InterviewSolution
Saved Bookmarks
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.
| 151. |
200 V ac source is fed to series LCR circuit having `X_(L)=50 Omega, X_(C )=50 Omega` and `R = 25 Omega`. Potential drop across the inductor isA. 100 VB. 200 VC. 400 VD. 10 V |
|
Answer» Correct Answer - C Here, `V_("rms")=200 V, X_(L) = 50 Omega, X_(C )= 50 Omega, R=25 Omega` Impedance of the circuit, `Z=sqrt(R^(2)+(X_(L)-X_(C ))^(2))=sqrt(25^(2)+(50-50)^(2))=25 Omega` Current in the circuit, `I_("rms")=(V_("rms"))/(Z)=(200 V)/(25 Omega)=8 A` Voltage drop across the inductor is `V_(L)=I_("rms")X_(L)=8 Axx50 Omega = 400 V` |
|
| 152. |
Initial voltage and input power of a transformer of effieciency `80%` are 100 V and 4 kW, respectively. IF the voltage of the secondary coil is 200 V, determine the currents flowing through the primary and the second ary coils? |
|
Answer» Power of the primary coil i.e., input power, `P_p=V_pI_p` or,`I_p=P_p/V_p=(4times1000)/100=40 A` Power of the secondary coil , `P_s=P_ptimes80/100`, Again, `P_s=V_sI_s` So, `I_s=P_s/V_s=80/100timesP_p/V_s=80/100times(4times1000)/200=16 A`. |
|
| 153. |
If the emf of an ac circuit be `E=E_0` sinomegatand current ` I=I_("cosomegat")``, what is the power dissipated in the circuit? |
|
Answer» `E=E_0sinomegat,I=I_0cosomegat=I_0sin(omegat+90^@)` So, phase difference , `theta=90^@` Therefore , power factor`=costheta=cos90^@=0` i.e., power dispated=0 |
|
| 154. |
Which of the two waveforms shown in Figure has a higher rms value? |
|
Answer» Correct Answer - Both have same value |
|
| 155. |
Statement I: A series LCR circuit when connected to an ac source gives the terminal potential difference 50 V across each of resistor R, inductor L and capacitor C. Then the terminal potential difference across LC is zero. Statement II: The terminal alternating voltages across the inductor and capacitor in a series LCR circuit are in opposite phase.A. Statement I is true, statement II is true, statement II is a correct explanation for statement I.B. Statement I is true, statement II is true, statement II is not a correct explanation for statement IC. Statement I is true, statement II is falseD. Statement I is false, statement II is true |
| Answer» Correct Answer - A | |
| 156. |
In an oscillating LC circuit the maximum charge on the capacitor is Q. When the charge is stored equally between the electric and magnetic fields, what is the charge on the capacitor? |
|
Answer» Maximum charge of the capacitor= Q So total energy of the circuit `=1/2Q^2/C` When the charge of the capacitor is q, then energy stored in the electric fields is `1/2q^2/C`.At this stage energy stored in the magnetic field is `1/2Li^2`. If it is equal to `1/2q^2/C`, then total energy `=1/2q^2/C+1/2Li^2=1/2q^2/C+1/2q^2/C=q^2/C` So, `q^2/C=1/2Q^2/C`or,`q^2=Q^2/2` or, `q=Q/sqrt2` |
|
| 157. |
Statement I: Form factor becomes different for different waveforms of alternating voltage and current. Statement II: The mean value of alternating voltage or current `=(2)/(pi)" rms value=1/(sqrt2)"timespeak value"` for any wave form.A. Statement I is true, statement II is true, statement II is a correct explanation for statement I.B. Statement I is true, statement II is true, statement II is not a correct explanation for statement IC. Statement I is true, statement II is falseD. Statement I is false, statement II is true |
| Answer» Correct Answer - C | |
| 158. |
Is the frequency of oscillation of magnetic energy or electrostatic energy is same as that of charge in LC oscillator ? |
