A series R-C combination is connected to an AC voltage of angular frequency -500 rad-s- If the impedance of the R-C circuit is R -1-25- the time constant of the circuit in milliseconds upto two digits after the decimal point is-

Z = 1/√(R^2 + (1/ω C)^2) R√(1.25)=√(R^2+1/(500C)^2) 1.25R^2=R^2 + 1/(500C)^2 0.25R^2=1/(500C)^2 ⇒R=2/(500C) ⇒RC=τ = 4*10^ 3seconds ...

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Byju's Answer

Standard XII

Physics

Impedance in RLC Circuit

A series R-C ...

Question

A series R-C combination is connected to an AC voltage of angular frequency $ω = 500 r a d / s$ . If the impedance of the R-C circuit is R $\sqrt{1.25}$ , the time constant of the circuit in milliseconds upto two digits after the decimal point is
.

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Solution

$Z = \frac{1}{\sqrt{R^{2} + (\frac{1}{ω C})^{2}}}$
$R \sqrt{1.25} = \sqrt{R^{2} + \frac{1}{(500 C)^{2}}}$
$1.25 R^{2} = R^{2} + \frac{1}{(500 C)^{2}}$
$0.25 R^{2} = \frac{1}{(500 C)^{2}}$
$\Rightarrow R = \frac{2}{(500 C)}$
$\Rightarrow R C = τ = 4 * 10^{- 3} s e c o n d s$

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A series R-C combination is connected to an AC voltage of angular frequency ω=500 rad/s. If the impedance of the R-C circuit is R √1.25, the time constant of the circuit in milliseconds upto two digits after the decimal point is .

Z=1√R2+(1ωC)2 R√1.25=√R2+1(500C)2 1.25R2=R2+1(500C)2 0.25R2=1(500C)2 ⇒R=2(500C) ⇒RC=τ=4∗10−3seconds

A series R-C combination is connected to an AC voltage of angular frequency $ω = 500 r a d / s$ . If the impedance of the R-C circuit is R $\sqrt{1.25}$ , the time constant of the circuit in milliseconds upto two digits after the decimal point is
.

$Z = \frac{1}{\sqrt{R^{2} + (\frac{1}{ω C})^{2}}}$
$R \sqrt{1.25} = \sqrt{R^{2} + \frac{1}{(500 C)^{2}}}$
$1.25 R^{2} = R^{2} + \frac{1}{(500 C)^{2}}$
$0.25 R^{2} = \frac{1}{(500 C)^{2}}$
$\Rightarrow R = \frac{2}{(500 C)}$
$\Rightarrow R C = τ = 4 * 10^{- 3} s e c o n d s$