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Find peak current in given circuit?

Option: 1

400mA


Option: 2

282.88mA


Option: 3

63.66mA


Option: 4

144.32mA


Answers (1)

best_answer

As we learnt,

 

Root mean square i_{rms} -

i_{rms}= \sqrt{\frac{i_{1}^{2}+i_{1}^{2}+----}{\eta }}= \sqrt{\overline{1}^{2}}

\sqrt{\frac{\int_{0}^{T}i^{2}dt}{\int_{0}^{T}dt}}= \frac{{i_{0}}'}{\sqrt{2}}= 0.707\, {i}'_{0}

                                        = 70.7 % {i}'_{0}

- wherein

{i}'_{0} - Peak Current

 

 

 

Peak Current = I_{rms}\sqrt2

                      = \frac{V_{rms}}{Z}\sqrt2 = \frac{\frac{200}{\sqrt2}\times \sqrt2}{\omega L}

                      = \frac{{200}}{2\pi\times 50\times 10^{-2}} = \frac{200}{\pi}

                      = 63.66\;mA

Posted by

Rakesh

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