| Answer» Correct Answer - No. The frequency of oscillation of kinetic energy or potential energy is double the frequency of oscillation of charge. | |
| 159. |
The instantaneous voltages at three terminals marked X, Y and Z are given by `V_(X)=V_(0)sin omegat, V_(Y)=V_(0)sin (omegat=(2pi)/(3))and V_(z)=V_(0)sin (omegat=(4pi)/(3))` An ideal voltmeter is configured to read runs value of the potential difference between its terminals. is connected between points X and Y and then between Y and Z. The reading the voltmeter will beA. `V_(X)=V_(0) sin omegat, V_(gamma)=V_(0)sin (omegat+(2pi)/(3))and V_(Z)=V_(0)sin (omegat+(4pi)/(3))`B. `V_(XY)^(rms)=V_(0)sqrt((3)/(2))`C. `V_(YZ)^(rms)=V_(0)sqrt((1)/(2))`D. indpependent of the choice of the two terminals |
| Answer» Correct Answer - A::C | |
| 160. |
When the rms voltages `V_(L), V_(C )` and `V_(R )` are measured respectively across the inductor `L`, the capacitor `C` and the resistor `R` in a series `LCR` circuit connected to an `AC` source, it is found that the ratio `V_(L) : V_(C ) : V_(R ) = 1 : 2 : 3`. If the rms voltage of the `AC` sources is `100 V`, the `V_(R )` is close to:A. `50 V`B. `70 V`C. `90 V`D. `100 V` |
|
Answer» Correct Answer - 3 Given, `V_(L) : V_(C ) : V_(R ) = 1 : 2 : 3` `V = 100 V` `V_(R ) = ?` As we know, `V = sqrt(V_(R )^(2) + (V_(L) - V_(C ))^(2))` Solving we get, `V_(R ) = 90 V` |
|
| 161. |
Statement I: If the total energy in an LC oscillator is equally distributed between the magnetic and electric fields then the charge stored in the capacitor is `1/sqrt2` fraction of the maximum charge stored in the capacitor during oscillation. Statement II: The charge stored in the capactior becomes maximum at a time when total energy of the LC circuit is stored in the electric field.A. Statement I is true, statement II is true, statement II is a correct explanation for statement I.B. Statement I is true, statement II is true, statement II is not a correct explanation for statement IC. Statement I is true, statement II is falseD. Statement I is false, statement II is true |
| Answer» Correct Answer - A | |
| 162. |
In an LCR series ac circuit `R=10Omega` , L=50 mH and `C=5muF`. Find out the resonant frequency and Q-factor. Find also the bandwidth and half-power frequencies. |
|
Answer» Hence, L=50mH =`5times10^-2H` `5 muF=5times10^-6F` `therefore` Resonant frequency, `f_0=1/(2pisqrt(Lc))=1/(2times3.4timessqrt((5times10^-2)times(5 times 10^-6)))` `=10^4/(2times3.14times5)=318.5Hz` and Q factor = `=1/Rsqrt(L/C)=1/10sqrt((5times10^-2)/(5times10^-6))=1/10times 100=10` Now, `Omega=omega_e/(Deltaomega)=f_0/(Deltaf)` `therefore` Bandwidth, `Deltaf=f_0/Q=318.5/10=31.85Hz` Half- power frequencies are, `f_1=f_0-(Deltaf)/2=318.5-31.85/2=302.6Hz` and `f_2=f_1+(Deltaf)/2=302.6+15.92=318.52 Hz` |
|
| 163. |
A series resonant LCR circuit has a quality factor (Q-factor)=0.4. If `R=2k Omega, C=0.1 mu F` then the value of inductance isA. `0.1 H`B. `0.064 H`C. 2 HD. 5 H |
|
Answer» Correct Answer - B Quality factor `Q = (1)/(R )sqrt((L)/(C ))` or `(L)/(C )=(QR)^(2)` Here, `Q = 0.4, R = 2 k Omega= 2xx10^(3)Omega` `C=0.1 mu F=0.1xx10^(-6)F` `therefore L=(QR)^(2)C` `therefore L =(0.4xx2xx10^(-3))^(2)xx0.1xx10^(-6)=0.064 H` |
|
| 164. |
Obtain the resonant frequency and Q-factor of a series LCR circuit with `L=3.0H, C=27muF, and R=7.4 Omega`. How will you improve the sharpness of resonance of the circuit by a factor of 2 by reducing its full width at half maximum? |
