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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.
| 1251. |
A varying current at the rate of `3A//s` in a coil generates an e.m.f. of `8mV` in a nearby coil. The mutual inductance of the two coils isA. `2.66 mH`B. `2.66xx10^(-3)mH`C. `2.66H`D. `0.266H` |
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Answer» Correct Answer - A `|e|=M(di)/(dt)implies8xx10^(-3)=Mxx3impliesA_(1)=2.66mH` |
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| 1252. |
Two coil are placed close to each other. The mutual inductance of the pair of coils depends upon.A. the rates at which currents are changing in the two coilsB. relative position and orientation of hte two coilsC. the materials of the wires of he coilsD. the currents in the two coils |
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Answer» Correct Answer - B (b) Mutual conductance depends on the relative position and orientation of the two coils. |
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| 1253. |
How is mutual inductance of a pair of coils affected when (i) separation between the coils in increased, (ii) the number of turns of each coil is increased , (iii) a thin iron sheet is placed between the two coils, other factors ramaining the same ? Explain your answer in each case. |
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Answer» (i) On increasing distance between the two coils, magnetic flux passing from one coil to the other decreases. Therefore, mutual inductance decreases (ii) When an iron sheet is placed between the two coils, mutual inductance increases because `M prop mu` (permeability). |
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| 1254. |
Compute the mutual inductance for a given pair of coils if increase in current from 2 A to 6 A in 0.1 s in one coil causes an induced e.m.f. of 1 V in the order coil. |
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Answer» Here, `A = 8 xx 2 = 16 sq cm = 16 xx 10^(-4) m^(-2)` `(dB)/(dt) = - 0.02 Ts^(-1)` Induced e.m.f. `e = (- d phi)/(dt) = - (A dB)/(dt)` ` = - 16 xx 10^(-4) (- 0.02) = 32 xx 10^(-6) V` Power loss `= (e^(2))/(R ) = ((32 xx 10^(-6))^(2))/(1.6)` `= 6.4 xx 10^(-10) W` |
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| 1255. |
The coefficient of mutual inductance of two coils is 10 mH. IF the current flowing in one coil is 4A then the induced e.m.f in the second coil will beA. 40 mVB. 20 mVC. zeroD. 10 mV |
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Answer» Correct Answer - C For steady currents, `(dI)/(dt)=0 :. e = M(dI)/(dt)=0`. |
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| 1256. |
The coefficient of mutual inductance of two coils is `6 mH`. If the current flowing in one is `2` ampere, then the induced e.m.f. in the second coil will beA. `3 mV`B. `2 mV`C. `3 V`D. `Zero` |
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Answer» Correct Answer - D In secondry e.m.f. induces only when current through primary changes. |
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| 1257. |
Find the mutual inductance between the two coils if a current of 10 ampere in primary coil changes the flus by 500 Wb per turn in the secondary coil of 200 turns. Also, find the induced e.m.f. acorss the ends of the secondary coil if this change occurs in 0.5 sec. |
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Answer» Correct Answer - `10^(4)H,2xx10^(5)` |
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| 1258. |
Two coils X and Y are placed in a circuit such that when the current changes by 2 A in coil X. The magnetic flux changes by 0.4 Wb in Y. The value of mutual inductance of the coils isA. 0.2 HB. 5 HC. 0.8 HD. 4 H |
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Answer» Correct Answer - A We have, `M=(Deltaphi_(2))/(Deltai_(1))` `M=(0.4)/(2)` `therefore` Mutual inductance, `M=0.2H` |
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| 1259. |
A coil and a bulb are connected in series with a `dc` source, a soft iron core is then inserted in the coil. ThenA. intensity of the bulb remains the sameB. intensity of the bulb decreaseC. intensity of the bulb increaseD. the bulb ceases to glow |
