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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.
| 301. |
When an inductor coil is connected to an ideal battery of emf `10V`, a constant current `2.5 A` flows. When the same inductor coil is connected to an ac source of `10V` and `50 HZ` then the current is `2A`. Find out the inductance of the coil. |
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Answer» When the coil is connected to dc source, the final current is decided by the resitance of the coil. `therefore r=(10)/(2.5)=4Omega` When the coil is connected to ac, the final current is decided by the impedance of the coil. `therefore Z=(10)/(2)=5Omega` `"But" Z=sqrt((r)^(2)+(X_(L))^(2))=X_(L)^(2)=5^(2)-4^(2)=9` `X_(L)=3Omega` `therefore omegaL=2pifL=3` `therefore 2pi 50 L =3 " " therefore L=3//100pi"Henry"` |
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| 302. |
In the above question if `V_(S)(t)=220sqrt2 sin (2pi 50 t)`, find (a)`l(t)` ,(b)`v_(R)` and (c )`v_(C)(t)` |
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Answer» (a)`i(t)=I_(m)sin(omegat+phi)` =`sqrt2sin(2pi50t+45^(@))` (b)`V_(R)=i_(R).R =i(t)R =sqrt2xx100 sin(100 pit+45^(@))` ( c)`V_(C)(t)=i_(C)X_(C)` (with a phase lag of `90^(@)`) =`sqrt2xx100 sin (100pit+45-90)` |
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| 303. |
A `9//(100 pi) H` inductor and a `12 Omega` resistance are connected in series to a `225 V`, `50 Hz` ac source. Calculate the current in the circuit and the phase angle between the current and the source voltage. |
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Answer» Here `X_(L)=omegaL=2pilL=2pixx50xx9/(100pi)=9Omega` So,`Z=sqrt(R^(2)+X_(L)^(2))=sqrt(12^(2)+9^(2))=15Omega` So (a)`I=V/Z=225/15=15A` and (b)`phi=tan^(-1)(X/L)=tan^(-1)(9/12)` i.e., the current will lag the applied voltage by `37^(@)`in phase. |
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| 304. |
A `(9)/(100pi)` H inductor and a 12ohm resitance are connected in series to a 225V, 50Hz ac source. Calculate the current in the circuit and the phase angle betweent the current and the source voltage. |
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Answer» `"Here" X_(L)=omegaL=2pif L=2pixx50xx(9)/(100pi)=9Omega` `"So", Z=sqrt(R^(2)+X_(L)^(2))=sqrt(12^(@)+9^(2))=15Omega` `"So" (a)I=(V)/(Z)=(225)/(15)=15A` `and (b) phi=tan^(-1)((X_(L))/(R))=tan^(-1)((9)/(12))=tan^(-1)3//4=37^(@)` i.e. the current will lag the applied voltage by `37^(@)` in phase. |
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| 305. |
In the above question if `V_(s)(t)=220sqrt2 sin (2pi 50t)` , find (a) i(t), (b), va and (c ) vc (t) |
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Answer» `(a) i(t)=I_(m)sin (omegat=phi)=2sin(2pi50t+45^(@))` `(b) V_(R)=i_(R).R=i(t) R=2xx100sin (100pit+45^(@))` `(c) V_(C)(t)=i_(c)X_(c)("with a phase lag of "90^(@))=2xx10sin(100pit+45-90)` |
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| 306. |
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. The ratio of reactance to resistance at the original frequency `omega` will be. |
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Answer» According to the given problems. `I=(V)/(Z)=(V)/([R^(2)+(1//Comega)^(2)]^(1//2))......(i)` `and (I)/(2)=(V)/([R^(2)+(3//Comega)^(2)]^(1//2))......(2)` Substituting the value of I from Equation (i) and (2). `4(R^(2)+(1)/(2C^(2)omega^(2)))=R^(2)+(9)/(C^(2)omega^(2)), i.e. (1)/(C^(2)omega^(2))=(3)/(5)R^(2)` `"So that ", (X)/(R)=((1//Comega))/(R)=((3)/(5)R^(2))^(1//2)/(R)=sqrt((3)/(5))` |
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| 307. |
Which of the following is the longest wave?A. `X-rays`B. `gamma-rays`C. `Microwaves`D. `Radiowaves` |
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Answer» Correct Answer - D |
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| 308. |
The structure of solids is invested by usingA. cosmic raysB. X-raysC. `gamma-rays`D. infrared waves |