|
Answer» Inductance, L=3.0H Capacitance, C=27`muF`=27`xx10^(-6) F` Resistance, R=`7.4Omega` at resonance, angular frequency of the source for the given LCR series circuit is given as: `omega_(r)=(1)/sqrt(LC)` `=(1)/sqrt(3xx27xx10^(-6))=(10^(3))/(9)=111.11` rad `s^(-1)` Q-factor of the series: `Q=(omega_(r)L)/(R)` `=(111.11xx3)/(7.4)=45.446` To improve the sharpness of the resonance by reduc ing its full width at half maximum. by a factor of 2 wihtout changing `omega_(r)`, we need to reduce R to half i.e., `(R)/(2)=(7.4)/(2)=3.7Omega` Resistance |
|
| 165. |
In `LCR` circuit current resonant frequency is `600Hz` and half power points are at `650` and `550 Hz`. The quality factor is |
|
Answer» Correct Answer - 6 `Q=("Resonant frequency")/("Band width")=(f_(0))/(f_(2)-f_(1))` |
|
| 166. |
A sinusoidal alternating current of peak value `(I_0)` passes through a heater of resistance R. What is the mean power output of the heater?A. `(l_(0)^(2)R)/(2)`B. `(l_(0)^(2)R)/(sqrt(2))`C. `2l_(0)^(2)R`D. `sqrt(2)l_(0)^(2)R` |
| Answer» Correct Answer - A | |
| 167. |
Statement I: Q-factor of a series LCR circuit is `1/Rsqrt(L/C)` Statement II: Resonant frequency of an LCR circuit does not depend on the resistance of the circuit.A. Statement I is true, statement II is true, statement II is a correct explanation for statement I.B. Statement I is true, statement II is true, statement II is not a correct explanation for statement IC. Statement I is true, statement II is falseD. Statement I is false, statement II is true |
| Answer» Correct Answer - B | |
| 168. |
Statement I: The peak values of alternating voltages and alternating current in a circuit are `V_0` and `I_0` respectively. The phase difference between voltage and current is `theta`. Then the power consumed is `V_0I_0costheta` Statement II: The consumed power in an alternating circuit depends on the phase difference between the emf and current.A. Statement I is true, statement II is true, statement II is a correct explanation for statement I.B. Statement I is true, statement II is true, statement II is not a correct explanation for statement IC. Statement I is true, statement II is falseD. Statement I is false, statement II is true |
| Answer» Correct Answer - D | |
| 169. |
The voltage time (V-t) graoh for triangular wave having peak value `(V_0)` is as shown in fig. A. `(V_(0))/(3)`B. `(V_(0))/(2)`C. `V_(0)/(sqrt2)`D. `V_(0)/(sqrt3)` |
|
Answer» Correct Answer - D |
|
| 170. |
In the circuit shown in figure : `R = 10 Omega , L = (sqrt(3))/(10) H, R_(2) = 20 Omega` and `C = (sqrt(3))/(2) mF`. Current in `L - R_(1)` circuit is `I_(1)` in `C - R_(1)` circuit is `I_(2)` and the main current is `I` Phase difference between `I_(1)` and `I_(2)` isA. `0^(@)`B. `90^(@)`C. `180^(@)`D. `60^(@)` |
|
Answer» Correct Answer - B `Tan phi_(1) = (omega L)/(R_(1))` , `Tan phi_(2) = (1)/(omega C R_(2))` `:. Delta phi = phi_(1) ~ phi_(2)` |
|
| 171. |
Assertion: In an `AC` circuit, potential difference across the capacitor may be greater than the applied voltage. Reason : `V_C=IX_C`, wheereas `V= IZ` and `X_C` can be greater than `Z` also.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. |
|
Answer» Correct Answer - A `Z=sqrt(R^2+(X_C-X_L)^2)` From this expression we can see that `X_C` may greater than `Z` also. |
|
| 172. |
In the circuit shown in figureure, the power consumed is .A. zeroB. `V_0^2/(2R)`C. `(V_0^2R)/(2(R^2+omega^2L^2))`D. none of these |
|
