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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.
1151. |
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^(3)rad//s`B. `10^(4)rad//s`C. `5000rad//s`D. `500rad//s` |
Answer» Correct Answer - C | |
1152. |
For an LCR circuit, the power transferred from the driving source to the driven oscillator is `P = I^(2) Z cos phi`.A. Hence, the power factor `cosphi gt- 0, P gt-0`B. The driving force can give no energy to the oscillator (P=0) in some cases.C. the driving force cannot syphon out`(Plt0)` the energy out of oscialltorD. The driving force can take away energy out of the oscillator. |
Answer» Correct Answer - A::B::C According to the question, power transferred, `P=I^(2)Zcosphi` Where I is the current, Z= impedance and `cosphi` is power factor. As power factor, `cosphi=R/Z` where, `Rgt0` and `Zgt0` `rArr cosphigt0 rArr Pgt0` |
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1153. |
An `ac` generator `G` with an adjustable frequency of oscillation is used in the circuit, as shown. Thermal energy produced by the resistance `R` in time duration `1 mus`, using the source at resonant condition, is A. `0 J`B. `1 muJ`C. `100 muJ`D. not possible to calculate from the given information |
Answer» Correct Answer - D | |
1154. |
For an LCR circuit, the power transferred from the driving source to the driven oscillator is `P = I^(2) Z cos phi`.A. the power factor `cos phi ge 0, P ge 0`B. the driving force can give no energy to the oscillator(P = 0) in some cases.C. the driving force cannot syphon out `(P lt 0)` the energy out of oscillator.D. all of these. |
Answer» Correct Answer - D Here, `P = I^(2)Z cos phi` (a) If power factor `cos phi ge 0 rArr P ge0` (b) For wattless component the driving force shall give no energy to the oscillator, so, at `phi = 90^(@), P = 0`. (c ) The driving force cannot syphon out the energy out of oscillator. i.e. P cannot be negative. Hence all options are correct. |
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1155. |
In the `LCR` circuit shown in figure unknown resistance and alternating voltage source are connected.When swith `S` is closed then there is a phase difference of `pi/4` between current and applied voltage and voltage across resister is `100/sqrt2 V`.When switch is open current and applied voltage are in same phase.Neglecting resistance of connecting wire answer the following questions : Peak voltage of applied voltage sources is: A. `200sqrt2 V`B. `100 V`C. `100sqrt2 V`D. `100/sqrt2 V` |
Answer» Correct Answer - C | |
1156. |
STATEMENT-1: By only knowing the power factor for a given LCR circuit it is not possible to tell whether the applied alternating emf leads or lags the current. STATEMENT-2: `cos theta=cos(-theta)`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. |
Answer» Correct Answer - A For a certain values of `cos theta` (power factor) two values of `theta` are possible. One is positive the other is much negative. Accordingly the applied e.m.f may lead or lag. |
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1157. |
In a series LCR circuit, the phase difference between the voltage and the current is `45^(@)`. Then the power factor will beA. `0.607`B. `0.707`C. `0.808`D. 1 |
Answer» Correct Answer - B Here, `phi = 45^(@)` In series LCR circuit, power factor `=cos phi` `therefore cos phi = cos 45^(@)=(1)/(sqrt(2))=0.707`. |
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1158. |
