In the circuit shown in figure, initially K₁is closed and K₂ is open. What are the charges on each capacitors? Then K₁ was opened and K₂ was closed (order is important), what will be the charge on each capacitor now? [C = 1 μF]

<p><span data-teams="true">This is a long answer type question as classified in NCERT Exemplar</span></p><p>When initially K₁&nbsp;is closed and K₂&nbsp;is open, the capacitors C₁&nbsp;and C₂&nbsp;connected in series with battery</p><p>Q=CV=C (C₁C₂/C₁+C₂) = 6<span contenteditable="false"> <math> <mo>×</mo> </math> </span>3/6+3=18<span contenteditable="false"> <math> <mi>μ</mi> </math> </span>C</p><p>Now K₁&nbsp;was opened and K₂&nbsp;was closed, the battery and capacitor C, are disconnected from the circuit .</p><p>Here charge being shared by both capacitor so a common potential will develop.</p><p>V=C₁V₁+C₂V₂/C₁+C₂ = 18/3+3=3V</p><p>Q₂=3CV=3<span contenteditable="false"> <math> <mo>×</mo> </math> </span>3=9μC</p><p>Q₃=3CV=3<span contenteditable="false"> <math> <mo>×</mo> </math> </span>3=9μC</p>

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