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Answer» Correct Answer - B There will be self-induction effect when soft iron core is inserted in the coil. So, intensity of bulb increases. |
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| 1260. |
The self-induced e.m.f in a `0.1H` coil when the current in it is changing at the rate of `200 ampere//second` isA. `8xx10^(-4)V`B. `8xx10^(-5)V`C. `20V`D. `125 V` |
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Answer» Correct Answer - C `e=(Ldi)/(dt)impliese=0.1xx200=20V` |
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| 1261. |
An air core solenoid has `1000`turns and is one metre long. Its cross-sectional area is `10cm^(2)`. Its self-inductance isA. `0.1256 mH`B. `12.56 mH`C. `1.256 mH`D. `125.6 mH` |
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Answer» Correct Answer - C `L=(mu_(0)N^(2)A)/(l)=(4pixx10^(-7)xx(1000)^(2)xx10xx10^(-4))/(1)` `=1.256 mH` |
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| 1262. |
Two pure inductors each of self-inductance `L` are connected in parallel but are well separted from each other. The total inductance isA. LB. 2 LC. L/2D. L/4 |
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Answer» Correct Answer - B `L_(5)=L_(1)+L_(2)=L+L=2L` |
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| 1263. |
Two pure inductors each of self-inductance `L` are connected in parallel but are well separted from each other. The total inductance isA. 2 LB. LC. `(L)/(2)`D. `(L)/(4)` |
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Answer» Correct Answer - C Equivalent inductance, `(1)/(L_(eq))=(1)/(L)+(1)/(L)` `therefore` The total inductance, `L_(eq)=(L)/(2)` |
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| 1264. |
Two pure inductors each of self-inductance `L` are connected in parallel but are well separted from each other. The total inductance isA. `2L`B. `L`C. `(L)/(2)`D. `(L)/(4)` |
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Answer» Correct Answer - C Inductors obey the laws of parallel and series combination of resistors. |
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| 1265. |
A `100mH` coil carries a current of `1` ampere. Energy stored in its magnetic field isA. `0.5J`B. `1J`C. `0.05J`D. `0.1J` |
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Answer» Correct Answer - C Energy `=(1)/(2)LI^(2)=(1)/(2)xx100xx10^(-3)xx1^(2)=0.05J` |
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| 1266. |
The self-inductance of a straight conductor isA. zeroB. very largeC. infinityD. very small |
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Answer» Correct Answer - A `Lpropn`(number of turns),for straight conductor `n=0`, hence `L=0`. |
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| 1267. |
Two circuits have coefficient of mutual induction of `0.09` henry. Average e.m.f. induced in the secondary by a change of current from `0` to `20` ampere in `0.006` second in the primary will beA. 120 VB. 80 VC. 200 vD. 300 V |
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Answer» Correct Answer - D Average emf, `e=M(di)/(dt)=0.09xx(20)/(0.006)=300V` |
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| 1268. |
Two circuits have coefficient of mutual induction of `0.09` henry. Average e.m.f. induced in the secondary by a change of current from `0` to `20` ampere in `0.006` second in the primary will beA. `240 V`B. `230 V`C. `100 V`D. `300 V` |
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Answer» Correct Answer - C `e=M(di)/(dt)impliese=0.1xx((20-0))/(0.02)=100V` |
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| 1269. |
Two circuits have coefficient of mutual induction of `0.09` henry. Average e.m.f. induced in the secondary by a change of current from `0` to `20` ampere in `0.006` second in the primary will beA. `120 V`B. `80 V`C. `200 V`D. `300 V` |
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Answer» Correct Answer - D `e=M(di)/(dt)=0.09xx(20)/(0.006)=300 V` |
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| 1270. |
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. `2500 V`B. `25000 V`C. `2510V`D. `Zero` |