Answer» Correct Answer - C `P=I_(max)^2R=(V_("rms")/Z)^2R` `=[((V_0//sqrt2)^2)/(R^2+omega^2L^2)]R` `=(V_0^2R)/(2(R^2+omega^2L^2))` |
|
| 173. |
Assertion : A step-up transformer changes a low voltage into a high voltage. Reason : This violate the law of conservation of energy.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 - C A step up transformer changes a low voltage into a voltage. This does not violate the law of conservation of energy, because the current is reduced by the same proportion. |
|
| 174. |
In the circuit shown if fig, the rims currents `(I_1), (I_2)` and` (I_3)` are altered by varying the freqency f of the oscillator. The output voltage of the oscillatro remains sinusoidal and has a fixed amplitude. When curves in figure correctly indicate the variation with frequency of the currents `(I_1), (I_2),and (I_3)`. A. (A)`(I_1)=Q`, `(I_2)=Q`, `(I_3)=Q`B. (B)`(I_1)=R`, `(I_2)=Q`, `(I_3)=Q`C. (C) `(I_1)=Q`, `(I_2)=P`, `(I_3)=R`D. (D)`(I_1)=Q`, `(I_2)=R`, `(I_3)=P` |
|
Answer» Correct Answer - D Rectancae of the inductor O is given by `X_(L)=2 pi f L`. Therefore, rms current through inductor L is `(I_2)` `=(V)/(2 pi f L) prop (1)/(f)` where V is the rms of the supply voltage . Reactance of the capacitor C is given by `(X_C)=(1)/(2 pi f C)` therefore, rms current through the capacitor is `(I_3)=(V)/(X_C) = 2 pi f CV prop f` The total current `(I_1)` is given by `(I_1)=(I_2)+(I_3)` `((V)/( 2 pi f L)- 2 pi f CV)` thus, `(I_2)` is best represented by curve R. `(I_3)` is best represented by curve P. `(I_1)` is best represented by curve Q. |
|
| 175. |
In a non -ideal transformer the primary and secondary voltages and currents are `V_1,I_1` and `V_2,I_2` respectively. The efficiency of the transformer isA. `V_2/V_1`B. `I_2/I_1`C. `(V_2I_2)/(V_1I_1)`D. `(V_1I_1)/(V_2I_2)` |
| Answer» Correct Answer - C | |
| 176. |
An alternating current is given by `I = i_1 cos omegat + i_2 sin omegat`. The rms current is given byA. `1/(sqrt(2))(i_(1)+i_(2))`B. `1/(sqrt(2))(i_(1)+i_(2))^(2)`C. `1/(sqrt(2))(i_(1)^(2)+i_(2)^(2))^(1//2)`D. `1/(2)(i_(1)^(2)+i_(2)^(2))^(1//2)` |
|
Answer» Correct Answer - C `i_(r.m.s)=sqrt((i_(1)^(2)+i_(2)^(2))/2)=1/(sqrt(2))(i_(1)^(2)+i_(2)^(2))^(1//2)` |
|
| 177. |
The peak value of an alternating emf E given by `E=(E_0) cos omega t` is 10V and frequency is 50 Hz. At time `t=(1//600)s` the instantaneous value of emf isA. `10V`B. `5sqrt(3)V`C. `5V`D. `1V` |
|
Answer» Correct Answer - B `E=E_(0)cos omegat=E_(0)cos (2pit)/T` `=10cos.(2pixx50xx1)/600=10cos(pi/6)=5sqrt(3)`volt. |
|
| 178. |
In an `LCR` circuit `R=100 ohm`. When capacitance `C` is removed, the current lags behind the voltage by `pi//3`. When inductance `L` is removed, the current leads the voltage by `pi//3`. The impedence of the circuit isA. `50 Omega`B. `100 Omega`C. `200 Omega`D. `400 Omega` |
|
Answer» Correct Answer - B |
|
| 179. |
In an `LCR` circuit `R=100 ohm`. When capacitance `C` is removed, the current lags behind the voltage by `pi//3`. When inductance `L` is removed, the current leads the voltage by `pi//3`. The impedence of the circuit isA. `50` ohmB. `100` ohmC. `200 ohm`D. `400` ohm |
|
Answer» Correct Answer - B When `C` is removed circuit becomes `RL` circuit hence `tan ((pi)/3)=(X_(L))/R.......(i)` When `L` is removed circuit becomes `RC` circuit hence `tan(pi/3) =(X_(C))/R.....(ii)` from equation `(i)` and `(ii)` we obtain `X_(L)=X_(C)`. This is the condition of resonance and in resonance `Z=R=100 Omega` |
|
| 180. |
