Question

six 7.5 - μf capacitors are connected in parallel. part b what is their equivalent capacitance if connected in series? express your answer to two significant figures and include the appropriate units.

Explanation:

Step1: Recall parallel - capacitance formula

For capacitors in parallel, the equivalent capacitance $C_{eq}$ is the sum of individual capacitances. Given $n = 6$ capacitors each of capacitance $C=7.5\ \mu F$. The formula is $C_{eq}=\sum_{i = 1}^{n}C_i$. Since $C_i = C$ for all $i$, we have $C_{eq}=nC$.
$C_{eq}=6\times7.5\ \mu F$

Step2: Calculate parallel equivalent capacitance

$C_{eq}=45\ \mu F$

Step3: Recall series - capacitance formula

For capacitors in series, the reciprocal of the equivalent capacitance $C_{eq}$ is the sum of the reciprocals of individual capacitances. For $n$ identical capacitors of capacitance $C$, the formula is $\frac{1}{C_{eq}}=\sum_{i = 1}^{n}\frac{1}{C_i}$. Since $C_i = C$ for all $i$, we have $\frac{1}{C_{eq}}=\frac{n}{C}$. Then $C_{eq}=\frac{C}{n}$.
$C_{eq}=\frac{7.5\ \mu F}{6}$

Step4: Calculate series equivalent capacitance

$C_{eq}=1.25\ \mu F$. Rounding to two significant figures, $C_{eq}=1.3\ \mu F$

Answer:

Part A:
Value: 45
Units: $\mu F$
Part B:
Value: 1.3
Units: $\mu F$