As shown in the figure an inductor of inductance $200 mH$ is connected to an $AC$ source of emf $220 V$ and frequency $50 Hz$ . The instantaneous voltage of the source is $0 V$ when the peak value of current is $\frac{\sqrt{a}}{π} A$ . The value of $a$ is _____ .

Question

Aicharya · Accepted Answer

The reactance of the given inductor is,

$X_{L} = ω L = 2 π f L = 2 π \times 50 \times 0.2 = 20 π Ω$ .

The maximum current in the given circuit will be,

$i_{0} = \frac{\sqrt{2} V_{r m s}}{X_{L}} = \frac{\sqrt{2} \times 220}{20 π} = \frac{\sqrt{242}}{π} A$

Therefore, $a = 242$

As shown in the figure an inductor of inductance 200   mH is connected to an AC source of emf 220 &#160…

Solution

As shown in the figure an inductor of inductance 200 &#160; mH is connected to an AC source of emf 220 &#160…

Solution

As shown in the figure an inductor of inductance 200 mH is connected to an AC source of emf 220 &#160…