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The peak current in given circuit is

Option: 1

77.73mA


 


Option: 2

87.46mA


Option: 3

49.38mA


Option: 4

64.91mA


Answers (1)

As we learned

 

Peak current -

{i}'_{0}= \frac{V_{0}}{Z}= \frac{V_{0}}{X_{L}-X_{c}}= \frac{V_{0}}{\omega L-\frac{1}{\omega c}}

-

 

 

Impedence, Z=X_{L}-X_{C} for L-C circuit

Now, X_{L}=wL

                    =2\pi f L

                    =2\times 50\times 10\times 10^{-3}\times \pi

                X_{L}=\pi \Omega

        X_{C}=\frac{1}{wc}

                =\frac{1}{2\pi fc} = \frac{1}{2\times \pi \times 50\times 10\times 10^{-3}}

            X_{c}=\frac{1}{\pi }

\Rightarrow Z=\pi -\frac{1}{\pi }

Z=2.83

Near peak current , i_{0}=\frac{V_{0}}{Z}

                        V_{rms}=\frac{V_{0}}{\sqrt{2}}

                        V_{0}=\sqrt{2}V_{rms}

                        V_{0}=\sqrt{2}\times \frac{220}{\sqrt{2}}

                    V_{0}=220mV

Now i_{0}=\frac{220}{2.83}

i_{0}=77.73 mA

Posted by

Sumit Saini

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