As shown in the figure,an inductor of inducta | vedclass

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(B) Given: Inductance $L = 200 \, mH = 0.2 \, H$,$V_{rms} = 220 \, V$,$f = 50 \, Hz$.The inductive reactance $X_L$ is given by:$X_L = 2 \pi f L = 2 \p

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$242$

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(B) Given: Inductance $L = 200 \, mH = 0.2 \, H$,$V_{rms} = 220 \, V$,$f = 50 \, Hz$.The inductive reactance $X_L$ is given by:$X_L = 2 \pi f L = 2 \pi 	imes 50 	imes 0.2 = 20 \pi \, \Omega$.The peak voltage $V_0$ is:$V_0 = V_{rms} \sqrt{2} = 220 \sqrt{2} \, V$.The peak current $i_0$ is:$i_0 = \frac{V_0}{X_L} = \frac{220 \sqrt{2}}{20 \pi} = \frac{11 \sqrt{2}}{\pi} \, A$.We can rewrite this as:$i_0 = \frac{\sqrt{11^2 	imes 2}}{\pi} = \frac{\sqrt{121 	imes 2}}{\pi} = \frac{\sqrt{242}}{\pi} \, A$.Comparing this with the given expression $\frac{\sqrt{a}}{\pi} \, A$,we get:$a = 242$.

As shown in the figure,an inductor of inductance $L = 200 \, mH$ is connected to an $AC$ source of emf $220 \, V$ and frequency $f = 50 \, Hz$. The peak value of current is given by $\frac{\sqrt{a}}{\pi} \, A$. The value of $a$ is..........

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