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Calculate peak current

Option: 1

1.3A


Option: 2

2.06A


Option: 3

0.8A


Option: 4

0.032A


Answers (1)

best_answer

As we learned

Peak Current -

{i}'_{0}= \frac{V_{0}}{z}= \frac{V_{0}}{\sqrt{R^{2}+4\pi ^{2}\nu ^{2}L^{2}}}

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 Peak current, i_{o}=\frac{V_{0}}{\sqrt{R^{2}+(\omega L)^{2}}}

i_{o}=\frac{V_{0}}{\sqrt{R^{2}+(2\pi f L)^{2}}}

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

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

V_{0}=\sqrt{\not{2}}\times \frac{100}{\sqrt{\not{2}}}mV

V_{o}=100mV

putting V_{0} in equation (1)

i_{o}=\frac{100\times 10^{-3}}{\sqrt{(0.03)^{2}+(2\times \pi \times 50\times 10\times 10^{-3})^{2}}}

i_{o}=0.032A

Posted by

avinash.dongre

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