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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.
| 1101. |
The power factor of an AC circuit having resistance (R) and inductance (L) connected in series and an angular velocity `omega` isA. zeroB. `(omegaL)/R`C. `R/(sqrt(R^(2)+Omega^(2)L^(2))`D. `R//omegaL` |
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Answer» Correct Answer - C Power factor, `cosphi = R/Z= R/((R^(2) + omega^(2)L^(2))` |
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| 1102. |
The power factor of an AC circuit having resistance (R) and inductance (L) connected in series and an angular velocity `omega` isA. `R//omegaL`B. `R//(R^(2)+omega^(2)L^(2))^(1//2)`C. `omegaL//R`D. `R//(R^(2)-omega^(2)L^(2))^(1//2)` |
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Answer» Correct Answer - B `cos varphi=R/Z=R/((R^(2)+omegaL^(2))^(1//2))` |
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| 1103. |
Assertion: An electric lamp connected in series with a variable capacitor and `AC` source, its brightness increases with increases in capacitance. Reason: Capacitive reactance decrease with increases in capacitance of capacitor.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 - A Capacitance reactance `X_(C)=1/(omegaC)`. When capacitance `(C)` increases, the capacitive reactance decreases. Due to decreases in its values, the current in the circuit will increases `(I=E/(sqrt(R^(2)+X_(C)^(2))))` and hence brightness of source (or electric lamp) will also increases. |
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| 1104. |
A coil of inductance 0.4 mH is connected to a capacitor of capacitance 400 pF. To what wavelength is this circuit tuned ? |
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Answer» `L=0.4 mH = 0.4 xx 10^(-3) H, C = 400 p f = 4 xx 10^(-10)F` Frequency `f=(1)/(2 pi sqrt(LC))=(1)/(2 xx 3.14)xx(1)/(sqrt(0.4xx10^(-3)xx4xx10^(-10)))=(10^(-7))/(6.28xx4)Hz` If the speed of electromagnetic wave is v, then `lambda = (v)/(l) = (3xx10^(8) ms^(-1))/((10^(7))/(6.28xx4 Hz))=753.6 m` |
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| 1105. |
Power factor is defined asA. apparent power`//` ture powerB. true power `//` apparent powerC. true power `("apparent power")^(2)`D. true power x apparent power |
| Answer» Correct Answer - 2 | |
| 1106. |
Assertion: An electric lamp connected in series with a variable capacitor and `AC` source, its brightness increases with increases in capacitance. Reason: Capacitive reactance decrease with increases in capacitance of capacitor.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 | |
| 1107. |
Assertion: An inductance and a resistance are connected in series with an `AC` circuit. In this circuit the current and the potential difference across the resistance lag behind potential difference across the inductance by an angle `pi//2`. Reason: In `LR` circuit voltage leads the current by phase angle which depends on the value of inductance and resistance both.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 - B We can use a capacitor of suitable capacitance as a chok coil, because average power consumed per cycle in an ideal capacitor is zero. Therefore, like a choke coil, a condenser can reduce `AC` without power dissipation. |
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| 1108. |
The capacitor offers zero resistance toA. `D.C` onlyB. `A.C` & `D.C`C. `A.C` onlyD. neither `A.C` nor `D.C` |
| Answer» Correct Answer - 4 | |
| 1109. |
Assertion: An inductance and a resistance are connected in series with an `AC` circuit. In this circuit the current and the potential difference across the resistance lag behind potential difference across the inductance by an angle `pi//2`. Reason: In `LR` circuit voltage leads the current by phase angle which depends on the value of inductance and resistance both.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 - b | |
