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Alternating Current question

2011 · Shift 2 · Q55
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Alternating Current question

2011 · Shift 2 · Q55

JEE AdvancedPhysicsAlternating CurrentNumerical+3 / −1
A series RC combination is connected to an AC voltage of angular frequency ω\omegaω = 500 rad/s. If the impedance of the RC circuit is R 1.25\sqrt{1.25}1.25​, the time constant (in millisecond) of the circuit is ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 4

  1. For a series RCRCRC circuit, the impedance is
Z=R2+XC2Z = \sqrt{R^2 + X_C^2}Z=R2+XC2​​

where

XC=1ωC.X_C = \frac{1}{\omega C}.XC​=ωC1​.
  1. Given in the question:
Z=R1.25.Z = R\sqrt{1.25}.Z=R1.25​.

So,

R2+(1ωC)2=R1.25.\sqrt{R^2 + \left(\frac{1}{\omega C}\right)^2} = R\sqrt{1.25}.R2+(ωC1​)2​=R1.25​.
  1. Squaring both sides:
R2+(1ωC)2=1.25R2.R^2 + \left(\frac{1}{\omega C}\right)^2 = 1.25R^2.R2+(ωC1​)2=1.25R2.

Thus,

(1ωC)2=0.25R2.\left(\frac{1}{\omega C}\right)^2 = 0.25R^2.(ωC1​)2=0.25R2.

Taking square root,

1ωC=0.5R.\frac{1}{\omega C} = 0.5R.ωC1​=0.5R.
  1. Rearranging:
ωC=2R\omega C = \frac{2}{R}ωC=R2​

so

RC=2ω.RC = \frac{2}{\omega}.RC=ω2​.
  1. The time constant of the circuit is
τ=RC=2500=0.004 s.\tau = RC = \frac{2}{500} = 0.004\text{ s}.τ=RC=5002​=0.004 s.
  1. Convert into milliseconds:
0.004 s=4 ms.0.004\text{ s} = 4\text{ ms}.0.004 s=4 ms.

Therefore, the required integer answer is:

4\boxed{4}4​
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