An `L-C-R` series circuit with `100Omega` resistance is connected to an `AC` source of `200V` and angular frequency `300rad//s`. When only the capacitance is removed, the current lags behind the voltage by `60^@`. When only the inductance is removed the current leads the voltage by `60^@`. Calculate the current and the power dissipated in the `L-C-R` circuitA. 50 WB. 100 WC. 200 WD. 400 W |
|
Answer» Correct Answer - D Phase angle , tan`phi` = `X_L`/R = `X_C`/R `implies` tan`60^(@)` = `X_L`/R = `X_C`/R `implies` `X_L` = `X_C` = `sqrt 3` i.e., Z = `sqrt R^(2) + (sqrt 3 R - sqrt 3 R)^(2) ` `implies` Z = R So, avaerage power , P = `V^(2)`/R = `200 xx 200 `/ 100 = 400 W |
|
| 181. |
The peak value of an alternating emf E given by `E=(E_0) cos omega t` is 10V and frequency is 50 Hz. At time `t=(1//600)s` the instantaneous value of emf isA. 10 VB. `5 sqrt(3) V`C. 5 VD. 1 V |
| Answer» Correct Answer - b | |
| 182. |
A current of 4 A flows in a coil when connected to a 12 V dc source. If the same coil is connected to a `12 V, 50 rad s^(-1)` ac source, a current of 2.4 A flows of the coil in the circuit. The inductance of the coil isA. 0.02HB. 0.04HC. 0.08HD. 1.0H |
|
Answer» Correct Answer - C When dc is applied , `I=V/R i.e, R=12/4 = 3 Omega` and when ac is applied , `Z+V/I = ((12)/(2.4))= 5 Omega` `R^(2) + X_(L)^(2) = 5^(2) (as Z= sqrt(R^(2)+X_(L)^(2)))` so, `X_(L)^(2) = 5^(2)-R^(2) = 5^(2)-3^(2)=4^(2), i.e. X_(L) = 4 Omega` but as , `X_(L) = omega L , L (X_L) /(omega) = (4)/(50)= 0.08 H`. |
|
| 183. |
A current of 4 A flows in a coil when connected to a 12 V dc source. If the same coil is connected to a `12 V, 50 rad s^(-1)` ac source, a current of 2.4 A flows of the coil in the circuit. The power developed in the circuit of a `2500 (mu)F` capacitor that is connected in series with the coil isA. 17.28 WB. 8.64 WC. 10 WD. 15 W |
|
Answer» Correct Answer - A Now when the capacitor is connected to the above circuit in series, `X_(C)=1/(omega C) = (1)/(50 xx 2500 xx 10^(-6)) =(10^3)/(125) = 8 Omega` so, `Z=sqrt(R^(2)+(X_(L)-X_(C ))^(2)) = sqrt(3^(2)+(4-8)^(2))= 5 Omega` and hence, `I_(rms) = V/Z 12/5 = 2.4 A` so, `P_(av) = V_(rms) I_(rms) cos phi = (I_(rms)xxZ) xx (I_(rms))xx((R )/(Z))` i.e., `P(av)=I_(rms)^(2)R=(2.4)^(2) xx 3 = 17.28 W`. |
|
| 184. |
A sereis R-C circuit is connected to AC voltage source. Consider two cases, (A) when C is without a dielectric medium and (B) when C is filled with dielectric of constant 4. The current `I_(R)` through the resistor and voltage `V_(c)` across the capacitor are compared in the two cases. Which of the following is/ are true?A. `I_(R )^(A) gt I_(R )^(B)`B. `I_(R )^(A) lt I_(R )^(B)`C. `V_(C )^(A) gt V_(C)^(B)`D. `V_(C )^(A) lt V_(C)^(B)` |
|
Answer» Correct Answer - B::C Case (A): `Z_(A)=sqrt(R^(2)+((1)/(omega C))^(2))` case (B) `Z_(B)=sqrt(R^(2)+((1)/(omega K C))^(2))` so, `Z_(B)ltZ_(A)` `I_(R )^(A)=V/(Z_A)` and `I_(R )^(B) = V/(Z_B) clearly `I_(R )^(A) lt I_(R )^(B)`. Since currnet in case (B) is greater so, p.d. across R will increases in case(B) and thus across capacitor will decrease. Hence `V_(C )^(A) gt V_(C)^(B)`. `(V_(R )^(2)+V_(C )^(2)=V_(0)^(2))` . |
|
| 185. |
In an `LCR` circuit `R=100 ohm`. When capacitance `C` is removed, the current lags behind the voltage by `pi//3`. When inductance `L` is removed, the current leads the voltage by `pi//3`. The impedence of the circuit isA. 50 ohmB. 100 ohmC. 200 ohmD. 400 ohm |
| Answer» Correct Answer - b | |
| 186. |