| 1110. |
A bulb and a capacitor are in series with an `AC` source. On increasing frequency how will glow of the bulb changeA. The glow decreasesB. The glow increasesC. The glow remain the sameD. The bulb quenches |
| Answer» Correct Answer - b | |
| 1111. |
Assertion: A capacitor of suitable capacitance can be used in an `AC` circuit in place of the choke coil. Reason: A capacitor blocks `DC` and allows `AC`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 - b | |
| 1112. |
The coil of choke in a circuitA. Increases the currentB. Decreases the currentC. Does not change the currentD. Has high resistance to dc circuit |
| Answer» Correct Answer - b | |
| 1113. |
A choke coil is preferred to a rheostate in `AC` circuit asA. It consumes almost zero powerB. It increases currentC. It increases powerD. It incresases voltage |
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Answer» Correct Answer - A A choke coil contains high inductance but negligible resistance, due to which power loss becomes appreciably small. |
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| 1114. |
Two alternating voltage generators produce emfs of the same amplitude`(E_0)` but with a phase difference of `(pi)//3`. The resultant emf isA. (A) `E_0 sin [ omega t+(pi)//3)]`B. (B) `E_0 sin [ omega t+(pi)//6)]`C. (C) `sqrt(3)E_0 sin [omega t+(pi)//6)]`D. (D) `sqrt(3)E_0 sin [ omega t+(pi)//2)]` |
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Answer» Correct Answer - C `E_(1)=(E_0) sin omega t, (E_2)=(E_0) sin [ omega t + (pi//3)]` `E=(E_2)+(E_1)` `=(E_0)sin (omega t +(pi//3)]+(E_0)sin omegat` `(E_0) [ 2 sin {omega t+(pi//6)}cos (pi//6)]` `= sqrt(3) (E_0) sin [omega t +(pi//6)]`. |
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| 1115. |
A choke coil has.A. high inductance and high resistanceB. high inductance and low resistanceC. low inductance and high resistanceD. low inductance and low resistance |
| Answer» Correct Answer - B | |
| 1116. |
A choke coil is preferred to a resistance for reducing current in an ac circuit because .A. choke coil is cheapB. there is no wastage of powerC. choke is compact in sizeD. choke is a good absorber of heat |
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Answer» Correct Answer - B In an AC circuit , the coil of high inductance and negligible resistance used to control current, is called the choke coil . The power factor of such a coil is given by cos `phi` = R/`sqrt R^(2) + omega^(2)L^(2) ` `approx` R `omega`L (as R ltlt `omega`L) |
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| 1117. |
A choke coil is preferred to a rheostate in `AC` circuit asA. It consumes almost zero powerB. It increases currentC. It increases powerD. It increases voltage |
| Answer» Correct Answer - a | |
| 1118. |
Assertion: `AC` generators are based upon `EMI` principle. Reason: Resistance offered by capacito for alternating current is zero.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 `X_(C)=1/(omegaC) omega ne0` area for `AC`. `I_(rms)[(int_(0)^(T) I^(2) dt)/(int_(0)^(T)dt)]^(1//2)=[(int_(0)^(2pi//omega) I_(0)^(2)sin^(2) omegat dt)/(int_(0)^(2pi//omega) dt)]^(1//2)=(I_(0))/(sqrt(2))`. |
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| 1119. |
A choke coil is preferred to a resistance for reducing current in an ac circuit because .A. choke coil is cheaperB. choke coil is easier to designC. choke coil consumers much less powerD. the eddy currents produced in a choke coil reduce the current . |
| Answer» Correct Answer - C | |
| 1120. |
Why a choke is needed with a fluorescent lamp with ac mains? Why a normal resistor can not be used in place of the choke? |
