Question

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?

Answer 1

`7.43 V`

Explanation:

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 muF` will be `34.92/4.7=7.43V`