In the `LCR` circuit shown in figure unknown resistance and alternating voltage source are connected.When swith `S` is closed then there is a phase difference of `pi/4` between current and applied voltage and voltage across resister is `100/sqrt2 V`.When switch is open current and applied voltage are in same phase.Neglecting resistance of connecting wire answer the following questions : Peak voltage of applied voltage sources is: A. `200sqrt2V`B. 100VC. `100sqrt2V`D. `(100)/(sqrt2)V` |
Answer» Correct Answer - C | |
1159. |
An emf of V=`V_0sin omegat` is applied on a series LCR circuit. IF there is no phase difference between the voltage and current then,A. `I=V_0/Rsinomegat`B. `omegaL=1/(omegaL)`C. effective power=`V_0^2/R`D. ratio of terminal potential difference across L and R =`1/(omegaCR)` |
Answer» Correct Answer - A::B::D | |
1160. |
What is the phase difference between voltage and current when the power factor in LCR series circuit is unity ? |
Answer» In LCR series circuit power factor `(cos phi) = 1` `therefore` Phase difference between voltage and current is zero. `(phi = 0)` |
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1161. |
The rms value of an ac of 50Hz is 10A. The time taken be an alternating current in reaching from zero to maximum value and the peak value will beA. `2xx10^(-2) s and 14.14 A`B. `1xx10^(-2) s and 7.07 A`C. `5xx10^(-3) s and 7.07 A`D. `5xx10^(-3) s and 14.14 A` |
Answer» Correct Answer - D Time for reaching maximum of peak value form 0 `(T)/(4) =(1)/(4)xx(1)/(500 s)=1/200 s =5xx10^(-3)s` `(I_0)=10 sqrt(2) A =14.14 A` |
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1162. |
Assertion: When capacitive reactance is smaler than the inductive reactance is `LCR` circuit, e.m.f. leads the current. Reason : The phase angle is the angle between the alternating e.m.f. and alternating current of the circuit.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 | |
1163. |
The rms value of an ac of 50Hz is 10A. The time taken by an alternating current in reaching from zero to maximum value and the peak value will beA. `2 xx 10^(-2)` sec and 14.14 ampB. `1 xx 10^(-2)` sec and 7.07 ampC. `5 xx 10^(-3)` sec and 7.07 ampD. `5 xx 10^(-3)` sec and 14.14 amp |
Answer» Correct Answer - D `l_("rms") = 10 A implies l_(o) = 10sqrt(2) A = 14.14` Amp Let `I = 0` when `t = 0`. `:. l = l_(o) sin omega t` `l_(o) = l_(o) sin omega t implies omega t = (pi)/(2)` `implies t = (pi)/(2 omega) = (pi)/(2 xx 2pi xx f) = (1)/(200) = 5 xx 10^(-3) sec` |
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1164. |
Assertion: In series `LCR` circuit resonance can take place. Reason: Resonance takes place if inductance and capacitive reactance are equal and opposite.A. Both Assertion and Reason are true and Reason is the correct explanation of Assertion.B. Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.C. Assertion is true but Reason is falseD. Assertion is false but Reason is true |
Answer» Correct Answer - 1 In a series `R-L-C` circuit resonance can takes place, because at a particlar frequency of ac source, the impedance can be minimum. `Z = sqrt(R^(2) + (X_(L) - X_(C ))^(2))` Resonance means current is maximum `implies` Inpdedance is minimum `implies X_(L) = X_(C )`. |
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1165. |
The r.m.s value of an a.c of `59 Hz` is `10 A`. The time taken by the alternating current in reaching from zero to maximum valuef and the peak value of current will beA. `2 xx 10^(-2) "sec"` and `14.14 A`B. `1 xx 10^(-2) "sec"` and `7.07 A`C. `5 xx 10^(-3) "sec"` and `7.07 A`D. `5 xx 10^(-3) "sec"` and `14.14 A` |
Answer» Correct Answer - 4 `i_(0) = sqrt(2) i_("rms")` , `T = (1)/(f), t = (T)/(4)` |
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1166. |
