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Alternating Current question

2023 · 29 Jan · Shift 2 · Q64
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  5. /2023 · 29 Jan · Shift 2 · Q64

Alternating Current question

2023 · 29 Jan · Shift 2 · Q64

JEE MainPhysicsAlternating CurrentNumerical+4 / −1
An inductor of inductance 2 μH\mathrm{\mu H}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 1xF\frac{1}{x}\mathrm{F}x1​F, where the value of xxx is ‾\underline{\hspace{2cm}}​. (Take π=227\pi=\frac{22}{7}π=722​)
Numerical answer
View written solutionFree

Correct answer: 3872

  1. Condition for maximum current in a series RLC circuit

For a series combination of RRR, LLL, and variable CCC, the current is maximum at resonance.

At resonance,

XL=XCX_L = X_CXL​=XC​

which gives

ωL=1ωC\omega L = \frac{1}{\omega C}ωL=ωC1​

So,

C=1ω2LC = \frac{1}{\omega^2 L}C=ω2L1​
  1. Given data
  • Inductance:
L=2 μH=2×10−6 HL = 2\,\mu H = 2\times 10^{-6}\,HL=2μH=2×10−6H
  • Frequency:
f=7 kHz=7×103 Hzf = 7\,kHz = 7\times 10^3\,Hzf=7kHz=7×103Hz
  • Angular frequency:
ω=2πf\omega = 2\pi fω=2πf

Using π=227\pi=\frac{22}{7}π=722​,

ω=2×227×7×103=44×103\omega = 2\times \frac{22}{7}\times 7\times 10^3 = 44\times 10^3ω=2×722​×7×103=44×103

Thus,

ω=4.4×104 rad/s\omega = 4.4\times 10^4\,rad/sω=4.4×104rad/s
  1. Calculate capacitance
C=1ω2LC = \frac{1}{\omega^2 L}C=ω2L1​

First,

ω2=(4.4×104)2=19.36×108=1.936×109\omega^2 = (4.4\times 10^4)^2 = 19.36\times 10^8 = 1.936\times 10^9ω2=(4.4×104)2=19.36×108=1.936×109

Now,

ω2L=1.936×109×2×10−6\omega^2 L = 1.936\times 10^9 \times 2\times 10^{-6}ω2L=1.936×109×2×10−6 =3.872×103=3872= 3.872\times 10^3 = 3872=3.872×103=3872

Therefore,

C=13872 FC = \frac{1}{3872}\,FC=38721​F
  1. Compare with the form given in the question

The capacitance is written as 1x F\dfrac{1}{x}\,Fx1​F. Hence,

x=3872x = 3872x=3872
  1. Comparison with stored answer

Stored correct answer = 387238723872

This matches our derived answer.

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