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Answer» Correct Answer - C |
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| 309. |
What is the cause of "Green house effect"?A. Infrared raysB. Ultraviolet raysC. X-raysD. Radio-waves |
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Answer» Correct Answer - A |
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| 310. |
The overall efficiency of a transformer is `90%`.The transformer is rated for an output of `9000` watt.The primary voltage is `1000` volt.The ratio of turns in the primary to the secondary coil is `5:1`.The iron losses at full load are `700` watt.The primary coil has a resistance of `1 ohm`. The voltage in secondary coil is:A. `1000` voltB. `5000` voltC. `200` voltD. zero volt |
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Answer» Correct Answer - C `E_(2)/E_(1)=N_(2)/N_(1)=1/5` `E_(2)=1000/5=200` volt. |
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| 311. |
The overall efficiency of a transformer is `90%`.The transformer is rated for an output of `9000` watt.The primary voltage is `1000` volt.The ratio of turns in the primary to the secondary coil is `5:1`.The iron losses at full load are `700` watt.The primary coil has a resistance of `1 ohm`. In the above, the copper loss in the primary coil isA. `100` wattB. `700` wattC. `200` wattD. `1000` watt |
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Answer» Correct Answer - A Cu losses in first coil `=I_(1)^(2)R_(1)=(10)^(2)xx1=100` `=E_(1)I_(1)-E_(2)I_(2)` `=10,000-9000` `=1000` |
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| 312. |
An electric bulb has a rated power of `50 W` at `100 V`. If it is used on `AC` source of `200 V`, `50 Hz`, a choke has to be used in series with it. This choke should have an inductance of |
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Answer» Here, `P = 50 W, V = 100` volt `I = (P)/(V) = (50)/(100) = 0.5 A, R = (V)/(I) = (100)/(0.5) = 200 Omega` Let `L` be the inductance of the choke coil `:. I_(v)= (E_(v))/(Z) =` or `Z = (E_(v))/(I_(v)) = (200)/(0.5) = 400 Omega` Now `X_(L) = sqrt(Z^(2) - R^(2)) = sqrt(400^(2) - 200^(2))` `omega L = 100 xx 2 sqrt(3)` `L = (200 sqrt(3))/(omega) = (200 sqrt(3))/(2 pi v) = (200 sqrt(3))/(100 pi) = (2 xx 1.732)/(3.14) = 1.1 H` |
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| 313. |
A transformer having efficiency 90% is working on `100 V` and at `2.0 kW` power. If the current in the seconary coil is `5 A`, calculate (i) the current in the primary coil and (ii) voltage across the secondary coil. |
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Answer» Here, `eta = 90% = (9)/(10), I_(s) = 5A, E_(p) = 100 V` (i) `E_(p) I_(p) = 2 kW = 2000 W` `I_(p) = (2000)/(E_(p))` or `I_(P) = (2000)/(100) = 20 A` (ii) `eta = ("Output power")/("Input power") = (E_(s) I_(s))/(E_(P) I_(P))` or `E_(s) I_(s) = eta xx E_(p) I_(P)` `= (9)/(10) xx 2000 = 1800 W` `:. E_(s) = (1800)/(I_(s)) = (1800)/(5) = 360` volt |
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| 314. |
In the above question, the capacitive reactance in the circuit isA. `100 Omega`B. `25 Omega`C. `sqrt(125xx75)Omega`D. `400 Omega` |
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Answer» Correct Answer - C Impadance `Z=(E_V)/(I_V) or Z = (200)/(2) = 100 Omega` But ` Z^(2) = R^(2)+((1)/(omega c))^(2)` or ` ((I)/(omega C))^(2) Z^(2) -R^(2) = (100)^(2) -(25)^(2) = 125 xx 75` or, `X_(C) =(1)/(omega C) = sqrt(125 xx 75 ) Omega`. |
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| 315. |
A series `AC` circuit has a resistance of `4 Omega` and a reactance of `3 Omega`. The impedance of the circuit isA. `5 Omega`B. `7 Omega`C. `12//7 Omega`D. `7//12 Omega` |
| Answer» `Z=sqrt(4^(2)+3^(2))=5 Omega` | |
| 316. |
A series `AC` circuit has a resistance of `4 Omega` and a reactance of `3 Omega`. The impedance of the circuit isA. `5Omega`B. `7Omega`C. `12//7Omega`D. `7//12Omega` |