| Answer» An inductor can introduce voltage drop in a cricuit without any loss of power put a resitor gets heated during the process and some power is lost. A choke acts as an inductor in a circuit, so a choke is used in place of resistor. | |
| 1121. |
A direct current of `5` amp is superimposed on an alternating current `I=10 sin omega t` flowing through a wire. The effective value of the resulting current will be:A. (A) `(15//2)A`B. (B) `5 sqrt(3)A`C. (C) `5 sqrt(5)A`D. (D) `15 A` |
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Answer» Correct Answer - B Given `l=5+10 sin omega t ` `I_(eff)=[(int_(0)^(T)(i^2)dt)/(int_(0)^(T)dt)] = [ 1/T int_(0)^(T) (5 + 10 sin omega t)^(2) dt ]^(1//2)` `[ 1/T int_(0)^(T) (25 + 100 sin omega t)+100 sin^(2)omega t)]^(1//2)` But as `(1)/(T) int_(0)^(T) sin omega t dt = 0` and ` 1/T int_(0)^(T) sin^(2) omega t dt = 1/2` so `I_(eff) = [ 25 + 1/2 xx 100 ]^(1//2) = 5 sqrt(3)A`. |
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| 1122. |
If a direct current of value a ampere is superimposed on an alternating current `1 = b sin omega t` flowing through a wire, what is the effective value of the resulting current in the circuit? A. `[(a^(2)+b^(2))/(2)]^(1/2)`B. `[a^(2)+(1)/(2)b^(2)]^(1/2)`C. `[a^(2)+b^(2)]^(1/2)`D. `[(1)/(2)a^(2)+b^(2)]^(1/2)` |
| Answer» Correct Answer - B | |
| 1123. |
If a direct current of value a ampere is superimposed on an alternating current `1 = b sin omega t` flowing through a wire, what is the effective value of the resulting current in the circuit? |
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Answer» Correct Answer - `I_(eff)=[a^(2)+1/2b^(2)]` `I_(rms)=[(underset(0)overset(T)int(a+b sin omegat)^(2)dt)/T]^(1/2)=I_(eff)=[a^(2)+1/2b^(2)]^(1//2)` |
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| 1124. |
An ideal choke coil takes a current fo 8 ampere when connected to an `AC` supply of `100` volt and `50 Hz`. A pure resistor under the same conditions takes a current of 10 ampere. If the two are connected to an `AC` supply of `150` volts and `40 Hz` then the current in a series combination of the above resistor and inductor is |
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Answer» For pure inductor, `X_(L) = (E_(0))/(I_(v)) = (100)/(8) = (25)/(2) Omega` `omega L = (25)/(2), L = (25)/(2 omega) = (25)/(2 xx 2 pi xx 50) = (1)/(8pi) H` `R = (V)/(1) = (100)/(10) = 10 Omega` For the combination, the supply is `150 v, 40 Hz`. `:. X_(L) = omega L = 2 pi xx 40 xx (1)/(8 pi) = 10 Omega` `Z = sqrt(X_(L)^(2) + R^(2)) = sqrt(10^(2) + 10^(2)) = 10 sqrt(2)` ohm `I_(v) = (E_(v))/(Z) = (150)/(10 sqrt(2)) A = (15)/(sqrt(2)) A` |
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| 1125. |
In the circuit shown, `R` is a pure resistor, `L` is an inductor of negligible resistance (as compared to `R`) and `S` is a`100V,50 Hz AC` source of negligible resistance. With eigther key `k_(1)` alone or `k_(2)` alone closed, the current is `I_(0)`. if the source is changed to `100 V, 100 Hz`, the current with `k_(1)` alone closed and with `k_(2)` alone closed will be respectively A. `I_(0),1/2I_(0)`B. `2I_(0),1/2I_(0)`C. `2I_(0),I_(0)`D. `2I_(0),1/2I_(0)` |
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Answer» Correct Answer - A With change in frequency resistance will not change but inductive reactance is reduced to half. |
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| 1126. |
An ideal choke takes a current fo `10 A` when connected to an ac supply of `125 V` and `50 Hz`. A pure resistor under the same conditions take a current of `12.5 A`. If the two are connected to an ac supply of `100 V` and `40 Hz`, then the current in series combination of above resistor and inductor is `(10^(x))/(sqrt(2))` what is the value of `x` |
