In a LR circuit discharging current is given by I-I0e-t- where - is the time constant of the circuit find the rms current from the period t-0 to t-

I _ rms ^ 2 #160; = 1 τ #160; ∫ _ 0 ^τ #160; I ^ 2 dt = 1 τ #160; ∫ _ 0 ^τ #160; I _ 0 ^ 2 e ^ 2t τ #160; #160; dt = 1 ...

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Byju's Answer

Standard XII

Physics

RLC Circuit

In a LR circu...

Question

In a LR circuit discharging current is given by $I = I_{0} e^{- t / τ}$ where $τ$ is the time constant of the circuit find the rms current from the period $t = 0$ to $t = τ$ .

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Solution

$I_{r m s}^{2}$ = $\frac{1}{τ} \int_{0}^{τ} I^{2} d t$
= $\frac{1}{τ} \int_{0}^{τ} I_{0}^{2} e^{\frac{- 2 t}{τ}} d t$
= $\frac{1}{τ} I_{0}^{2} (- \frac{τ}{2} e_{τ}^{- t})$ (add $0$ to $τ)$
= $\frac{I_{0}^{2}}{τ} (\frac{τ}{2} (1 - e^{- 1}))$
$= I_{r m s}^{2}$ $= \frac{I_{0}^{2}}{2} (1 - e^{- 2})$
$I_{r m s}^{2}$ $= \frac{I_{0}}{\sqrt{2}} \sqrt{(1 - e^{- 2})}$

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In a LR circuit discharging current is given by I=I0e−t/τ where τ is the time constant of the circuit find the rms current from the period t=0 to t=τ.

I2rms = 1τ∫τ0I2dt= 1τ∫τ0I20e−2tτdt= 1τI20(−τ2e−tτ) (add 0 to τ)=I20τ(τ2(1−e−1))=I2rms =I202(1−e−2)I2rms=I0√2√(1−e−2)

In a LR circuit discharging current is given by $I = I_{0} e^{- t / τ}$ where $τ$ is the time constant of the circuit find the rms current from the period $t = 0$ to $t = τ$ .