answer - 2 | |
| 125. |
Assertion : A given transformer can be used to step-up ot step-down the voltage. Reason : The output voltage depends upon the ratio of the number of turns of the two coils of the transformer.A. If both assertion ans reason are true ans reaason 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 istrue but reason is false.D. If both assertion and reason are false. |
| Answer» Correct Answer - A | |
| 126. |
For an ideal transformer ratio of output ot the input power is alwaysA. greater than oneB. equal ot oneC. less than oneD. zero |
| Answer» Correct Answer - 2 | |
| 127. |
The magnitude of the e.m.f. across the secondary of a transformer does not depend onA. The number of thef turns in the primaryB. The number of the turns in the secondaryC. The magnitude of the e.m.f applied across the primaryD. The resistance of the primary and the secondary |
| Answer» Correct Answer - 4 | |
| 128. |
Transformers are usedA. d.c circuit onlyB. a.c. circuit onlyC. Both a.c and d.c circuitsD. Integrated circuits |
| Answer» Correct Answer - 2 | |
| 129. |
Assertion (A) : If changing current is flowing through a machine with iron parts, results in loss of energy. Reason (R ) : Changing magnetic flux through an area of the iron parts causes eddy currents.A. Both `A` and `R` are individually true and `R` is the correct explanation of `A`B. Both `A` and `R` are individually true but `R` is not the correct explanation of `A`C. `A` is true but `R` is falseD. Both `A` and `R` are false |
| Answer» Correct Answer - 1 | |
| 130. |
Two impedances `Z_1` and `Z_2` when connected separately across a `230 V, 50 Hz` supply consume `100 W` and `60 W` at power factor of `0.5` lagging and `0.6` leading respectively. If these impedances are now connected in series across the same supply, find (a) total power absorbed and overall power factor (b) the value of reactance to be added in series so as to raise the overall power factor to unity. |
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Answer» Correct Answer - A::B::D (a) `0.5=R_1/Z_1` Further, `P=V_("rms")i_("rms")cos phi` or `100=230xx230/Z_1xx0.5` `:.Z_1=264.5Omega` and `R_1=132.25Omega` Further `X_L=sqrt(Z_1^2-R_1^2)=sqrt3/2Z_1` `=229Omega` In second case `0.6=R_2/Z_2` and `60=(230xx230)/Z_2xx0.6` `:. Z_2=529Omega` and `R_2=317.4Omega` Further, `X_C=sqrt(Z_2^2-R_2^2)` `=423.2Omega` When connected in series, `R=R_1+R_2=449.65Omega` `X_C-X_L=194.2` `:. Z=sqrt((449.65)^2+(194.2)^2)` `=489.79Omega` Power factor, `cosphi=R/Z=0.92` (leading) `P=V_("rms")i_("rms")cosphi` `=(230)(230/489.79)(0.92)` `=99W` (b) Since `X_C-X_L=194.C Omega` Therefore, if `194.2Omega` inductive reactance is to be added in series, then it will become only `R` circuit and power factor will come unity. |
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| 131. |
(a) What is the reactance of a `2.00 H` inductor at a frequency of `50.0 Hz`? (b) What is the inductance of an inductor whose reactance is `2.00 Omega` at `50.0 Hz`? (c) What is the reactance of a `2.00 muF` capacitor at a frequency of `50.0 Hz`? (d) What is the capacitance of a capacitor whose reactance is `2.00 Omega ` at `50.0 Hz`? |
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Answer» Correct Answer - A::B::C::D (a) `X_L=2pifL` (b) `L=X_L/(2pif)` (c) `X_c=1/(2pi f C)` (d) `C=1/(2pi f X_(C))` |
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| 132. |
Find the average value in the following cases (i) `i=4+3 cos omegat` (ii) `5sinomegat + 2sin2omegat + 3sin3omegat` (iv) `V= cosomegat + 3cos2omegat + 3cos3omegat + 2` |
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Answer» (i) Average value of current `I_(av) = I_(0) = 4` unit (ii) Average value of the voltage `V_(av) =0` (iii) Average value of current `I_(av) =0` (iv) Average value of voltage `V_(av) = 2`unit |
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| 133. |
A coil has a iductance of 0.7H and is joined in series with a resistance of `220 Omega`. When an alternating emf of 220V at 50 cps is applied to it, then the wattless component of the current in the circuit is `(take 0.7 = 2.2)`A. 5AB. 0.5AC. 0.7AD. 7A |
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Answer» Correct Answer - B Wattless component of ac `-(I_v) sin phi =(E_v)/(Z) (X_L)/(Z) =(E_(v)X_(L))/(Z^2) =(220 xx omegaL)/(R^(2)+omega^(2)L^(2))` As `omega L =0.7 xx 2 pi xx 50 =220 Omega` Hence wattless component of ac `(200 xx(220))/((220)^(2)_(220)^(2))=0.5 A`. |
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| 134. |
