NEET-XII-Physics
39: Alternating Current
- #7The current in a discharging LR circuit is given by i = i0 e-t/τ , where τ is the time constant of the circuit. Calculate the rms current for the period t = 0 to t = τ.Ans : As per the question,
`` i={i}_{0}{e}^{\frac{-t}{\tau }}``
We need to find the rms current. So, taking the average of i within the limits 0 to τ and then dividing by the given time period τ, we get:
`` {{i}_{\,\mathrm{\,rms\,}}}^{2}=\frac{1}{\tau }{\int }_{\mathit{0}}^{\tau }{i}_{0}^{2}{\,\mathrm{\,e\,}}^{\mathit{-}\mathit{2}t\mathit{/}\tau }dt``
`` \Rightarrow {{i}_{rms}}^{2}=\frac{{i}_{0}^{2}}{\tau }{\int }_{0}^{\,\mathrm{\,\tau \,}}{\,\mathrm{\,e\,}}^{\mathit{-}\mathit{2}t\mathit{/}\tau }dt``
`` =\frac{{i}_{0}^{2}}{\tau }\times {\left[\frac{-\tau }{2}{\,\mathrm{\,e\,}}^{\mathit{-}\mathit{2}t\mathit{/}\tau }\right]}_{0}^{\tau }``
`` =-\frac{{i}_{0}^{2}}{\tau }\times \frac{\tau }{2}\times \left[{\,\mathrm{\,e\,}}^{-2}-1\right]``
`` =\frac{{{i}_{0}}^{2}}{2}(1-\frac{1}{{\,\mathrm{\,e\,}}^{2}})``
`` {i}_{\,\mathrm{\,rms\,}}=\frac{{i}_{0}}{e}\sqrt{\frac{{e}^{2}-1}{2}}``
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