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Answer» Correct Answer - B `e=-M(di)/(dt)=-5xx((-5))/(10^(-3))=25000V` |
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| 1271. |
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. `5 V`B. `1 V`C. `25 V`D. `10 V` |
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Answer» Correct Answer - B `e=M(di)/(dt)=0.2xx5=1V` |
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| 1272. |
When the current in a coil changeg from `8` amperes to `2` amperes in `3xx10^(-2)` seconds, the e.m.f. induced in the coil is `2` volt. The self-inductance of the coil (in millihenry) isA. `1`B. `5`C. `20`D. `10` |
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Answer» Correct Answer - D `e=-L(di)/(dt)implies2=-L((8-2)/(3xx10^(-2)))` `impliesL=0.01H=10mH` |
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| 1273. |
How is capacitative reactance affected when frequency of a.c. supply is tripled ? |
| Answer» As `X_(C ) = (1)/(omega C) = (1)/(2 pi v C)`, therefore when v is tripled, `X_(C )` becomes one third | |
| 1274. |
How does inductive reactance vary when frequency of a.c. source in the circuit is halved ? |
| Answer» As `X_(L) = omega L = 2 pi v L`, therefore, when v is halved, `X_(L)` becomes half. | |
| 1275. |
An e.m.f. of `12 "volts"` is induced in a given coil when the current in it changes at the rate of `48` amperes per minute. The self-inductance of the coil isA. `0.25 henry`B. `15 henry`C. `1.5 henry`D. `9.6 henry` |
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Answer» Correct Answer - B `L=(e)/(di//dt)=(12)/(48//60)=15H` |
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| 1276. |
If the current is halved in a coil, then the energy stored is how much times the previous valueA. `(1)/(2)`B. `(1)/(4)`C. `2`D. `4` |
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Answer» Correct Answer - B `U=(1)/(2)Li^(2)i.e.(U_(2))/(U_(1))=((i_(2))/(i_(1)))^(2)=((1)/(2))^(2)=(1)/(4)` `impliesU_(2)=(1)/(4)U_(1)` |
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| 1277. |
A current of 1 A through a coil of inductance of 200 mH is increasing at a rate of `0.5As^(-1)`. The energy stored in the inductor per second isA. `0.5Js^(-1)`B. `5.0Js^(-1)`C. `0.1Js^(-1)`D. `2.0Js^(-1)` |
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Answer» Correct Answer - C The energy stored in an inductor is `U=(1)/(2)LI^(2)` The energy stored in the indutor per second is `(dU)/(dt)=LI(dI)/(dt)=200xx10^(-3)Hxx1Axx0.5 As^(-1)=0.1 J s^(-1)` |
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| 1278. |
Calculate the energy stored in an inductor of inductance 50 mH when a current of 2.0 A is passed through it.A. `1`B. `0.1`C. `0.05`D. `0.5` |
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Answer» Correct Answer - B Energy stored `E=(1)/(2)Li^(2)=(1)/(2)x50xx10^(-3)xx4=0.1J` |
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| 1279. |
An inductor of inductance L = 400 mH and resistors of resistance `R_(1) = 2Omega` and `R_(2) = 2Omega` are connected to a battery of emf 12 V as shown in the figure. The internal resistance of the battery is negligible. The switch S is closed at `t = 0`. The potential drop across L as a function of time is A. `(12)/(t)e^(-3t)V`B. `6(1-e^(-t//0.2))V`C. `12e^(-5t)V`D. `6e^(-5t)V` |
| Answer» Correct Answer - C | |
| 1280. |
An inductor-coil of inductance 20 mH having resistance `10 Omega` is joined to an ideal battery of emf 5.0 V. Find the rate of change of the induced emf at t=0, (b) t = 10 ms and (c ) t = 1.0s. |
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Answer» Correct Answer - A::B::C `epsilon=L(dl)/(dt)=epsilon_(v)e^(-Rt//L) rArr (depsilon)/(dt)=-epsilon_(0)R//L e^(-Rt//L)` |
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| 1281. |
The average e.m.f. induced in a coil in which the current changes from `2` amperes to `4` amperes in `0.05` seconds is `8` volts. What is the self-inductance of the coil?A. `0.1 H`B. `0.2 H`C. `0.4 H`D. `0.8 H` |
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Answer» Correct Answer - B `e=-L((di)/(dt))implies8=-Lxx(-(2)/(0.05)) impliesL=0.2H` |
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| 1282. |
