Question

24.53 • in fig. p24.53, c1 = c5 = 8.4 μf and c2 = c3 = c4 = 4.2 μf. (a) what is the equivalent capacitance of the network between points a and b?

Explanation:

Step1: Identify parallel - connected capacitors

Capacitors $C_3$, $C_2$ and $C_4$ are in parallel. The formula for equivalent capacitance of capacitors in parallel is $C_{eq - p}=C_2 + C_3+C_4$.
$C_{eq - p}=4.2\ \mu F+4.2\ \mu F + 4.2\ \mu F=12.6\ \mu F$

Step2: Identify series - connected capacitors

Now, $C_1$, $C_{eq - p}$ and $C_5$ are in series. The formula for equivalent capacitance of capacitors in series is $\frac{1}{C_{total}}=\frac{1}{C_1}+\frac{1}{C_{eq - p}}+\frac{1}{C_5}$.
Substitute $C_1 = C_5=8.4\ \mu F$ and $C_{eq - p}=12.6\ \mu F$ into the formula:
$\frac{1}{C_{total}}=\frac{1}{8.4}+\frac{1}{12.6}+\frac{1}{8.4}$
First, find a common denominator, which is 25.2.
$\frac{1}{C_{total}}=\frac{3 + 2+3}{25.2}=\frac{8}{25.2}$
Then, $C_{total}=\frac{25.2}{8}=3.15\ \mu F$

Answer:

$3.15\ \mu F$