A series R-C combination is connected to an AC voltage of angular frequency omega=500 radian / s . I

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7.Alternating Current

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A series $R-C$ combination is connected to an $AC$ voltage of angular frequency $\omega=500 \ radian / s$. If the impedance of the $R-C$ circuit is $R \sqrt{1.25}$, the time constant (in millisecond) of the circuit is

A

$5$

B

$6$

C

$4$

D

$8$

(IIT-2011)

Solution

Given : $\omega=500 \ radian / s$

Let the capacitance of the capacitor be $C$.

Thus resistance of capacitor $X _{ C }=\frac{1}{\omega C }=\frac{1}{500 C }$

Impedance of the circuit $Z=R \sqrt{1.25}$

Using $Z ^2= R ^2+ X _{ C }^2$

$\therefore 1.25 R ^2= R ^2+\frac{1}{(500)^2 C ^2}$

Or $0.25 R^2=\frac{1}{.25 \times 10^6 C ^2}$

$\Rightarrow R ^2 C ^2=\frac{10^{-6}}{(0.25)^2}$

We get time constant of the circuit $R C=\frac{10^{-3}}{0.25}=0.004 s=4 \ ms$

Standard 12

Physics

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