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Asked by Prashant DIGHE | 11 Mar, 2020, 22:25: PM
answered-by-expert 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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