Question #c823d

1 Answer
P dilip_k
Jan 21, 2017

Capacitance of 1st capacitor #C_1=19muF#
Capacitance of 2nd capacitor #C_2=4muF#
When connected in series their equivalent total or capacitance becomes #C_T#

So #C_T=(C_1xxC_2)/(C_1+C_2)=(19xx4)/(19+4)=76/23muF#

The combination is connected in series across #10V# . So each capacitor will have same charge and that will be

#Q=C_Txx10=76/23muFxx10V=760/23muC#

So P.D across 1st capacitor #Q/C_1=(760/23muC)/(19muF)=40/23V#

And P.D across 2ndcapacitor #Q/C_2=(760/23muC)/(4muF)=190/23V#

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