Eddy currents are produced in a matterial when it isA. heatedB. placed in a time varying magnetic fieldC. placed in an electric fieldD. placed in a unifrom magnetic field |
Answer» Correct Answer - B |
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1167. |
An iron rod is inserted in the inductro, what will be the effect on the brightness of the bulb? |
Answer» First decreases, thaen becomes same. Beacause as the rod is inserted, L increases From `f=LI, I decreases. But soon the current will become same as original. | |
1168. |
A soft iron rod is inserted in the solenoid. After this what will be the effect on the brightness of the bulb? |
Answer» The brightness will decreases, because L increases on inserting the iron rid and impedence `Z= sqrt(R^(2)+(omega L)^(2))` increases. This decreases the current, hence brightness decreases. | |
1169. |
A lamp L is connected to an ac source along with an air core inductor as shown in the Figure. How is the brilliance of the lamp affected if core made of following material is inserted inside the inductor? (a) Iron (b) Copper (c) Iron sheets pasted together with insulation in between [laminated Iron core] In which case the lamp will be least bright? |
Answer» Correct Answer - (i) Brilliance of lamp decreases (ii) Brilliance increase (iii) Brilliance decrease The lamp is least bright with laminated iron core. |
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1170. |
In a step-up transformer the turn ratio is `1:10`. A resistance of `200` ohm connected across the secondary is drawing a current of `0.5A`. What is the primary voltage and current?A. `50V,1A`B. `10V,5A`C. `25V, 4A`D. `20 V, 2A` |
Answer» Correct Answer - B `V_(s)=I_(s)R=0.5xx200=100` volts `(V_(p))/(V_(s))=(n_(p))/(n_(s))=1/10` or `V_(s)=(V_(s))/10=100/10 V=10V` Again `(I_(p))/(I_(s))=(n_(p))/(n_(s))=10` or `I_(p)=10 I_(s)=10xx0.5 A=5A` |
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1171. |
A `50 Hz` alternating current of peak value `1` ampere flows through the primary coil of a transformer. If the mutual inductance between the primary secondary be `1.5` henry, then the peak value of the induced voltage isA. `75 V`B. `150 V`C. `225 V`D. `300 V` |
Answer» Correct Answer - D `T=1/50 s` For the current to rise from zero to peak value, required time is `T/4` i.e. `1/200 s`. Now `E=1.5 (1-0)/(1//200) V=1.5xx200V=300 V`. |
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1172. |
A coil of `40 H` inductance is connected in series with a resistance of `8` ohm and the combination is joined to the terminals of a `2 V` battery. The time constant of the circuitA. `5s`B. `1//5s`C. `40s`D. `20s` |
Answer» Correct Answer - A Time constant of `L-R` circuit is, `tau=L/R` Here, `L=40 H, R=8 Omega, E=2V` `:. tau =40/8=5s` |
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1173. |
When 100 V dc is applied across a coil, a current of 1A flows through it and when 100 V ac of 50 Hz is applied to the same coil, only 0.5 flows The inductance of coil isA. 5.5 HB. `3//(pi)H`C. `sqrt(3)//(pi)H`D. `2.5 H` |
Answer» Correct Answer - C when dc is applied , `R=V/I =(100)/(1) = 100 Omega` and when ac of 50 Hz is applied `I=V/Z i.e., Z=V/I = 100/0.5 = 200 Omega`. `Z=sqrt(R^(2)+omega^(2)L^(2))` i.e., `omega^(2)L^(2)=Z^(2)-R^(2)` i.e., `(2 pi fL)^(2)=(200^2)-(100^2)=3xx10^(4)` (`as o,ega = 2 pi f`) so,` L=(sqrt(3)xx10^(2))/(2 pi xx 50 ) =(sqrt(3))/(pi)H=0.55H`. |
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1174. |
