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

2023 · 13 Apr · Shift 1 · Q68
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. /2023 · 13 Apr · Shift 1 · Q68

Alternating Current question

2023 · 13 Apr · Shift 1 · Q68

JEE MainPhysicsAlternating CurrentNumerical+4 / −1
In the given figure, an inductor and a resistor are connected in series with a battery of emf E volt. Ea2b J/s\frac{E^{a}}{2 b} \mathrm{~J} / s2bEa​ J/s represents the maximum rate at which the energy is stored in the magnetic field (inductor). The numerical value of ba\frac{b}{a}ab​ will be ‾\underline{\hspace{2cm}}​. JEE Main 2023 (Online) 13th April Morning Shift Physics - Alternating Current Question 44 English
Numerical answer
View written solutionFree

Correct answer: 25

  1. Current growth in an RLRLRL circuit

For a resistor RRR and inductor LLL in series connected to a battery of emf EEE, the current is

I(t)=ER(1−e−Rt/L).I(t)=\frac{E}{R}\left(1-e^{-Rt/L}\right).I(t)=RE​(1−e−Rt/L).
  1. Rate of storage of magnetic energy in the inductor

Energy stored in the inductor at any instant is

U=12LI2.U=\frac{1}{2}LI^2.U=21​LI2.

So the rate at which energy is stored is

dUdt=LIdIdt.\frac{dU}{dt}=LI\frac{dI}{dt}.dtdU​=LIdtdI​.

Now,

dIdt=ELe−Rt/L.\frac{dI}{dt}=\frac{E}{L}e^{-Rt/L}.dtdI​=LE​e−Rt/L.

Hence,

dUdt=L(ER(1−e−Rt/L))(ELe−Rt/L)\frac{dU}{dt}=L\left(\frac{E}{R}(1-e^{-Rt/L})\right)\left(\frac{E}{L}e^{-Rt/L}\right)dtdU​=L(RE​(1−e−Rt/L))(LE​e−Rt/L) ⇒dUdt=E2R(1−e−Rt/L)e−Rt/L.\Rightarrow \frac{dU}{dt}=\frac{E^2}{R}(1-e^{-Rt/L})e^{-Rt/L}.⇒dtdU​=RE2​(1−e−Rt/L)e−Rt/L.

Let

x=e−Rt/L.x=e^{-Rt/L}.x=e−Rt/L.

Then 0<x<10<x<10<x<1, and

dUdt=E2Rx(1−x).\frac{dU}{dt}=\frac{E^2}{R}x(1-x).dtdU​=RE2​x(1−x).
  1. Find maximum value

The expression x(1−x)x(1-x)x(1−x) is maximum at

x=12,x=\frac{1}{2},x=21​,

with maximum value

x(1−x)=14.x(1-x)=\frac{1}{4}.x(1−x)=41​.

Therefore,

(dUdt)max⁡=E24R.\left(\frac{dU}{dt}\right)_{\max}=\frac{E^2}{4R}.(dtdU​)max​=4RE2​.
  1. Compare with given form

Given,

(dUdt)max⁡=Ea2b J/s.\left(\frac{dU}{dt}\right)_{\max}=\frac{E^a}{2b}\,\text{J/s}.(dtdU​)max​=2bEa​J/s.

From our result,

E24R.\frac{E^2}{4R}.4RE2​.

To match the form, we identify

a=2,a=2,a=2,

and

2b=4R⇒b=2R.2b=4R \Rightarrow b=2R.2b=4R⇒b=2R.

Since the asked numerical value is ba\dfrac{b}{a}ab​, we need the resistor value from the figure. From the figure, R=25 ΩR=25\,\OmegaR=25Ω.

Thus,

ba=2R2=R=25.\frac{b}{a}=\frac{2R}{2}=R=25.ab​=22R​=R=25.
  1. Final answer
25\boxed{25}25​
PreviousNext

More from Alternating Current

  • Given below are two statements: Statement I : An AC circuit undergoes electrical resonance if it contains either a capacitor or an inductor. Statement II : An AC circuit containing a pure capacitor or a pure inductor consumes high power…2023 · MCQ
  • In an LC oscillator, if values of inductance and capacitance become twice and eight times, respectively, then the resonant frequency of oscillator becomes x times its initial resonant frequency ω0​. The value of x is :2023 · MCQ
  • An LCR series circuit of capacitance 62.5 nF and resistance of 50 Ω, is connected to an A.C. source of frequency 2.0 kHz. For maximum value of amplitude of current in circuit, the value of inductance is ​…2023 · Numerical
  • A series LCR circuit is connected to an AC source of 220 V, 50 Hz. The circuit contains a resistance R = 80 Ω, an inductor of inductive reactance XL​=70Ω, and a capacitor of capacitive reactance XC​=130Ω…2023 · Numerical
  • For the given figures, choose the correct options : Includes diagram2023 · MCQ
  • An inductor of inductance 2 μH is connected in series with a resistance, a variable capacitor and an AC source of frequency 7 kHz. The value of capacitance for which maximum current is drawn into the circuit x1​F…2023 · Numerical
  • In a series LR circuit with XL​=R, power factor P1. If a capacitor of capacitance C with XC​=XL​ is added to the circuit the power factor becomes P2. The ratio of P1 to P2 will be :2023 · MCQ
  • In the given circuit, rms value of current (Irms​) through the resistor R is: Includes diagram2023 · MCQ