A 5.0- $\mu \mathrm { F }$ and a 12.0- $\mu \mathrm { F }$ capacitor are connected in series, and the series arrangement is connected in parallel to a 29.0- $\mathrm { F }$ capacitor. How much capacitance would a single capacitor need to replace this combination of three capacitors?
A) $13 \mu \mathrm { F }$
B) $33 \mu \mathrm { F }$
C) $38 \mu \mathrm { F }$
D) $16 \mu \mathrm { F }$

Question

Quizplus · Accepted Answer

First, calculate the equivalent capacitance of the series combination: $C_{\text{series}} = \left(\frac{1}{5.0} + \frac{1}{12.0}\right)^{-1} \mu F = \left(\frac{12 + 5}{60}\right)^{-1} \mu F = \frac{60}{17} \mu F = 3.53 \mu F$. Then, add this to the parallel capacitor: $C_{\text{total}} = 3.53 \mu F + 29.0 \mu F = 32.53 \mu F$, which rounds to $33 \mu F$.

A 50- μF\mu \mathrm { F }μF And a 120- μF\mu \mathrm { F }μF

A 50- $\mu \mathrm { F }$ And a 120- $\mu \mathrm { F }$