A circuit containing an inductance and a resistance connected in series, has an AC source of 200V, 50Hz connected across it. An AC current of 10A rms flows through th ecircuit and the power loss is measured to be 1W.A. The inductance of the circuit is `(sqrt3)/(10pi)H`B. The frequency of the AC when the phase difference between the current and emf becomes `pi/4`, with the above components is `(50)/(3)Hz`C. The frequency of the AC when the phase difference between the current and emf becomes `pi//3`, with the above components is `(25)/(sqrt3)Hz`D. The frequency of the AC when the phase difference between the current and emf becomes `pi//4`, with the above components is `(25)/(sqrt3)Hz`. |
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Answer» Correct Answer - A::B |
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| 135. |
The given arrangement carries a capacitor with capacitance 40mF and two inductors `L_(1)=25H and L_(2)=100H`. If the capacitor initially carries a charge of 10mC, then A. the maximum current through the inductor `L_(1)` when key `K_(1)` is closed is 20mAB. the maximum current through the inductor `L_(2)` when key `K_(2)` is closed is 5mAC. the maximum current through the inductor `L_(2)` when both the keys are closed is `sqrt5A`D. |
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Answer» Correct Answer - B::C::D |
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| 136. |
The peak value of an alternating emf E given by `E=(E_0) cos omega t` is 10V and freqency is 50 Hz. At time `t=(1//600)s` the instantaneous value of emf isA. 10VB. `5sqrt(3)V`C. `5 V`D. 1 V |
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Answer» Correct Answer - B Given that `E_(0) =10 V, t=1/600 s` `:. E=(E_0) cos 2 pi f t ` `=10 cos [ 2 pi xx 50 xx 1/600]` `=10 cos (pi//6)=10 (sqrt(3)///2)=5 sqrt(3)V`. |
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| 137. |
In the given circuit, then AC source has `omega=50 rad//s` Considering the indcutor and capacitor and capacitor to be ideal, the correct choice(s) is (are): A. The voltage across `100Omega` resistor `20sqrt2V`B. The voltage across `50Omega` resistor `20sqrt2V`C. The current through the circuit, l is `(2)/(sqrt10)A`D. The current through the circuit, l is 1.2A |
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Answer» Correct Answer - A::B::C |
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| 138. |
In the circuit of fig, the source freqency is `omega=2000 rad s^(-1)`. The current in the will be A. (A) `2A`B. (B) `3.3A`C. (C) `2 //sqrt(5)A`D. (D) `sqrt(5) A` |
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Answer» Correct Answer - A `(X_L) = L omega = 5 xx 10 ^(-3) xx 2000 = 10 Omega` `(X_C) = (1)/(C omega)=(1)/(50 xx 10^(-6) xx 2000) =(100)/(10) Omega = 10 Omega` since `(X_L)=(X_C)` therefore, `Z=R = 5.9 + 0.10 +4 = 10 Omega` `(I_V) =(E_V)/(Z)=(20)/(10) A =2A`. |
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| 139. |
In the given circuit, Fig., the reading of voltmeter `V_(1)` and `V_(2)` 300 votls each. The reading of the voltemeter `V_(3)` and ammeter `A` are respectively A. `800V, 2A`B. `300V, 2A`C. `220V, 2.2A`D. `100V, 2A` |
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Answer» Correct Answer - C Voltage across `L` and `C` cancel out. So, voltage across `k` is `220 V` Again, `I_(0)=220/100A=2.2A` |
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| 140. |
In the circuit of fig, the source frequency is `omega=2000 rad s^(-1)`. The current in the will be A. `2A`B. `3.3 A`C. `2//sqrt(5)A`D. `sqrt(5)A` |
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Answer» Correct Answer - A `L omega=5xx10^(-3)xx2000=10 Omega` `1/(C omega)=1/(50xx10^(-6)xx2000)=100/10 Omega=10 Omega` since `L omega=1/(L omega)`, `:. Z=R=0.10+4=10.1 Omega` `I_(0)=E/Z=20/10.1 A=2A` |
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| 141. |
A group of electric lamps having a total power rating of `1000` watt is supplied by an `AC` voltage `E=200sin(310t+60^(@))`. Then the r.m.s value of the circuit current isA. `10 A`B. `10 sqrt(2) A`C. `20 A`D. `20 sqrt(2) A` |
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Answer» Correct Answer - B `P=1/2V_(0)i_(0)cos varphi implies 1000=1/2xx200xxi_(0)cos 60^(@)` `implies i_(0)=20A implies i_(rms)=(I_(0))/(sqrt(2))=20/(sqrt(2))=10sqrt(2)A`. |
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| 142. |
A group of electric lamps having a total power rating of `1000` watt is supplied by an `AC` voltage `E=200sin(310t+60^(@))`. Then the r.m.s value of the circuit current isA. 0.41666666666667B. `5sqrt(2)`AC. 20 AD. `10sqrt(2)`A |
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Answer» Correct Answer - B `P=V_(rms)I_(rms) = V_(0)/sqrt(2)I_(rms)` `therefore I_(rms) = (1000sqrt(2))/(200) = 5sqrt(2)` |
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| 143. |
The armature of a `DC` motor has `20Omega` resistance. It draws a current of `1.5 A` when run by `200 V DC` supply The value of back emf induced in it will beA. 150 VB. 170 VC. 180 VD. 190V |