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Answer» Correct Answer - 1 `2pixx50L=(125)/(10) , therefore L=(12.5)/(100pi)` and `R=(125)/(12.5)=10Omega` In the series combination of `L` and `R` `i=(100)/(sqrt(R^(2)+(2pixx40L)^(2)))=(100)/(sqrt(10^(2)+10^(2)))=(10)/(sqrt2)` |
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| 1127. |
An ideal choke takes a current of 10 A when connected to an ac supply of 125 V and 50 Hz. A pure resistor under the same conditions takes a current of 12.5 A. If the two are connected to an ac supply of 100 V and 40 Hz, then the current in series combination of above resistor and inductor isA. `10` ampB. `12.5` ampC. `20` ampD. `25` amp |
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Answer» Correct Answer - A `R=125/12.5 =10 Omega` `X_(L)=omegaL=2pinL=V/I=125/10=12.5` `:. 2pinL=12.5` or `2piL=12.5/50=0.25` `:. X_(L)=2piLxxn=0.25xx40=10 Omega` Impedence of the circuit `Z=sqrt(R^(2)+X_(L)^(2))=10sqrt(2)` ohm `:.` Current `=(100sqrt(2))/(10sqrt(2))=10` amp |
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| 1128. |
In the circuit shown in fig. R is a pure resistor, L is an inductor of ngligibe resistance (as compared to R), S is a 100 V , 50 Hz ac source of negligible resistnce. With either kiy `(K-1) `alone or `(K_2)` alone closed, the current is `(I_0)`. If the source os changed to 100 V, 100 Hz the current with `(K_1)` alone closed and with `(K_2)` alone closed will be, respectively. A. `(I_0), (I_0)/(2)`B. `(I_0), 2(I_0)`C. `2(I_0), (I_0)`D. `2(I_0),(I_0)/(2)` |
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Answer» Correct Answer - A Current remains unchanged in R. However, it becomes half in L, because reactance is doubled on doubling the frequencuy. |
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| 1129. |
An ideal inductor takes a current of `10 A` when connected to a `125 V, 50 Hz AC` supply, A pure resistor across the same source takes `12.5 A`. If the two are connected in series across a `100 sqrt(2) V, 40 Hz` supply, the current through the circuit will beA. `10 A`B. `12.5 A`C. `20 A`D. `25 A` |
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Answer» Correct Answer - 1 For `50 Hz` and `125 V` supply `X_(L) = omega L = (V)/(i_(L)) implies L = (1)/(8pi), R = (V)/(i_(R)) = 10 Omega` For `40 Hz`, `100 sqrt(2) V` supply `i= (V)/(sqrt(R^(2) + X_(L)^(2))) = (1)/(sqrt(R^(2) + 4 pi^(2) f^(2) L^(2)))` |
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| 1130. |
In an `A.C` circuit the instantaneous values of current and voltage are `I = 120 sin omega t` ampare and `E = 300 sin (omega t + pi//3)` volt respectively. What will be the inductive reactance of series `LCR` circuit if the resistance and capacitve reactrance are 2 ohm and 1 ohm respectively?A. 4.5 ohmsB. 2 ohmsC. 2.5 ohmsD. 3 ohms |
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Answer» Correct Answer - 1 `I = 120 sin omega t, E = 300 sin (omega t + pi //3)` Clearly, `phi = pi//3` Now, `cos phi = (R )/(Z) = cos 60^(@) = (1)/(2) :. Z = 2R` As `R = 2 Omega, :. Z = 2 xx 2 = 4 Omega , X_(C ) = 1 Omega` Now `(X_(L) - X_(C ))^(2) = Z^(2) - R^(2) = 4^(2) - 2^(2) = 12` `X_(L) - X_(C ) = +- sqrt(12) = +- 2sqrt(3)` `X_(L) = X_(C ) +- 2 sqrt(3) = 1 +- 3.464` Taking + value, `X_(L) = 1 + 3.464 = 4.465 Omega` |
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| 1131. |
A coil has an inductance of `0.7 H` and is joined in series with a resistance of `220 Omega`. When an alternating e.m.f of `220 V` at 50 c.p.s. is applied to it, then the wattless component of the current in the circuit isA. 5 ampereB. 0.5 ampereC. 0.7 ampereD. 7 ampere |
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Answer» Correct Answer - 2 Watt less component of `A.C = I_(V) sin theta = (E_(v))/(Z) sin theta` `=(200)/(sqrt(R^(2) + L^(2) omega^(2))) xx (L omega)/(sqrt(R^(2) + L^(2) omega^(2)))` `:. L 0.7 xx 2 pi xx 50` `=(220 xx L omega)/((R^(2) + L^(2) omega^(2))) = 0.7 xx 2 xx (22)/(7) xx 50` `= (220 xx (0.7 xx 2pi xx 50))/((220^(2) + 220^(2))) = 220 Omega` `= (220 xx 220)/(220^(2) (2)) = (1)/(2) = 0.5` |