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Answer» Correct Answer - A `Z=sqrt(4^(2)+3^(2))"=5Omega` |
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| 317. |
An AC source producing `emf` `epsilon = epsilon_0 [cos(100 pi s^(-1)) t + cos (500 pi s^(-1))t]` is connected in series with a capacitor and a resistor. The steady-state current in the circuit is found to be `I = i_1 cos [(100 pi s^(-1) t + varphi_1] + i_2 cos [(500 pi s^(-1))t +phi_2]`.A. `I_(1) gt i_(2)`B. `i_(1) =i_(2)`C. `i_(1) lt i_(2)`D. the information is insufficient to find the relation between is found to be |
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Answer» Correct Answer - C `"Impedence z is give by z "sqrt(((1)/(omegaC))^(2)+R^(2))` For higher `omega`, z will be lower so current will be higher |
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| 318. |
A capasitor of capacitance `1 (mu)F` is charged to a potential of 1 V, it is connected in parallel to an inductor of inductance `10^(-3)H`. The maximum current that will flow in the circuit has the valueA. (A) `sqrt(1000)mA`B. (B) `1mA`C. (C) `1(mu)A`D. (D) `1000 mA` |
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Answer» Correct Answer - A Change on the capacitor, ltbRgt `(q_0)=CV = 1xx10^(-6) xx1 =10^(-6)C` or `(I_0) = omega (q_0)=`maximum current Now `omega =(1)/(sqrt(LC))=(1)/(sqrt(10^(-9)) =(10^(9))^(1//2)` `:. I_0=(10^(9))^(1//2) xx(1 xx 10^(-6)) = sqrt(10^(-3)) A=sqrt(1000)mA`. |
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| 319. |
The output of a step-down transformer is measured to be `24 V` when connected to a 12 watt light bulb. The value of the peak current isA. `1/sqrt(2)A`B. `sqrt(2)A`C. 2AD. `2sqrt(2)A` |
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Answer» Correct Answer - A Secondary voltage, `V_(S) = 24V` Power associated with seconday, `P_(S) = 12W` `I_(S) = P_(S)/V_(S) = 12/24=1/2A = 0.5A` Peak value of the current in the secondary `I_(0)=I_(s)sqrt(2)` `(0.5)(1.414) = 0.707=1/sqrt(2)A` |
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| 320. |
The output of a step-down transformer is measured to be `24 V` when connected to a 12 watt light bulb. The value of the peak current isA. `(1)/(sqrt(2))A`B. `sqrt(2)A`C. `2A`D. `2sqrt(2)A` |
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Answer» Correct Answer - A Here, `V_(s)=24 V, P_(s) =12 W` `I_(s)=(P_(s))/(V_(s))=(12)/(24)=0.5 A , I_(m)=sqrt(2) I_(s)=sqrt(2)xx0.5=(1)/(sqrt(2)) A` |
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| 321. |
When a voltage measuring device is connected to a.c. mains the meter shows the steady input voltage of `220 V`. This meansA. input voltage cannot be AC voltage, but a DC voltageB. maximum input voltage is 220V.C. the meter reads not v but `lt v^(2) gt` and is calibrated to read `sqrt(lt v^(2) gt)`D. the pointer of the meter is stuck by some mechanical defect. |
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Answer» Correct Answer - C The voltmeter connected to AC mains reads mean value `(lt v^(2) gt)` and is calibrated in such a way that it gives value of `lt v^(2)gt`, which is multiplied by form factor to give rms value. |
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| 322. |
When a voltage measuring device is connected to a.c. mains the meter shows the steady input voltage of `220 V`. This meansA. input voltage cannot be ac voltage, but a dc voltage.B. maximum input voltage is 220 VC. the meter reads voltage not V but `lt V^(2)gt` and is calibrated to read `sqrt(lt V^(2)gt)`.D. the pointer of the meter is stuck by some mechanical defect. |
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Answer» Correct Answer - C The voltmeter connected to ac mains is calibrated to read root mean square value or virtual value of ac voltage. |
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| 323. |
A inductor of reactance `1 Omega` and a resistor of `2 Omega` are connected in series to the terminals of a 6 V (rms) a.c. source. The power dissipated in the circuit isA. 8WB. 12 WC. 14.4 WD. 18 W |
