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  • Help me understand! - Electromagnetic Induction and alternative Currents - NEET

A 100 \Omega resistance and a capacitor of 100 \Omega reactance are connected in series across a 220 V source. When the capacitor is 50% charged, the peak value of the displacement current is

  • Option 1)

    2.2 A

  • Option 2)

    11 A

  • Option 3)

    4.4 A

  • Option 4)

    11\sqrt 2

 

Answers (1)

best_answer

As we discussed

Impedence -

Z= sqrt{R^{2}+left ( omega L -frac{1}{omega c} ight )^{2}}

-

 

 

Peak current -

{i}'_{0}= \frac{v_{0}}{R}

-

 

 Z=\sqrt{R^{2}+X^{2}_{c}}

=\sqrt{100^{2}+100^{2}}

=100\sqrt{2}

i_{max}=\frac{v_{max}}{2}

=\frac{220\sqrt{2}}{100\sqrt{2}}

=2.2


Option 1)

2.2 A

Option is correct

Option 2)

11 A

Option is incorrect

Option 3)

4.4 A

Option is incorrect

Option 4)

11\sqrt 2

Option is incorrect

Posted by

divya.saini

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