Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Alternating Current question

2021 · 25 Jul · Shift 1 · Q64
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Physics
  4. /Alternating Current
  5. /2021 · 25 Jul · Shift 1 · Q64

Alternating Current question

2021 · 25 Jul · Shift 1 · Q64

JEE MainPhysicsAlternating CurrentNumerical+4 / −1
An inductor of 10 mH is connected to a 20V battery through a resistor of 10 k Ω\OmegaΩ and a switch. After a long time, when maximum current is set up in the circuit, the current is switched off. The current in the circuit after 1 μ\muμ s is x100{x \over {100}}100x​ mA. Then x is equal to ‾\underline{\hspace{2cm}}​. (Take e −-− 1 = 0.37)
Numerical answer
View written solutionFree

Correct answer: 74

  1. Given data
  • Inductance: L=10 mH=10−2 HL = 10\,\text{mH} = 10^{-2}\,\text{H}L=10mH=10−2H
  • Resistance: R=10 kΩ=104 ΩR = 10\,\text{k}\Omega = 10^4\,\OmegaR=10kΩ=104Ω
  • Battery voltage: V=20 VV = 20\,\text{V}V=20V
  • Time after switching off: t=1 μs=10−6 st = 1\,\mu s = 10^{-6}\,st=1μs=10−6s
  1. Maximum current before switch-off

After a long time in an RLRLRL circuit connected to a DC battery, the inductor behaves like a short circuit.

So the steady current is

I0=VR=20104=2×10−3 A=2 mAI_0 = \frac{V}{R} = \frac{20}{10^4} = 2\times 10^{-3}\,\text{A} = 2\,\text{mA}I0​=RV​=10420​=2×10−3A=2mA
  1. Current decay after switching off

When the current is switched off, the current in an RLRLRL circuit decays as

I(t)=I0e−Rt/LI(t) = I_0 e^{-Rt/L}I(t)=I0​e−Rt/L

First compute the exponent:

RtL=104⋅10−610−2=1\frac{Rt}{L} = \frac{10^4 \cdot 10^{-6}}{10^{-2}} = 1LRt​=10−2104⋅10−6​=1

Hence,

I(t)=I0e−1I(t) = I_0 e^{-1}I(t)=I0​e−1

Given e−1=0.37e^{-1} = 0.37e−1=0.37,

I(t)=2 mA×0.37=0.74 mAI(t) = 2\,\text{mA} \times 0.37 = 0.74\,\text{mA}I(t)=2mA×0.37=0.74mA
  1. Match with the required form

Given current after 1 μs1\,\mu s1μs is

x100 mA\frac{x}{100}\,\text{mA}100x​mA

So,

x100=0.74\frac{x}{100} = 0.74100x​=0.74 x=74x = 74x=74
  1. Comparison with stored answer

Derived answer: 747474

Stored correct answer: 747474

They match.

PreviousNext

More from Alternating Current

  • A 10 Ω resistance is connected across 220V − 50 Hz AC supply. The time taken by the current to change from its maximum value to the rms value is :2021 · MCQ
  • Two circuits are shown in the figure (a) & (b). At a frequency of ​ rad/s the average power dissipated in one cycle will be same in both the circuits. Includes diagram2021 · Numerical
  • A series LCR circuit driven by 300 V at a frequency of 50 Hz contains a resistance R = 3 k Ω, an inductor of inductive reactance XL = 250 πΩ and an unknown capacitor. The value of capacitance to maximize the average power…2021 · MCQ
  • In the given circuit the AC source has ω = 100 rad s-1. Considering the inductor and capacitor to be ideal, what will be the current I flowing through the circuit? Includes diagram2021 · MCQ
  • An alternating current is given by the equation i = i1 sin ω t + i2 cos ω t. The rms current will be :2021 · MCQ
  • In a series LCR resonant circuit, the quality factor is measured as 100. If the inductance is increased by two fold and resistance is decreased by two fold, then the quality factor after this change will be ​.2021 · Numerical
  • Find the peak current and resonant frequency of the following circuit (as shown in figure). Includes diagram2021 · MCQ
  • The alternating current is given by i={42​sin(T2π​t)+10}A The r.m.s. value of of this current is ................. A.2021 · Numerical