# Question #c823d

Jan 21, 2017

Capacitance of 1st capacitor ${C}_{1} = 19 \mu F$
Capacitance of 2nd capacitor ${C}_{2} = 4 \mu F$
When connected in series their equivalent total or capacitance becomes ${C}_{T}$

So ${C}_{T} = \frac{{C}_{1} \times {C}_{2}}{{C}_{1} + {C}_{2}} = \frac{19 \times 4}{19 + 4} = \frac{76}{23} \mu F$

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

$Q = {C}_{T} \times 10 = \frac{76}{23} \mu F \times 10 V = \frac{760}{23} \mu C$

So P.D across 1st capacitor $\frac{Q}{C} _ 1 = \frac{\frac{760}{23} \mu C}{19 \mu F} = \frac{40}{23} V$

And P.D across 2ndcapacitor $\frac{Q}{C} _ 2 = \frac{\frac{760}{23} \mu C}{4 \mu F} = \frac{190}{23} V$