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What will be the current though an ideal capacitor. if it is connected across 2 v battery.​

Answer» <html><body><p><strong><a href="https://interviewquestions.tuteehub.com/tag/explanation-455162" style="font-weight:bold;" target="_blank" title="Click to know more about EXPLANATION">EXPLANATION</a>:</strong></p><p>Toppr</p><p>Question</p><p>An ideal capacitor of capacitance 0.2mu F is <a href="https://interviewquestions.tuteehub.com/tag/charged-7263949" style="font-weight:bold;" target="_blank" title="Click to know more about CHARGED">CHARGED</a> to a potential difference of 10V . The charging <a href="https://interviewquestions.tuteehub.com/tag/battery-893842" style="font-weight:bold;" target="_blank" title="Click to know more about BATTERY">BATTERY</a> is then disconnected. The capacitor is then connected to an ideal inductor of self inductance 0.5mH . The current at a time when the potential difference across the capacitor is 5V , is:</p><p>Answer · 3 votes</p><p>Using Energy <a href="https://interviewquestions.tuteehub.com/tag/conservation-929810" style="font-weight:bold;" target="_blank" title="Click to know more about CONSERVATION">CONSERVATION</a>, Initial energy stored in capacitor = Final energy stored in capacitor + Energy stored in inductor 12CVi^2 = 12CVf^2 + 12Li^2 C( Vi^2 - Vf^2) = Li^2 i = √( C(Vi^2- V_f^2)<a href="https://interviewquestions.tuteehub.com/tag/l-535906" style="font-weight:bold;" target="_blank" title="Click to know more about L">L</a>) Here, Vi = 10 V Vf = 5 V C = 0.2 mu F L = 0.5 mH On putting these values, we will get i = 0.17 A</p><p>More</p></body></html>


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