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Alternating Current — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsAlternating Current2 Marks
Question
The current in a discharging LR circuit is given by $\text{i}=\text{i}_0\text{e}^{\frac{-\text{t}}{\tau}}$ where $\tau$ is the time constant of the circuit. Calculate the rms current for the period $\text{t}=0$ to $\text{t}=\tau.$
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Answer

$\text{i}=\text{i}_0\text{e}^{\frac{-\text{t}}{\tau}}$$\text{i}^2=\frac{1}{\tau}\int\limits_0^\tau\text{i}_0{^2}\text{e}^{\frac{-2\text{t}}{\tau}}\text{dt}$
$=\frac{\text{i}_0{^2}}{\tau}\int\limits_0^\tau\text{e}^{\frac{-2\text{t}}{\tau}}\text{dt}$
$=\frac{\text{i}_0{^2}}{\tau}\times\Big[\frac{\tau}{2}\text{e}^{\frac{-2\text{t}}{\tau}}\Big]_0^\tau$
$=-\frac{\text{i}_0{^2}}{\tau}\times\frac{\tau}{2}\big[\text{e}^{-2}-1\big]$
$\sqrt{\text{i}^2}=\sqrt{-\frac{\text{i}_0^2}{2}\Big(\frac{1}{\text{e}^2}-1\Big)}$
$=\frac{\text{i}_0^2}{2}\sqrt{\Big(\frac{\text{e}^2-1}{2}\Big)}$

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