A box P and a coil Q are connected in series with an ac source of variable frequency. The emf of the source is constant at 10 V. Box P xontains a capacitance of `32 Omega`. Coil Q has a self inductance of 4.9 mH and a resistance of `68 Omega` in series. The frequency is adjusted so that maximum current flows in P and Q. The impedance of P at this frequency isA. `77 Omega`B. `36 Omega`C. `40 Omega`D. `125 Omega` |
Answer» Correct Answer - A As this circuit is a series LCR circuit, cuttent will be maximum at reasonance , i.e. `omega = (1)/(sqrt(LC))v =(1)/(sqrt((4.9 xx10^(-3))(10^(-6)))=(10^5)/(7) rad s^(-1)` with `I=(V)/(R ) =(10)/((32+68))=(1)/(10)A` so the impedance, `Z_(P)=[R_(1)^(2)+(1//omegaC)^(2)]^(1//2) = sqrt(5924)=77 Omega`. |
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1175. |
A sinusoidal voltage of frequency 60 Hz and peak value 150 V is applied to a series L-R circuit, where `R=20Omega` and L=40 mH. a) Compute T, `omega, X |
Answer» `As I_(rms) = V_(rms)/X_(L)` Where, `X_(L)=omegaL=2pifL` is the reactance of the inductor Here, f=50H, L=44mH =`44xx10^(3)`H, `V_(rms)=220V` `therefore X_(L)=2pi xx 50 xx 44 xx 10^(-3)= 13.82Omega` `therefore` rms value of current in the circuit `I_(rms) = 220/13.82=15.9A` |
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1176. |
A sinusoidal voltage of peak value `283 V` and frequency `50 Hz` is applied to a series `LCR` circuit in which `R = 3 Omega, L = 25.48 mH`, and `C = 796 mu F`. Find the impdedance of the circuit. |
Answer» Correct Answer - 5 To find the impedance of the circuit, we first calculate `X_(L)` and `X_(C)`. `X_(L)=2pifL` `=2xx3.14xx50xx25.48xx10^(-3)Omega=8Omega` `X_(C)=(1)/(2pifC)=(1)/(2xx3.14xx50xx796xx10^(-6))=4Omega` Therefore, `Z=sqrt(R^(2)+(X_(L)-X_(C))^(2))=sqrt(3^(2)+(8-4)^(2))=5Omega` |
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1177. |
A sinusoidal voltage of peak value 293 V and frequency 50 Hz is applie to a series LCR circuit in which `R=6 Omega, L=25 mH` and `C=750mu F`. The impedance of the circuit isA. `7.0 Omega`B. `8.9 Omega`C. `9.9 Omega`D. `10.0 Omega` |
Answer» Correct Answer - A Here, `R = 6 Omega, L=25 mH=25xx10^(-3)H`, `C=750 mu F = 750xx10^(-6)F, upsilon = 50 Hz` `X_(L)=2pi upsilon L =2xx3.14xx50xx25xx10^(-3)=7.85 Omega` `X_(C )=(1)/(2pi upsilon C)=(1)/(2xx3.14xx50xx750xx10^(-6))=4.25 Omega` `therefore X_(L)-X_(C )=7.85-4.25=3.6Omega` Impedence of the series LCR circuit is `Z=sqrt(R^(2)+(X_(L)-X_(C ))^(2))` `therefore Z=sqrt((6)^(2)+(3.6)^(2))=sqrt(36+12.96)=7.0 Omega` |
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1178. |
A sinusoidal voltage of peak value 283 V and frequency 50 Hz is applied to a series LCR circuit in which R = 3 Ω . L = 25.48 m H . And C = 796 μ F .Suppose the frequency of the source in the previous example can be varied. Calculate the impedance, the current, and the power dissipated at the resonant condition. |
Answer» The impedance Z at resonant condition is equal to the resistance `Z = R = 3Omega` The rms current at resonance is, as `V=(upsilon_(m))/(sqrt(2))` `I=(V)/(Z)=(V)/(R )=((283)/(sqrt(2)))(1)/(3)=66.7A` The power dissipated at resonance is `P=I^(2)xxR=(66.7)^(2)xx3=13.35 kW` You can see that in the present case, power dissipated at resonance is more than the power dissipated. |
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1179. |
In a transformer , the coefficient of mutual inductance between the primary and the secondary coil is `0.2` henry. When the current changes by `5` ampere//second in the primary, the induced e.m.f. in the secondary will beA. 5VB. 1 VC. 25 VD. 10 V |
Answer» Correct Answer - B Induced emf in the circuit , e = M di/dt = (0.2)(5) = 1 V |
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1180. |
