Question

consider the combination of capacitors shown in the diagram, where $c_1 = 3.00 mu f$, $c_2 = 11.0 mu f$, $c_3 = 3.00 mu f$, and $c_4 = 5.00 mu f$. (figure 1)

part a
find the equivalent capacitance $c_a$ of the network of capacitors. express your answer in microfarads.

part b
two capacitors of capacitance $c_5 = 6.00 mu f$ and $c_6 = 3.00 mu f$ are added to the network. find the equivalent capacitance $c_b$ of the new network of capacitors.

Explanation:

Step1: Analyze $C_2$ and $C_3$ parallel

For parallel - connected capacitors, the equivalent capacitance $C_{23}$ of $C_2$ and $C_3$ is given by the sum of their capacitances.
$C_{23}=C_2 + C_3$
Substitute $C_2 = 11.0\ \mu F$ and $C_3=3.00\ \mu F$ into the formula:
$C_{23}=11.0\ \mu F+3.00\ \mu F = 14.0\ \mu F$

Step2: Analyze $C_{23}$ and $C_4$ in series

For series - connected capacitors, the equivalent capacitance $C_{234}$ of $C_{23}$ and $C_4$ is given by $\frac{1}{C_{234}}=\frac{1}{C_{23}}+\frac{1}{C_4}$.
Substitute $C_{23}=14.0\ \mu F$ and $C_4 = 5.00\ \mu F$ into the formula:
$\frac{1}{C_{234}}=\frac{1}{14.0\ \mu F}+\frac{1}{5.00\ \mu F}=\frac{5 + 14}{70\ \mu F}=\frac{19}{70\ \mu F}$
$C_{234}=\frac{70}{19}\ \mu F\approx3.68\ \mu F$

Step3: Analyze $C_1$ and $C_{234}$ in series

The equivalent capacitance $C_A$ of the whole network (where $C_1$ and $C_{234}$ are in series) is given by $\frac{1}{C_A}=\frac{1}{C_1}+\frac{1}{C_{234}}$.
Substitute $C_1 = 3.00\ \mu F$ and $C_{234}=\frac{70}{19}\ \mu F$ into the formula:
$\frac{1}{C_A}=\frac{1}{3.00\ \mu F}+\frac{19}{70\ \mu F}=\frac{70+57}{210\ \mu F}=\frac{127}{210\ \mu F}$
$C_A=\frac{210}{127}\ \mu F\approx1.65\ \mu F$

Answer:

$1.65\ \mu F$