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Answer» Correct Answer - C Given, `X_(L)=1Omega, R=2Omega` `E_(rms) = 6V, P_(av)=?` Average power dissipated in the circuit, `P_(av) = E_(rms) I_(rms) cosphi`...........(i) `I_(rms) = I_(0)/sqrt(2) = E_(rms)/Z` `Z=sqrt(R^(2) + X_(L)^(2))` `=sqrt(4+1) = sqrt(5)` `I_(rms) = 6/sqrt(5) A` `cosphi = R/Z=2/sqrt(5)` `P_(av) = 6 xx 6/sqrt(5) xx 2/sqrt(5)` {from Eq.(i)} `=72/(sqrt(5)sqrt(5))=73/5=14.4` W |
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| 324. |
When a voltage measuring device is connected to a.c. mains the meter shows the steady input voltage of `220 V`. This meansA. input voltage cannot be a.c. voltage, but a.d.c voltage.B. maximum input voltage is `220 V`C. the meter reads not `v` but `lt v^(2) gt` and is calibrated to read `sqrt(lt v^(2) gt)`.D. the pointer of the meter is stuck by some mechanical defect |
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Answer» Correct Answer - 3 The voltmeter connected to a.c. mains is calibrated to read rood mean square value of virtual value of a.c. voltage. |
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| 325. |
Suppose the circuit in Exercise 18 has a resistance of `15 Omega`. Obtain the average power transferred to each element of the circuit, and the total power absorbed. |
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Answer» Given, the rms value of voltage `V_("rms")=230 V` Resistance `R = 15 Omega` Frequency `f = 50 Hz` Average power across inductor and capacitor is zero as the phase difference between current and voltage is `90^(@)`. Total power absorbed = power absorbed in resistor, `P_("av")` `= V_("rms").I_("rms")` So, `Z=sqrt(R^(2)+(X_(L)-X_(C ))^(2))` `= sqrt((15)^(2)+(2xx3.14xx50xx80xx10^(-13)-(1)/(2xx3.14xx50xx60xx10^(-6))))` `= sqrt(1002.85)=31.67 Omega` `I_("rms")=(V_("rms"))/(Z)=(230)/(31.67)=7.26 A` Total power absorbed `P_("av")` `= V_("rms").I_("rms")` `= I_("rms").R.I_("rms")` `= (7.26)^(2)xx15=790.6W` Total power absorbed = 790.6W |
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| 326. |
A circuit draws a power of 550 watt from a source of 220 volt, `50 Hz`. The power factor of the circuit is 0.8 and the current lags in phase behind the potential difference. To make the power factor of the circuit as 1.0, The capacitance should be connected in series with it isA. `(1)/(42 pi)xx10^(-2)F`B. `(1)/(41pi)xx10^(-2)F`C. `(1)/(5pi)xx10^(-2)F`D. `(1)/(84pi)xx10^(-2)F` |
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Answer» Correct Answer - A As the current lags behind the potential difference, the circuit contains resistance and inductance. Power, `P=V_("rms")xxi_("rms")xxcos phi` Here, `I_("rms")=(V_("rms"))/(Z)`, where `Z=sqrt([(R )^(2)+(omega L)^(2)])` `therefore P = (V_("rms")^(2))/(Z)xxcos phi` or `Z=(V_("rms")^(2))/(P)xx cos phi` So, `Z=((220)^(2)xx0.8)/(550)=70.4 Omega` Now, Power factor `cos phi =(R )/(Z)` or `R = Z cos phi` `therefore R = 70.4xx0.8=56.32 Omega` Further, `Z^(2)=R^(2)+(omega L)^(2)` or `omega L = sqrt((Z^(2)-R^(2)))` or `omega L = sqrt((70.4)^(2)-(56.32)^(2))=42.2 Omega` When the capacitor is connected in the circuit, `therefore Z = sqrt([R^(2)+(omega L-(1)/(omega C))^(2)])` According to the question, Power factor `cos phi = 1` Hence, `R = Z cos phi = Zxx1, R=Z` `therefore Z=sqrt(Z^(2)+(omega L-(1)/(omega C))^(2)) , (omega L-(1)/(omega C))^(2)=0` `therefore C = (1)/(omega(omega L))=(1)/(2pi upsilon(42.2))=(1)/(2pixx50xx42.2)~~(1)/(42pi)xx10^(-2)F` |
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| 327. |
A series `R-L-C` circuit has `R = 100` ohm. `L = 0.2 mH` and `C = (1)/(2) mu F`. The applied voltage `V = 20 sin omega t`. Then At resonant frequency `omega_(0), ((V_(R ))_("max"))/((V_(L))_("max")) =`A. 2B. 5C. 3D. 4 |
| Answer» Correct Answer - B | |
| 328. |
The potential difference a `2 H` inductor as a function of time is shown in figure. At time `t = 0`, current is zero Current versus time graph across the inductor will beA. B. C. D. |
