An ideal 15Vdc source, an ideal 10uF capacitor and an ideal 5uF capacitor are connected in series.?

The above connection is in series hence the effective capacitance would be 50/15 uF.

C= (C1xC2)/(C1+C2)

in series the charge stored by both the capacitor is same so the total charge stored is

Q= CXV

= 50

The voltage drop across 5uF is hence 50/5 =10V

and across another is 5V

There is one case is assume to be correct, Since the capacitor is an ideal one, once fully charged it wont allow any current to pass. When you connect a resistance parralel to it, it discharges and thus voltage is zero.

Steady state Voltage drops across capacitors connected in series are inversely proportional to capacitance values and the the sum of the Voltage drops equals the source Voltage.

1.

(V of 5uF) + 5/10(V of 5uF) = 15V

1.5 (V of 5uF) = 15V

V of 5uF = 10 Volts

2.

After adding the 10 Ohm resistor the Voltage across the 10uF capacitor charges up an additional 5 Volts through the 10 Ohm resistor.to a total Voltage of 10V + 5V = 15 Volts.

Vof 5uF + V of 10uF must = source Voltage = 15V

Vof 5uF = 15V - (V of 15V) = (15 - 15)V = 0 Volts

1) When the circuit is initially connected the same current flows through all of the components. Since charge is current i times time, both caps have the same charge. Also C=Q/V so the voltage on the 5 uF cap is 6.67 volts and the voltage on the 10 uF cap is 3.33 volts.

2) The voltage on the 5 uFcap will discharge to zero.