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Answer» Correct Answer - D For DC motor, `I=(V_(S) - V_(B))/R` and `V_(B) propto omega` `1.5 = (220-V_(R))/(20) rArr V_(B) = 190 V` |
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| 144. |
In a dc motor if `E` is the applied emf and `e` is the back emf then the efficiency is .A. `(E -e)/(E)`B. `(e)/(E)`C. `((E -e)/(E))^(2)`D. `((e)/(E))^(2)` |
| Answer» Correct Answer - B | |
| 145. |
Assertion : An electric motor will have maximum efficiency when back emf becomes equal to half of applied emf. Reason : Efficiency of electric motor depends only on magnitude of back emf.A. If both the assertion and reason are true and reason is a true explantion of the assertion.B. If both the assertion and reason are true but the reason is not true the correct explantion of the assertion.C. If the assertion is true but reason falseD. If both the assertion and reason are false. |
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Answer» Correct Answer - C Since the effective of an electric motor is given by `eta=("output power")/("input power")` From the above relation, it is quite clear that maximum output power corresponds maximum efficiency of motor. Now, output power is given by `eI=(e(E-e))/R.....(i)` To obtain maximum output power differentiating Eq. (i) with repect to `e` which will be equal to zero. so, `d/(de)[(e(E-e))/R]=0 implies e=E/2` Thus, when back emf becomes equal to half of the applied emf, the efficiency of motor will be maximum. |
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| 146. |
Assertion: Eddy current is produced in any metallic conductor when magnetic flux is changed aroundn it. Reason: Electric potential determines the flow of chargesA. If both the assertion and reason are true and reason is a true explantion of the assertion.B. If both the assertion and reason are true but the reason is not true the correct explantion of the assertion.C. If the assertion is true but reason falseD. If both the assertion and reason are false. |
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Answer» Correct Answer - B When a metallic conductor is moved in a magnetic field, magnetic flux is varied. It disturbs the free electrons of the metal and set up an induced amf in it. As there are no free ends of the metal i.e., it will be closed in itself so there will be induced current. |
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| 147. |
Find the avarage value of current (in A) shown graphically in fig. From `t=0 to t=2 s`. . |
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Answer» From the i-t graph, area from t-0 to t=2 =`(1)/(2) xx2 xx 10=10 "Amp. Sec"`. `therfore "Average Current"=(10)/(2)=5"Amp"` |
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| 148. |
In series LCR circuit voltage drop across resistance is 8V, across inductor is 6V and across capacitor is 12V. ThenA. voltage of the source will be leading current in the circuitB. voltage drop across each element will be less than the applied voltage.C. power factor of circuit will be `4//3`D. none of these. |
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Answer» Correct Answer - D Since `cos phi =(R )/(Z) =(IR)/(IZ) =8/10=4/5` (also `cos phi` can never be greater than 1) Hence, (c ) is wrong . Also, `IX_(C ) gtIX_(L) implies (X_C) gt(X_L)` `:.` Current will be leading In and LCR circuit, `V=sqrt((v_(L)-v_(C ))^(2)+c_(R )^(2))=sqrt((6-12)^(2)+8^(2))` `V=10,` which is less than voltage drop across capacitor. |
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| 149. |
A transformer having effieciency of 90% is working on 200 V and 3kW power supply. If the current in the secondary coil is 6A, the voltage across the secondary coil and the current in the primary coil respectively areA. 300V,15AB. 450V,15AC. 450V,13.5AD. 600V,15A |
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Answer» Correct Answer - B Power of the primary coil, `P_p=V_pI_p` `thereforeI_p=(3times1000)/200=15A` Power of the secondary coil , `P_s=V_sI_s` `thereforeV_s=P_s/I_s=(P_ptimes90/100)/6=3times1000times9/100times1/6=450V` |
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| 150. |
The number of turns in the primary and secondary coils of an ideal transformer are 140 and 280, respectively. If the current through the primary coils is 4 A, what will be the current in the secondary coil? |
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Answer» In an ideal transformer, the powers of the secondary and primary coild are equal, i.e., `V_sI_s=V_pI_p` `therefore I_s=I_p.p/V_s=I_P.N_P/N_s=4times140/280=2A`. |
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