24.53 • in fig. p24.53, $c_1 = c_5 = 8.4 mu f$ and $c_2 = c_3 = c_4 = 4.2 mu f$. (a) what is the equivalent…

Explanation:

Step1: Analyze capacitors in parallel

Capacitors $C_3$, $C_4$ are in parallel. The formula for equivalent - capacitance of parallel capacitors is $C_{eq - p}=C_3 + C_4$. Given $C_3 = C_4=4.2\ \mu F$, then $C_{34}=C_3 + C_4=4.2\ \mu F+4.2\ \mu F = 8.4\ \mu F$.

Step2: Analyze capacitors in series

Now, $C_1$, $C_2$, $C_{34}$, $C_5$ are in series. The formula for equivalent - capacitance of series capacitors is $\frac{1}{C_{eq - s}}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_{34}}+\frac{1}{C_5}$. Given $C_1 = C_5=8.4\ \mu F$, $C_2 = 4.2\ \mu F$, and $C_{34}=8.4\ \mu F$. Substitute the values: $\frac{1}{C_{eq - s}}=\frac{1}{8.4}+\frac{1}{4.2}+\frac{1}{8.4}+\frac{1}{8.4}=\frac{1 + 2+1 + 1}{8.4}=\frac{5}{8.4}$.

Step3: Calculate the equivalent capacitance

Then $C_{eq}=\frac{8.4}{5}=1.68\ \mu F$.

Answer:

$1.68\ \mu F$