NEET Class neet Answered
please answer this
Asked by Prashant DIGHE | 11 Mar, 2020, 10:25: PM
Expert Answer
It is assumed initially all the capacitors are fully charged, because battery is connected to the combination of capacitors.
Given capacitor combination is , a capacitor of capacitance C is in series with parallel combination of 3 capacitors
of each having capacitance C.
Hence effective capacitance Ceff is C in series with 3C ,
i.e effective capacitance is given by (1/Ceff ) = (1/C) + [ 1 / (3C) ]
Hence, we get Ceff = (3/4)C , charge on effective capacitance Q = (3/4) C V
Hence initially left most side of capacitor is having charge Q = ( 3 C V )/4
when switch is closed , capacitors connected in parallel are getting shorted .
Hence after closing the switch, effective capacitance is C due to left most side capacitor in the given circuit
and this capacitor will acquire charge Q = CV.
Since this capacitor is having charge (3/4)CV already, excess charge (1/4) CV will flow from battery
Answered by Thiyagarajan K | 12 Mar, 2020, 09:16: AM
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