In an `LCR` circuit `R=100 ohm`. When capacitance `C` is removed, the current lags behind the voltage by `pi//3`. When inductance `L` is removed, the current leads the voltage by `pi//3`. The impedence of the circuit isA. 50ohmB. 100ohmC. 200ohmD. 400ohm |
|
Answer» Correct Answer - B |
|
| 187. |
An `L-C-R` series circuit with `100Omega` resistance is connected to an `AC` source of `200V` and angular frequency `300rad//s`. When only the capacitance is removed, the current lags behind the voltage by `60^@`. When only the inductance is removed the current leads the voltage by `60^@`. Calculate the current and the power dissipated in the `L-C-R` circuitA. `1A,200` wattB. `1A,400` wattC. `2A,200` wattD. `2A,400` watt |
| Answer» Correct Answer - D | |
| 188. |
In the given figure, the instantaneous value of alternating e.m.f. is `e = 14.14 sin omega t`. The reading of voltmeter in volt will be A. `141.0`B. 10C. 200D. 70.7 |
|
Answer» Correct Answer - 2 Reading of voltmeter in rms value `E = (14.14)/(sqrt(2)) = 10 V` |
|
| 189. |
A current of 4 A flows in a coil when connected to a 12 V dc source. If the same coil is connected to a `12 V, 50 rad s^(-1)` ac source, a current of 2.4 A flows of the coil in the circuit. The power developed in the circuit of a `2500 (mu)F` capacitor that is connected in series with the coil is |
|
Answer» Correct Answer - `08 H; 17.28` watt Given that `R=V/I=12/4=3 Omega` `Z=12/2.4=5 Omega` `Z^(2)=R^(2)+(omegaL)^(2)` `25=p+(50xxL)^(2) rArr L=0.08 H` Now for `LCR` circuit `Z=sqrt((R^(2):(X_(L)-X_(C))^(2)` ) `=sqrt(3^(2)+(50xx0.08-1/(50xx2500xx10^(-5)))^(2))=5Omega`. So Power=`(12/Z)^(2)xxR=(12/5)^(3)xx3=17.28` watt. |
|
| 190. |
A sereis R-C circuit is connected to AC voltage source. Consider two cases, (A) when C is without a dielectric medium and (B) when C is filled with dielectric of constant 4. The current `I_(R)` through the resistor and voltage `V_(c)` across the capacitor are compared in the two cases. Which of the following is/ are true?A. `I_(R)^(A) gtI_(R)^(B)`B. `I_(R)^(A) ltI_(R)^(B)`C. `V_(R)^(A) gt V_(R)^(B)`D. `V_(R)^(C) gt V_(C)^(B)` |
| Answer» Correct Answer - B::C | |
| 191. |
In `LCR` circuit at resonance current in the circuit is `10 sqrt(2) A`. If tnow frequency of the source is changed such that now current lags by `45^(@)` that applied voltage in the circuit . Which of the following is correct.A. Frequency must be increased and current after the change is 10 AB. Frequency must be decreased and current after the change is 10 AC. Frequency must be decreased and current is same as that of initial valueD. The given information is insufficient to conclude anything |
| Answer» Correct Answer - A | |
| 192. |
A current of `4A` flows in a coil when connected to a `12V DC` source. If the same coil is connected to a `12V, 50 rad//s AC` source, a current of `2.4A` flows in the circuit. Determine the inductance of the coil. Also, find the power developed in the circuit if a`2500muF` capacitor is connected in series with the coil.A. `(pi/sqrt3)H`B. `(sqrt3/pi)H`C. `(2/pi)H`D. none of these |
|
Answer» Correct Answer - B `I_(DC)=V_(DC)/R` `I=100/RimpliesR=100Omega` `I_(AC)=V_(AC)/(sqrt(R^2+X_L^2))` `0.5=100/(sqrt((100)^2+X_L^2))` or `X_L=100sqrt3Omega=(2pifL)` `:. L=(100sqrt3)/(2pif)=(100sqrt3)/(2pi(50))` `=(sqrt3/pi)H` |
|
| 193. |
A current of `4A` flows in a coil when connected to a `12V DC` source. If the same coil is connected to a `12V, 50 rad//s AC` source, a current of `2.4A` flows in the circuit. Determine the inductance of the coil. Also, find the power developed in the circuit if a`2500muF` capacitor is connected in series with the coil. |