Calculate the energy stored in an inductor of inductance 50 mH when a current of 2.0 A is passed through it.A. `0.4 J`B. `4.0J`C. `0.8J`D. `0.04J` |
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Answer» Correct Answer - A `U=(1)/(2)Li^(2)=(1)/(2)xx(50xx10^(-3))xx(4)^(2)=400xx10^(-3)0.4J` |
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| 1283. |
The mutual inductance between a primary and secondary circuit is `0.5H`. The resistance of the primary and the secondary circuits are `20 ohms` and `5ohms` respectvely. To genrate a current of `0.4A` in the secondary,current in the primary must be changed at the rate ofA. `4.0 A//s`B. `16.0 A//s`C. `1.6 A//s`D. `8.0 A//s` |
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Answer» Correct Answer - A `e_(e)=M(di_(1))/(dt)=0.4xx5=0.5xx(di_(1))/(dt)` `implies(di_(1))/(dt)=4 A//sec`. |
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| 1284. |
When the number of turns in a coil is doubled without any change in the length of the coil, its self-inductance becomesA. `(1)/(4)L`B. `L`C. `4L`D. `16 L` |
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Answer» Correct Answer - D `LpropN^(2)` |
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| 1285. |
Two different coils have self-inductances `L_(1) = 8 mH and L_(2) = 2 mH`. The current in one coil is increased at a constant rate. The current in the second coil is also increased at the same constant rate. At a certain instant of time, the power given to the two coil is the same. At that time, the current, the induced voltage and the energy stored in the first coil are `i_(1), V_(1) and W_(1)` respectively. Corresponding values for the second coil at the same instant are `i_(2), V_(2) and W_(2)` respectively. Then:A. `i_(1)li_(2)=1//4`B. `i_(1)//i_(2)=4`C. `w_(2)//w_(1)=4`D. `v_(2)//v_(1)=1//4` |
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Answer» Correct Answer - A |
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| 1286. |
Two different coils have self-inductances `L_(1) = 8 mH and L_(2) = 2 mH`. The current in one coil is increased at a constant rate. The current in the second coil is also increased at the same constant rate. At a certain instant of time, the power given to the two coil is the same. At that time, the current, the induced voltage and the energy stored in the first coil are `i_(1), V_(1) and W_(1)` respectively. Corresponding values for the second coil at the same instant are `i_(2), V_(2) and W_(2)` respectively. Then:A. `(i_(1))/(i_(2))=(1)/(4)`B. `(i_(1))/(i_(2))=4`C. `(W_(1))/(W_(2))=(1)/(4)`D. `(V_(2))/(V_(1))=(1)/(4)` |
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Answer» Correct Answer - B `V=L(di)/(dt)or V propL(as(di)/(dt)rarr"same")` `therefore" "(V_(1))/(V_(2))=(L_(1))/(L_(2))=4," "P = Vi="constant"` `therefore" "iprop(1)/(V)or (i_(1))/(i_(2))=(V_(2))/(V_(1))=(1)/(4)rArrW=(1)/(2)Li^(2)` `therefore" "(W_(1))/(W_(2))=(L_(1))/(L_(2))((i_(1))/(i_(2)))^(2)=(4)((1)/(46))=(1)/(4)` `because" "(i_(1))/(i_(2))ne 4` Thus, wrong option is (b). |
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| 1287. |
Calculate the induced emf in a coil of 10 H inductance in which the current changes from 8A to 3A in 0.2 s. |
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Answer» Correct Answer - 250 V |
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| 1288. |
How much is the self-induced voltage across a 4 H inductances, produced by a current change of 12 `As^(-1)` ? |
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Answer» Correct Answer - 48 V |
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| 1289. |
The magnetic flux passing through a metal ring varies with time `t` according to `phi_(B)=6t^(3)-I8t^(2)` , `t` is in second. The resistance of the ring is `3Omega` . Determine the maximum current induced in the ring during the interval from `t=0` to `t=2sec` . |
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Answer» `phi=6t^(3)-18t^(2)` `e=-(dphi)/(dt)=-18t^(2)+36t` `I=(e)/(R)=(-18t^(2)+36t)/(3)=-6t^(2)+12t` I will be maximum when `(dI)/(dt)=-12t+12=0 implies t=1sec` At `t=1sec` `I=-6(1)^(2)+12(1)=6A` |
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| 1290. |
