NEET Class neet Answered
Asked by doc.anjali6 | 02 May, 2019, 07:38: AM
Expert Answer
Charge Q on the initial capacitor is givn by, Q = C×V
where C is capacitance and V is potential difference, hence Q = 2×200 = 400 μC
when a second capacitor is added parallel to initial capacitor after disconnecting the battery, common potential difference becomes 40 V
hence Q = Ceq × 40 or Ceq = 400 / 40 = 10 μF.
Since capacitors are connected in parallel, capacitance of second capacitor is 8 μF
Answered by Thiyagarajan K | 02 May, 2019, 12:55: PM
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