In an inductor of inductance `L=100mH`, a current of `I=10A` is flowing. The energy stored in the inductor isA. 5 JB. 10 JC. 100 JD. 1000 J |
Answer» Correct Answer - A | |
1181. |
A step-down transformer is used on a `1000 V` line to deliver `20A` at `120V` at the secondary coil. If the efficiency of the transformer is `80%` the current drawn from the line is.A. `3 A`B. `30A`C. `0.3A`D. `2.4 A` |
Answer» Correct Answer - A `eta=(Output)/(Input) implies 80/100 =(20xx20)/(1000xxi_(l))` `implies i_(l)=(20xx120xx100)/(1000xx80)=3A`. |
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1182. |
A step-down transformer is used on a `1000 V` line to deliver `20A` at `120V` at the secondary coil. If the efficiency of the transformer is `80%` the current drawn from the line is.A. 3AB. 30 AC. 0.3AD. 2.4 V |
Answer» Correct Answer - A `eta = ("Output Power")/("Input Power")` `rArr 80/100 = (20 xx 120)/(1000 xx I)` `rArr I=(20 xx 120xx100)/(1000 xx I)` `rArr I = (20 xx 120 xx 100)/(1000 xx 80) = 3A` |
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1183. |
Transformer is a device used to increase or decrease the voltage in the transmission line according to requirement, Generally the input line voltage is fed in a primary coil and the output line voltage is obtained from the terminals of another coil In an ideal transformer, the primary and secondary coils are linked is such a way that there is no loss of magnetic flux and electrical energy. In an ideal transformer, if the number of turns and input voltage across the terminals of the primary coils be `N_1` and `V_1`, then the output voltage at the two terminals of the secondary coil `V_2=V_1.N_2/N_1`, where `N_2` is the number of turns in the secondary coil . The ratio of number of turns of the primary and secondary coils of an ideal transformer is 2:1 IF the input voltage is 440 V, then output voltage isif the input power of the transformer be 44 W, then output power isIn the above mentioned transformer the input and output currents are respectively.A. 220 VB. 440 VC. 880VD. None of these |
Answer» Correct Answer - A | |
1184. |
Transformer is a device used to increase or decrease the voltage in the transmission line according to requirement, Generally the input line voltage is fed in a primary coil and the output line voltage is obtained from the terminals of another coil In an ideal transformer, the primary and secondary coils are linked is such a way that there is no loss of magnetic flux and electrical energy. In an ideal transformer, if the number of turns and input voltage across the terminals of the primary coils be `N_1` and `V_1`, then the output voltage at the two terminals of the secondary coil `V_2=V_1.N_2/N_1`, where `N_2` is the number of turns in the secondary coil . In question (i) if the input power of the transformer be 44 W, then output power isA. 22 WB. 44 WC. 88 WD. None of these |
Answer» Correct Answer - B | |
1185. |
Transformer is a device used to increase or decrease the voltage in the transmission line according to requirement, Generally the input line voltage is fed in a primary coil and the output line voltage is obtained from the terminals of another coil In an ideal transformer, the primary and secondary coils are linked is such a way that there is no loss of magnetic flux and electrical energy. In an ideal transformer, if the number of turns and input voltage across the terminals of the primary coils be `N_1` and `V_1`, then the output voltage at the two terminals of the secondary coil `V_2=V_1.N_2/N_1`, where `N_2` is the number of turns in the secondary coil . In the above mentioned transformer the input and output currents are respectively.A. 100 mA,100mAB. 200 mA,200mAC. 100mA,200mAD. 200mA,100mA |