A circular loop is placed in magnetic field `B=2t.` Find the direction of induced current produced in the loop. |
| Answer» `B=2t`, means `ox` magnetic field (or we can also say cross magnetic flux) passing through the loop is increasing. So, induced current will produced dot magnetic field. To produce `o.` magnetic field induced current from our side should be anti clockwise. | |
| 1291. |
The magnetic flux passing through a metal ring varies with time `t` as: `phi_(B)=3(at^3-bt^2)T-m^2` with `a=2.00s^-3` and `b=6.00s^-2`. The resistance of the ring is `3.0Omega`. Determine the maximum current induced in the ring during te interval from `t=0` to `t=2.0s`. |
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Answer» Given `phi_B=3(at^3-bt^2)` `:. |e|=|(dphi_B)/(dt)|=9at^2-6bt` `:.` Induced current, `i=(|e|)/R=(9at^2-6bt)/3=3at^2-2bt` For cuurrent to be maximum, `(di)/(dt)=0` `:. 6at-2b=0` or `t=b/(3a)` i.e. at `t=b/(3a)`, current is maximum. This maximum current is `i_max.=3a(b/(3a))^2-2b(b/(3a))` `= b^2/(3a)-(2b^2)/(3a)=b^2/(3a)` Substituting the given values of `a` and `b` we have `i_max=((6)^2)/(3(2))=6.0A` |
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| 1292. |
Find the current passing through battery immediately after key `(K)` is closed. It is given that initially all the capacitors are uncharged. (Given that `R=6Omega` and `C=4muF)` A. `1A`B. `5A`C. `3A`D. `2A` |
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Answer» Correct Answer - A At time `t=0`, resistance offered by a capacitor `=0` and resistance offered by an inductor `=alpha` `R_("net")=R/2+R/3=(5R)/6=5Omega` `:.` Current from the battery, `i=E/R_("net")=5/5=1A` |
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| 1293. |
The transformation ratio in the step-down transformer isA. `1`B. greater than oneC. less than oneD. the ratio greater ot less than one depends on the other factors |
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Answer» Correct Answer - B Transformation ratio `k=(N_(s))/(N_(p))=(v_(s))/(V_(p))` for step-up transformations, `N_(s)gtN_(p)i.e.V_(s)gtV_(p)`, hence `kgt1. |
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| 1294. |
A transformer is employed toA. obtain a suitable dc voltageB. convert dc into acC. obtain a suitable ac voltageD. convert ac into dc |
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Answer» Correct Answer - C A transformer is a device to convert alternating current at high voltage into low volltage and vice-versa. |
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| 1295. |
A step down transformer of efficiency `80%` is used on a 1000V line to deliver a current of 20 A at 120 V at the secondary coil. What is the current drawn from the line?A. 0.3 AB. 30 AC. 2.4 AD. `2 A` |
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Answer» Correct Answer - D `V_(p) = 1000 V , V_(s) = 120 V and I_(s) = 20 A` For an ideal transfomer, `V_(p) xx I_(p) = V_(s) xx I_(s)` `:. I_(p) = (120 xx 20)/(1000) = 2.4 A` But for the given transformer, the efficiency is `80%` `:. I_(p)` should be `2.4 xx 100/80 = 3A`. |
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| 1296. |
What is increase in step-down transformer?A. currentB. voltageC. powerD. frequency |
| Answer» Correct Answer - A | |
| 1297. |
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. `3A`B. `30A`C. `0.3A`D. `2.4A` |
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Answer» Correct Answer - A `eta=(P_(S))/(P_(P))=(I_(S)*e_(S))/(I_(P)*e_(P))` `I_(P)=(I_(S)*e_(S))/(eta*e_(P))=(20xx120)/(0.8xx1000)=3A` |
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| 1298. |
A transfomer is employed toA. obtain a suitble dc voltageB. convert dc into acC. obtain a suitable ac voltageD. convert ac into dc |
| Answer» Correct Answer - C | |
| 1299. |
Why are electromagnets made of soft iron?A. hysteresis lossesB. eddy current lossesC. force opposing electric currentD. none of these |
| Answer» Correct Answer - A | |
| 1300. |
A device which converts low ac voltage at high current into high ac voltage at low current isA. electrochemical cellB. photo cellC. transformerD. rectifier |
| Answer» Correct Answer - C | |