|
Answer» (i) A coil consists of an inductance `(L)` and a resistance `(R)` In `DC`, only resistance is effective. Hence, `R=V/i=12/4=3Omega` In `AC`, `i_("rms")=V_("rms")/Z=V_("rms")/(sqrt(R^2+omega^2L^2))` `:.L^2=1/omega^2[(V_("rms")/i_("rms"))^2-R^2]` `:. L=1/omega[(V_("rms")/i_("rms"))^2-R^(2)` Substituting the values, we have `L=1/50sqrt((12/2.4)^2-(3)^(2))` `=0.08H` (ii) When capacitor is connected to the circuit, the impedance is `Z=sqrt(R^2+(X_L-X_C)^(2))` Here `R=3Omega` `X_L=omegaL=(50)(0.08)=4Omega` and `X_C=1/(omegaC)=1/((50)(2500xx10^-6))=8Omega` `:. Z=sqrt((3)^2+(4-8)^(2))=5 Omega` Now, `(:P:)=V_("rms")i_("rms")cosphi` `=V_("rms")xxV_("rms")/ZxxR/Z` `=(V_("rms")/Z)^2xxR` Substituting the values, we have `(:P:)=(12/5)^2xx3` `=17.28W`. |
|
| 194. |
An AC voltage source of variable angular frequency `(omega)` and fixed amplitude `V_(0)` is connected in series with a capacitance C and an electric bulb of resistance R (inductance zero). When `(omega)` is increasedA. the bulb glows dimmerB. the bulb glows brighterC. total impendance of the circuits is unchangedD. total impendance of the circuit increases |
| Answer» Correct Answer - 2 | |
| 195. |
An AC voltage source of variable angular frequency `(omega)` and fixed amplitude `V_(0)` is connected in series with a capacitance C and an electric bulb of resistance R (inductance zero). When `(omega)` is increasedA. the bulb dimmerB. the bulb glows brigherC. total impedience of the circuit is unchangedD. total impednece of the circuit increases |
| Answer» Correct Answer - B | |
| 196. |
An AC voltage source of variable angular frequency `(omega)` and fixed amplitude `V_(0)` is connected in series with a capacitance C and an electric bulb of resistance R (inductance zero). When `(omega)` is increasedA. The bulb glows dimmerB. The bulb glows brighterC. Total impedance of the circuit is unchangedD. Total impedance of the circuit increases |
|
Answer» Correct Answer - B (b) `I_(rms)= V_(rms)/sqrt (R^(2) + 1/(omega^(2)c^(2))` when `omega increases, I_(rms)` increases so thwe bulb glows brighter . |
|
| 197. |
The phase difference between V and I of an LCR circuit in series resonance isA. `pi`B. `pi/2`C. `pi/4`D. 0 |
| Answer» Correct Answer - D | |
| 198. |
A series LCR circuit containing a resistance of 120 ohm has angular resonance frequency `4 xx 10^(3)rad s^(-1)`. At resonance, the voltage across resistance and inductance are 60V and 40 V respectively. The values of Land C are respectivelyA. `20mH, 25//8muF`B. `20mH,1//35muF`C. `20mH, 1//40muF`D. `2mH, 25//8muF` |
| Answer» Correct Answer - A | |
| 199. |
An `AC` circuit consists of an inductor of inductance `0.5H` and a capacitor of capacitance `8muF` in series. The current in the circuit is maximum when the angular frequency of `AC` source isA. `500 rad//sec`B. `2xx10^(5) rad//sec`C. `4000 rad//sec`D. `5000 rad//sec` |
|
Answer» Correct Answer - A Current will be maximum at the condition of resonance. So resonant frequency `omega_(0)=1/(sqrt(LC))=1/(sqrt(0.5xx8xx10^(-6)))` `=500rad//s` |
|
| 200. |
The phase difference between the current and voltage of `LCR` circuit in series combination at resonance isA. `0`B. `pi//2`C. `pi`D. `-pi` |
|
Answer» Correct Answer - A At resonance `LCR` series circuit behaves as pure resistive circuit. For resistive circuit `varphi=0^(@)`. |
|