Question

Question #c823d

Answer 1

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`