| Answer» Correct Answer - B | |
| 329. |
An ac source is connected to a resistive circuits. Which of the following statements are false ?A. Current leads the voltage and both are in same phaseB. Current lags behind the voltage and both are in same phaseC. Current and voltage are in same phaseD. Any of the above may be true depending upon the value of resistance |
| Answer» Correct Answer - c | |
| 330. |
The potential difference across a `2 H` inductor as a function of time is shown in figure. At time `t=0`, current is zero Current `t = 2` second is A. `1 A`B. `3 A`C. `4 A`D. `5 A` |
| Answer» Correct Answer - D | |
| 331. |
The potential difference across a `2 H` inductor as a function of time is shown in figure. At time `t=0`, current is zero Current `t = 2` second is A. `1 A`B. `3 A`C. `4 A`D. `5 A` |
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Answer» Correct Answer - D |
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| 332. |
Average energy stored in a pure inductance `L` when current `i` flows through it, isA. `1/2 LI^(2)`B. `1/4 LI^(2)`C. `2 Li^(2)`D. Zero |
| Answer» Correct Answer - d | |
| 333. |
In oscillating Lc circuit, the total stored energy is U and maximum charge upon capacitor is `(Q)/(2)`, the energy stored in the inductor isA. `(U)/(2)`B. `(U)/(4)`C. `(4)/(3) U`D. `(3U)/(4)` |
| Answer» Correct Answer - D | |
| 334. |
The resonance frequency of a certain RLC series circuit is `omega_(0)` . A source of angular frequency ` 2 omega_(0)` is inserted into the circuit. After transients die out, the angular frequency of current oscillation isA. `(omega_(0))/(2)`B. `omega_(0)`C. `2omega_(0)`D. `1.5omega_(0)` |
| Answer» Correct Answer - C | |
| 335. |
A resonant `AC` circuit contains a capacitor of capacitance `10^(-6)F` and an inductor of `10^(-4)H`. The frequency of electrical oscillation will beA. `10^(5) Hz`B. `10 Hz`C. `10^(5)/(2pi) Hz`D. `10/(2pi) Hz` |
| Answer» Correct Answer - c | |
| 336. |
Self inductance of a soleniod can be increased byA. increaing the current passing through the solenoidB. decreasing the current passing through the solenoidC. inserting an iron core in the solenoidD. incresing number of turns per unit length |
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Answer» Correct Answer - C::D |
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| 337. |
Comparing the L-C oscillations with the oscillations of a spring-block systemA. L is equivalent to mB. C is equivalent to KC. current is equivalent to speedD. rate of change of current is equivalent to accelerate |
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Answer» Correct Answer - A::C::D |
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| 338. |
Comparing L-C oscillation with the oscillation if spring-block-system, match the following table. Comparing L-C oscillations with the oscillations of spring-block sytem, match the following table `{:(,,underset(("LC oscillations"))"Table-1",,underset(("Spring-block oscillations"))"Table-2"),(,(A),L,(P),k),(,(B),C,(Q),m),(,(C),i,(R),v),(,(D),(di)/(dt),(S),x),(,,,(T),"None"):}` |
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Answer» Correct Answer - (A) Q, (B) T, (C) R, (D) T |
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| 339. |
In previous question the resistance of coil `B` isA. `0.58 Omega`B. `5.8Omega`sC. `1.16 Omega`D. `11 .6Omega` |
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Answer» Correct Answer - B `cos phi = (R )/(Z) = (R_(A) + R_(B))/(Z)` `(3)/(4) = (5 +R_(B))/(14.4) implies R_(B) =5.8 Omega` |
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| 340. |
The electromagnetic radiations are caused byA. A stationary chargeB. uniformly moving chargesC. accelerated chargesD. All of the above |
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Answer» Correct Answer - C |
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| 341. |