Answer» Correct Answer - C | |
1186. |
An ac source of angular frequency `omega` is fed across a resistor R and a capacitor C in series. The current registered is I. If now the frequency of source is changed to `omega//3` (but maintaining the same voltage), the current in the circuit is found to be halved. Calculate the ratio of the reactance to resistance at the original frequency `omega`.A. `sqrt(2/5)`B. `sqrt(3/5)`C. `sqrt(1/5)`D. `sqrt(4/5)` |
Answer» Correct Answer - B At angular frequency `omega`, the current in `RC` circuit is given by `I_(rms)=(V_(rms))/(sqrt(R^(2)+(1/(omegaC))^(2)))......(i)` Also, `(I_(rms))/2=(V_(rms))/(sqrt(R^(2)+[1/((omega c)/3)]^(2))) =(V_(rms))/(sqrt(R^(2)+9/(omega^(2)C^(2)))).....(ii)` From eq(i), and (ii) we get `3R^(2)=5/(omega^(2)C^(2))` `implies (1/(omegaC))/R=sqrt(3/5) implies (X_(C))/R=sqrt(3/5)` |
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1187. |
A resistance of `300Omega` and an inductance of `1/(pi)` henry are connected in series to an `AC` voltage of `20 "volts"` and `200Hz` frequency. The phase angle between the voltage and current isA. `tan^(-1) (4/3)`B. `tan^(-1) (3/4)`C. `tan^(-1) (3/2)`D. `tan^(-1) (2/5)` |
Answer» Correct Answer - A Phase angle `tan varphi=(omegaL)/R=(2pixx200)/300xx1/(pi)=4/3` `:. varphi=tan^(-1)(4/3)` |
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1188. |
The rms value of current in an ac circuit is 10A. What is the peak current ? |
Answer» Using the formula `I_("rms")=(I_(0))/(sqrt(2))` The prak value of current `I_(0)=sqrt(2)xx I_("rms")` `= sqrt(2)xx10 =1.414xx10` `I_(0)=14.14 A` |
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1189. |
An `AC` source of variable frequency `f` is connected to an `LCR` series circuit. Which one of the graphs in figure represents the variation of current of current `I` in the circuit with frequecy `f`?A. B. C. D. |
Answer» Correct Answer - C As explainded in solution `(1)` for frequency `0-f_(r),Z` decrease hence `(i=V//Z)` increases and for frequency `f_(r)-oo,Z` increases hence `i` decreases. |
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1190. |
Obtain the resonant frequency `(omega_(r))` of a series LCR circuit withL = 2.0 H, C = 32 `muF` and R = 10 ohm. What is the Q value of this circuit ? |
Answer» Inductive, L=2.0H Capacitance, C=`32muF=32xx10^(-6)F` Resistance, R=`10Omega` Resonant frequency is given by the relation, `omega_(r)=(1)/sqrt(LC)` `=(1)/(10)sqrt((2)/(32xx10^(-6)))=(1)/(10)xx(1)/(4xx10^(-3))=25 ` ltbr gt Hence, the Q-value of this circuit is 25. |
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1191. |
A capacitor and the resistor are connected in series with an ac source. IF the potential differences acorrs C,R are 120 V, 90 V respectively and if the rms current of the circuit is 3A , calculate the power factor of the circuit. |
Answer» Power factor of the circuit,`costheta=V_R/V=90/150=0.6` | |
1192. |
An `AC` source of variable frequency `f` is connected to an `LCR` series circuit. Which one of the graphs in figure represents the variation of current of current `I` in the circuit with frequecy `f`?A. B. C. D. |
Answer» Correct Answer - d | |
1193. |
Obtain the resonant frequency `omega_(r )` of a series LCR circuit with `L = 2.0H. C = 32 mu F` and `R = 10 Omega`. What is the Q-value of this circuit ? |
Answer» Given, `L = 2H, C = 32muF, R = 10 Omega` Resonant angular frequency `omega_(r )=(1)/(sqrt(LC))=(1)/(sqrt(2xx32xx10^(-6)))=125` rad/s Q-factor of this circuit, `Q = (1)/(R )sqrt((L)/(C ))=(1)/(10)sqrt((2)/(32xx10^(-5)))=(10^(3))/(40)=25`. |
