If a 0.047 $\mu F$, 0.047 $\mu \mathrm { F }$ and 0.47 $\mu \mathrm { F }$ capacitor are connected in series across a 25 V source, the total capacitance equals ____________and each 0.047 $\mu F$ capacitor drops___
A) 0.564 $\mu F$,1.2V
B) 0.022 $\mu \mathrm { F }$,25 V
C) 0.022 $\mu F$,11.9 V
D) 0.564 $\mu F$,11.9 V

Question

If a  0.047 $\mu F$, 0.047 $\mu \mathrm { F }$ and  0.47 $\mu \mathrm { F }$ capacitor are connected in series across a  25 V source, the total capacitance equals ____________and each  0.047 $\mu F$ capacitor drops___
A)  0.564 $\mu F$,1.2V
B)  0.022  $\mu \mathrm { F }$,25 V
C)  0.022 $\mu F$,11.9 V
D)  0.564  $\mu F$,11.9 V

Quizplus · Accepted Answer

The total capacitance in series is given by $1/C_{total} = 1/C_1 + 1/C_2 + 1/C_3$. Substituting the given values, we find $C_{total} = 0.022 \mu F$. The voltage drop across each capacitor in a series is proportional to its capacitance, so the 0.047 $\mu F$ capacitors will each drop 11.9 V.

If a 0047 μF\mu FμF , 0047 μF\mu \mathrm { F }μF

If a 0047 $\mu F$ , 0047 $\mu \mathrm { F }$