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Alternating Current question

2023 · 31 Jan · Shift 2 · Q64
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Alternating Current question

2023 · 31 Jan · Shift 2 · Q64

JEE MainPhysicsAlternating CurrentNumerical+4 / −1
A series LCR\mathrm{LCR}LCR circuit consists of R=80Ω,XL=100Ω\mathrm{R}=80 \Omega, \mathrm{X}_{\mathrm{L}}=100 \OmegaR=80Ω,XL​=100Ω, and XC=40Ω\mathrm{X}_{\mathrm{C}}=40 \OmegaXC​=40Ω. The input voltage is 2500 cos⁡(100πt)V\cos (100 \pi \mathrm{t}) \mathrm{V}cos(100πt)V. The amplitude of current, in the circuit, is ‾\underline{\hspace{2cm}}​ A.
Numerical answer
View written solutionFree

Correct answer: 25

  1. Given data

    For the series LCRLCRLCR circuit: R=80 Ω,XL=100 Ω,XC=40 ΩR=80\,\Omega,\quad X_L=100\,\Omega,\quad X_C=40\,\OmegaR=80Ω,XL​=100Ω,XC​=40Ω

    Input voltage: V=2500cos⁡(100πt) VV=2500\cos(100\pi t)\,\text{V}V=2500cos(100πt)V

    So, the amplitude of applied voltage is V0=2500 VV_0=2500\,\text{V}V0​=2500V

  2. Net reactance

    In a series LCRLCRLCR circuit, net reactance is X=XL−XC=100−40=60 ΩX=X_L-X_C=100-40=60\,\OmegaX=XL​−XC​=100−40=60Ω

  3. Impedance of the circuit

    The impedance is Z=R2+(XL−XC)2Z=\sqrt{R^2+(X_L-X_C)^2}Z=R2+(XL​−XC​)2​ Z=802+602Z=\sqrt{80^2+60^2}Z=802+602​ Z=6400+3600Z=\sqrt{6400+3600}Z=6400+3600​ Z=10000=100 ΩZ=\sqrt{10000}=100\,\OmegaZ=10000​=100Ω

  4. Current amplitude

    Current amplitude is given by I0=V0ZI_0=\frac{V_0}{Z}I0​=ZV0​​ I0=2500100=25 AI_0=\frac{2500}{100}=25\,\text{A}I0​=1002500​=25A

  5. Final answer

    The amplitude of current in the circuit is 25\boxed{25}25​

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