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| 1132. |
What will be the approximate resistance offered by a capacitor of 10`mu`F and frequency 100Hz?A. `160 Omega`B. `1600 Omega`C. `16 Omega`D. None of these |
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Answer» Correct Answer - A `underset(C )(X)` = 1/2`pi`fC = 1/`2pi xx 100 xx 10^(-5)` = `160 Omega` |
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| 1133. |
A coil of inductance `0.1 H` is connected to `50 V, 100Hz` generator and current is found to be `0.5A`. The potential difference across resistance of coil is:A. `15 V`B. `20 V`C. `25 V`D. `39 V` |
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Answer» Correct Answer - 4 `I = (E)/(Z) , 0.5 = (50)/(Z) = 100 Omega` `Z^(2) = R^(2) + omega^(2) L^(2)`, then `R = 78 Omega` Now `V_(R ) = sqrt(V_(LR)^(2) = V_(L)^(2)) = 39 V , [V_(R)^(2) + V_(L)^(2) = V_(LR)^(2)]` |
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| 1134. |
A solenoid with inductance `L=7 m H` and active resistance `R=44 Omega` is first connected to a source of direct voltage `V_(0)` and then to a source of sinusoidal voltage with effective value `V=V_(0)` . At what frequency of the oscillator will be power consumed by the solenoid be `eta=5.0 ` times less than in the former case ? |
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Answer» Correct Answer - 2 In case of direct current, `P_(DC)=V_(M)^(2)/R` The impedance of `LR` circuit is `Z=sqrt(R^(2)+X_(L)^(2))=sqrt(R^(2)+omega^(2)L^(2))` `:. I_(M)=V_(M)/Z=V_(M)/sqrt(R^(2)+omega^(2)L^(2))` `:. P_(AC)=I_(M)^(2)R=(V_(M)^(2)R)/(R^(2)+omega^(2)L^(2))` According to the problem, `P_(DC)=eta P_(AC)` or `V_(M)^(2)/R=eta (V_(M)^(2)R)/(R^(2){1+((omegaL)/R)^(2)})` or `1/R=eta/(R{1+((omegaL)/R)^(2)})` or `omega=R/L sqrt(eta^(2)-1) rArr 2piv=R/L sqrt(eta^(2)-1)` `v=R/(2pi L)sqrt(eta^(2)-1)=2kHz` |
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| 1135. |
In a series R-C circuit the supply voltage (Vs) is kept constant at 2V and the frequency f of the sinusoidal voltage is varied from 500 Hz to 2000 Hz. The voltage across the resistance R=1000 ohm is measured each time as `V_(R)`. For the determination of the C student wants to draw a linear graph and try to get C from the slope. Then she may draw a graph ofA. `f^(2)"against"V_(R)^(2)`B. `(1)/(f^(2))"Against"(V_(S)^(2))/(V_(R)^(2))`C. `(1)/(f^(2))"Against"(1)/(V_(R)^(2))`D. `f" Against" `(V_(R))/(sqrt(V_(s)^(2)-V_(R)^(2)))` |
| Answer» Correct Answer - B::C::D | |
| 1136. |
A series LCR circuit containing a resistance of `120 Omega` has angular resonance frequency `4 xx 10^(5) rad s^(-1)`. At resonance the vlotage across resistance and inductance are 60V and 40 V, repectively, At what frequency, the current in the circuit lags the voltage bu `45^(@)` ?A. `16xx10^(5)"rad "s^(-1)`B. `8xx10^(5)"rad "s^(-1)`C. `4xx10^(5)"rad "s^(-1)`D. `2xx10^(5)"rad "s^(-1)` |
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Answer» Correct Answer - B At resonance as X = 0, `I = (epsilon_(0))/(R )=(60)/(120)=(1)/(2)A=0.5 A` As `V_(L)=IX_(L)=I omega L rArr L = (V_(L))/(I omega)` `therefore L = (40)/((0.5)xx4xx10^(5))=0.2 mH` Also, `omega_(0)=(1)/(sqrt(LC)) therefore C=(1)/(L omega_(0)^(2))` `=(1)/(0.2xx10^(-3)xx(4xx10^(5))^(2))=(1)/(32)mu F` Now in case of series LCR circuit. `therefore tan phi = (X_(L)-X_(C ))/(R )` So current will lag the applied voltage by `45^(@)` if, `tan 45^(@)=(omega L-(1)/(omega C))/(R )` ....(i) or `R = omega L -(1)/(omega C) " " (because tan 45^(@)=1)` or `120=omegaxx2xx10^(4)-(1)/(omega(1//32)xx10^(-6))` (using (i)) `rArr omega^(2)-6xx10^(5)omega-16xx10^(10)=0` `therefore omega = (6xx10^(5)pm sqrt((6xx10^(5))^(2)+64xx10^(10)))/(2)` `therefore omega = (6xx10^(5)+10xx10^(5))/(2)=8xx10^(5)"rad "s^(-1)` |