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1194. |
A `200muF` capacitor is in series with a `50Omega` resistor and is connected to a 220 V, 50 Hz ac source(i) What is the maximum current in the circuit (ii) What is the difference in time when the current and the voltage attain maximum values? |
Answer» Angular frequency of the source, `omega=2pif=2times3.14times50Hz` `C=200muF=2times10^-4F` Maximum current through the circuit , `I_0=V_0/(sqrt(R^2+1/(C^2omega^2)))` `=(sqrt2times220)/(sqrt((50)^2+1/((2times10^-4)^2times4times(3.14)^2times(50)^2)))` Now if`theta` is the phase angle then `tantheta=1/(omegaCR)=1/(2pifCR)=1/(2times3.14times50times2times10^-4times50)` =0.3185 or, `theta=tan^-1(0.3185)=17.67^@=(17.67timespi)/180rad` If the voltage and the current attain maximum value at a time difference of t, then `theta=omegat` or,`t=theta/omega=(17.67timespi)/(180times2pitimes50)=9.82times10^-4s` |
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1195. |
If a LC circuit is considered analogous to a harmonically osicallting spring block system, which energy of the LC circuit would be analogous to potential energy and which one analogous to kinetic energy ? |
Answer» Correct Answer - A If we consider a L-C circuit analogous to a harmonically osciallting springblock system. The electrostatic energy `1/2CV^(2)` is analogous to potential energy and energy associated with moving charges (current) that is magnetic energy `(1/2LI^(2))` is analogous to kinetic energy. |
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1196. |
Study the circuit (a) and (b) shown in Fig. and answer the following questions. (a) Under which conditions would the rms currents in the two circuits be the same ? Can the rms curent in circuit (b) be larger than that in (a) ? |
Answer» Correct Answer - A Let, `(I_("rms"))a`=rms current in circuit (a) `(I_("rms"))b`=rms current in circuit (b) `(I_("rms"))a=V_(rms)/R =V/R` `(I_("rms"))b=V_("rms")/Z=V/sqrt(R^(2)+(X_(L)-X_(C))^(2)` a) When `(I_("rms"))a=(I_("rms"))b` `R=sqrt(R^(2) + (X_(L)-X_(C))^(2)` `rArr X_(L) = X_(C)`, resonance condition. b) As `Z gt-R` `rArr (I_("rms")a)/(I_("rms")b)=sqrt(R^(2)+(X_(L)-X_(C))^(2)/R` `=Z/R ` `rArr (I_("rms"))agt-(I_("rms"))b` No, the rms current in circuit (b), cannot be larger than that in a). |
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1197. |
A generator produces a time varying voltage given by `V=240 sin 120 t,` where `t` is in second. The rms voltage and frequency areA. `60Hz` and `240Hz`B. `19Hz` and `120Hz`C. `19Hz` and `170Hz`D. `754Hz` and `70Hz` |
Answer» Correct Answer - C `v=(omega)/(2pi)=(120xx7)/(2xx22)=19Hz` `V_(r.m.s)=240/(sqrt(2))=120sqrt(2)~~170V` |
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1198. |
A generator produces a time varying voltage given by `V=240 sin 120 t,` where `t` is in second. The rms voltage and frequency areA. `170V` and `19 Hz`B. `240 V` and `60 Hz`C. `170V` and `60 Hz`D. `120 V` and `19 Hz` |
Answer» Correct Answer - A `V_0=240V` `:. V_("rms")=V_0/sqrt2=240/sqrt2=170V` `omega=120rad//s` `f=omega/(2pi)=120/(2 xx 3.14)=19hZ` |
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1199. |
A generator produces a time varying voltage given by `V=240 sin 120 t,` where `t` is in second. The rms voltage and frequency areA. 60 Hz and 240 VB. 19 Hz and 120 VC. 19 Hz and 170 VD. 754 Hz and 70 V |
Answer» Correct Answer - c | |
1200. |
Is the same true for rms voltage? |
Answer» No, it is not true for rms voltage because the potential differences across various parts of the circuit may not be in the same phase. | |