# If C1, a 4.7 µF capacitor, and C2, a 3.3 µF capacitor are in series with 18 Vdc applied, what is the voltage across C1?

$7.43 V$
As the two capacitors are in series,their equivalent capacitance is (4.7×3.3)/(4.7+3.3)=1.94 muF
So,total charge flowing in the circuit is Q=CV=1.94×18=34.92 muC
So,voltage drop across $4.7 \mu F$ will be $\frac{34.92}{4.7} = 7.43 V$