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| 1137. |
The power factor of the circuit shown in the figure is A. 0.4B. 0.2C. 0.8D. 0.6 |
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Answer» Correct Answer - D Total resistance `R = 20 + 40 = 60 Omega` The impedance of `LCR` circuit is given by `Z = sqrt(R^(2) + (X_(L) - X_(C ))^(2))` `= sqrt(60^(2) + (100 + 20)^(2)) = 100` Power factor, `cos phi = R//Z = 60//100 = 0.6` |
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| 1138. |
A series R-C combination is connected to an AC voltage of angular frequency `omega=500 radian//s`. If the impendance of the R-C circuit is `Rsqrt(1.25)`, the time constant (in millisecond) of the circuit is |
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Answer» Correct Answer - B,C `R sqrt(1.25) = sqrt(R^(2) + X_(C )^(2))` `X_(C ) = (R )/(2)`, hence `RC = (2)/(omega) = 4 ms` |
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| 1139. |
For a given `AC` source the average emf during the positive half cycleA. depends on `E_(0)`B. depends on shape of waveC. both 1 and 2D. depends only on peak value of `E_(0)` |
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Answer» Correct Answer - 4 `E_(av) = (2 E_(0))/(T) int_(0)^(T//2) sin (omega t) dt = (2 E_(0))/(pi)` |
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| 1140. |
Statement (A) : The reactance offered by an inductance in `A.C`. Circuit decreases with increase of `AC` frequency Statement (B) : The reactance offered by capacitor in `AC` circuit increases with increase of `AC` frequency.A. `A` is ture but `B` is falseB. Both `A` and `B` are trueC. `A` is false but `B` is trueD. Both `A` and `B` are false |
| Answer» Correct Answer - 4 | |
| 1141. |
As the frequency of an ac circuit increases, the current first increases and then decreases. What combination of circuit elements is most likely to comprise the circuit ?A. Inductor and capacitorB. Resistor and inductorC. Resistor and capacitorD. Resistor, inductor and capacitor |
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Answer» Correct Answer - 1,4 We know that, `I = (V)/(sqrt(R^(2) + (X_(L) - X_(C ))^(2)))` `X_(L) = 2 pi upsilon L` and `X_(C ) = (1)/(2 pi upsilon C)` So with increase in frequency, `R` remains constant, inductive reactance increases and capacitive reactance decreases. |
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| 1142. |
An ac generator `G` with an adjustable frequency of oscillation is used in the circuit, as shown. To increase resonant frequency of the circuit, some of the changes in the circuit are carried out. Which changes would certainly result in the increase in resonnatn frequency ?A. `R` is increasedB. `L_(1)` is increased and `C_(1)` is decreasedC. `L_(2)` is decreased and `C_(2)` is increasedD. `C_(3)` is removed from the circuit |
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Answer» Correct Answer - D `c_(eq)` decreases thereby increasing resonant frequency. |
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| 1143. |
A voltage source `V=V_(0) sin (100 t)` is connected to a black box in which there can be either one element out of `L,C,R` or any two of them connected in series. At steady state, the variation of current in the circuit and the source voltage are plotted together with time.using an oscilloscope, as shown Values of the parameters of the elements present in the black box are- A. `R=50Omega , C=200 muf`B. `R=50Omega , L=2mmu`C. `R=400 Omega, C=50 muf`D. None of these |
| Answer» Correct Answer - A | |
| 1144. |