An L-C-R series circuit , connected to a source E, is at resonance. Then,A. the voltage across R is zeroB. the voltage across R equals applied voltageC. the volatage across C is zeroD. the voltage across C equals applied voltage |
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Answer» Correct Answer - B Equation of voltage , E = `sqrt ` At resonance (series circuit) `underset(L)(V)` = `underset(C )(V)` `implies` E = `underset(R )(V)` i.e., whole applied voltage appeared across |
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| 342. |
In the circuit shown in figure switch S is closed at time t=0, which statement is true after one time constant of L-R circuit? A. `(2Lepsilon)/(R)(1-e^(-1))`B. `(epsilon)/(2R)(1-e^(-1))`C. `(Lepsilon)/(R)(1-e^(-1))`D. Current through battery is `(2epsilon)/(R )` |
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Answer» Correct Answer - C |
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| 343. |
If `i= t^(2), 0 lt t lt T` then `r.m.s.` value of current isA. `(T^(2))/(sqrt2)`B. `(T^(2))/(2)`C. `(T^(2))/(sqrt5)`D. none of these |
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Answer» Correct Answer - C |
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| 344. |
`2.5/(pi) muF` capacitor and `3000-`ohm resistance are joined in series to an `AC` source of `200 "volts"` and `50sec^(-1)` frequency. The power factor of the circuit and the power dissipated in it will respectivelyA. 0.6, 0.06WB. 0.06, 0.6WC. 0.6, 4.8WD. 4.8, 06W |
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Answer» Correct Answer - C |
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| 345. |
A direct current of 2 A and an alternating current having a maximum value of 2 A flow through two identical resistances. The ratio of heat produced in the two resistances will beA. `1:1`B. `1:2`C. `2:1`D. `4:1` |
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Answer» Correct Answer - C |
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| 346. |
A direct current of 2 A and an alternating current having a maximum value of 2 A flow through two identical resistances. The ratio of heat produced in the two resistances will beA. 0.042361111111111B. 0.043055555555556C. 0.084027777777778D. 0.16736111111111 |
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Answer» Correct Answer - C `underset(DC)(H) / underset(DC)(H) = ( underset(DC)(i)/underset(rms)(i))^(2) (sinceH proptoi^(2) = 2/sqrt 2)^(2) = 2` |
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| 347. |
For a resistance R and capacitance C in series the impedence is twice that of a parallel combinations of the same elements. The frequency of the applied emf shall beA. `(2pi)/(RC)`B. `(1)/(2piRC)`C. `(2pi)/(sqrtRC)`D. `(1)/(2pisqrtRC)` |
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Answer» Correct Answer - B |
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| 348. |
What is the phenomenon involved in the working of transformer ? |
| Answer» Transformer works on the principle of mutual induction. | |
| 349. |
What is transformer ratio ? |
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Answer» The ratio of secondary e.m.f to the primary e.m.f. (or) number of turns in secondary to the number of turns in the primary is called the transformer ratio. Transformer ratio `= (V_(S))/(V_(P))=(N_(S))/(N_(P))` |
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| 350. |
A 50 Hz AC signal is applied in a circuit of inductance of `(1//pi)`H and resistance `2100Omega.` The impedance offered by the circuit isA. `1500 Omega`B. `1700 Omega`C. `2102 Omega`D. `2500 Omega` |
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Answer» Correct Answer - C Impedance, `Z= sqrt(R^(2)+X_(L)^(2)), X_(L)=omegaL=2pifL` Given, `R=2100Omega`, f=50Hz, `L=1/piH` `rArr Z= sqrt((2100)^(2) + (2 xx 50)^(2)) = sqrt((2100)^(2) + (100)^(2))` `rArr Z=2102 Omega` |
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