A voltage source `V=V_(0) sin (100 t)` is connected to a black box in which there can be either one element out of `L,C,R` or any two of them connected in series. At steady state, the variation of current in the circuit and the source voltage are plotted together with time.using an oscilloscope, as shown The element (s) present in black box is/are: A. only `C`B. `L` and `C`C. `L` and `R`D. `R` and `C` |
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Answer» Correct Answer - D As current is leading the source voltage, so circuit should be capacitive in nature and as phase difference is not `pi/2`, it must contain resistor also. |
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| 1145. |
An ac generator `G` with an adjustable frequency of oscillation is used in the circuit, as shown. To increase resonant frequency of the circuit, some of the changes in the circuit are carried out. Which changes would certainly result in the increase in resonnatn frequency ?A. `R` is increasedB. `L_(1)` is increased and `C_(1)` is decreased.C. `L_(2)` is decreased and `C_(2)` is increasedD. `C_(3)` is removed from the circuit. |
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Answer» Correct Answer - D `C_(eq)` decreases thereby increasing resonant frequency. |
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| 1146. |
Electrical energy is transmitted over large distances at high alternating voltages. Which of the following statements is (are) correct?A. For a given power level, there is a lower current.B. Lower current implies less power less.C. Transmission lines can be made thinner.D. It is easy reduce the voltage at the receving end using step-down transformers |
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Answer» Correct Answer - 1,2,4 According to relation, `P = EI` when `I` is low, powe loss `(= I^(2) R)` is also low. A step down transformer lowers voltage by increasing current. |
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| 1147. |
A voltage source `V=V_(0) sin (100 t)` is connected to a black box in which there can be either one element out of `L,C,R` or any two of them connected in series. At steady state, the variation of current in the circuit and the source voltage are plotted together with time.using an oscilloscope, as shown If `AC` source is removed, the circuit is shorted for some time so that capacitor is fully discharge and then a battery of constant `EMF` is connected across the black box, at `t=0`.The current in the circuit will- A. increase exponentially with time constant = 0.02 sec.B. decrease exponentially with time constant = 0.01 sec.C. oscillate with angular frequency 20 rad/secD. first increase and then decrease |
| Answer» Correct Answer - B | |
| 1148. |
An ac generator `G` with an adjustable frequency of oscillation is used in the circuit, as shown. Current drawn from the ac souce will be maximum if its angular frequency isA. `10^(5) rad//s`B. `10^(4) rad//s`C. 5000 rad/sD. 500 rad/s |
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Answer» Correct Answer - C |
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| 1149. |
An ac generator `G` with an adjustable frequency of oscillation is used in the circuit, as shown. Current drawn from the ac souce will be maximum if its angular frequency isA. `10^(5) "rad"//s`B. `10^(4) "rad"//s`C. `5000 "rad"//s`D. `500 "rad"//s` |
| Answer» Correct Answer - C | |
| 1150. |
An ac generator `G` with an adjustable frequency of oscillation is used in the circuit, as shown. Current drawn from the ac souce will be maximum if its angular frequency isA. `10^(5) "rad"//s"`B. `10^(4) "rad"//s"`C. `5000 "rad"//s"`D. `50 "rad"//s"` |
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Answer» Correct Answer - C Current drawn is maximum at resonant angular frequency. `L_(eq) = 4 mH C_(eq) = 10 mF` `omega = (1)/(sqrt(LC)) = 5000 "rad"//s` |
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