Capacitors in series and parallel

Capacitors, like resistors, can be placed in series and in parallel. Increasing levels of capacitance can be obtained by placing capacitors in parallel, while decreasing levels can be obtained by placing capacitors in series.
For capacitors in series, the charge is the same on each capacitor (Fig. 1):
$$\bbox[10px,border:1px solid grey]{Q_T = Q_1 = Q_2 = Q_3} \tag{1}$$ Applying Kirchhoff's voltage law around the closed loop gives $$ E = V_1 + V_2 + V_3 $$ However, $$ V = Q/C $$ so that, $$ {Q_T \over C_T} = {Q_1 \over C_1} + {Q_2 \over C_2}+ {Q_3 \over C_3}$$
Fig. 1: Series capacitors
Using Eq. (1) and dividing both sides by Q yields $$ \bbox[10px,border:1px solid grey]{{1 \over C_T} = {1 \over C_1} + {1 \over C_2}+ {1 \over C_3}} \tag{2}$$ which is similar to the manner in which we found the total resistance of a parallel resistive circuit. The total capacitance of two capacitors in series is $$ \bbox[10px,border:1px solid grey]{C_T = {C_1 C_2 \over C_1+C_2}}\tag{3}$$ The voltage across each capacitor of Fig. 1 can be found by first recognizing that $$Q_T = Q_1$$ or $$ C_T E = C_1 V_1$$ Solving for V1: $$V_1 = {C_TE \over C_1}$$ A similar equation will result for each capacitor of the network.
Fig. 2: Parallel capacitors
For capacitors in parallel, as shown in Fig. 2, the voltage is the same across each capacitor, and the total charge is the sum of that on each capacitor: $$\bbox[10px,border:1px solid grey]{Q_T = Q_1 + Q_2 + Q_3} \tag{4}$$ However, $$Q = CV$$ Therefore, $$C_TE = C_1V_1 + C_2V_2 + C_3V_3$$ but $$E = V_1 = V_2 = V_3 $$ Thus, $$\bbox[10px,border:1px solid grey]{C_T = C_1+ C_2 + C_3}$$ which is similar to the manner in which the total resistance of a series circuit is found.
Example 1: For the circuit of Fig. 3:
a. Find the total capacitance.
b. Determine the charge on each plate.
c. Find the voltage across each capacitor.
Fig. 3: Example 1
Solution:
a. $$ \begin{split} {1 \over C_T} &= {1 \over C_1} + {1 \over C_2}+ {1 \over C_3}\\ &= {1 \over 200 \times 10^{-6} F} + {1 \over 50 \times 10^{-6} F}+ {1 \over 10 \times 10^{-6} F}\\ &= 0.005 \times 10^{6} + 0.02 \times 10^{6}+ 0.1 \times 10^{6}\\ &= 0.125 \times 10^{6}\\ {1 \over C_T} &= 0.125 \times 10^{6}\\ C_T &= {1 \over 0.125 \times 10^{6}} = 8 \mu F \end{split}$$ b.In series capacitors $$ \begin{split} Q_T &= Q_1 = Q_2 = Q_3 \\ &= C_T E = (8 \times 10^{-6} F)(60V) \\ &= 480 \mu F \\ \end{split}$$ c. $$V_1 = {Q_1 \over C_1} = {480 \mu F \over 200 \mu F } = 2.4 V$$ $$V_2 = {Q_2 \over C_2} = {480 \mu F \over 50 \mu F } = 9.6 V$$ $$V_3 = {Q_3 \over C_3} = {480 \mu F \over 10 \mu F } = 48 V$$ reverse check: $$E = V_1+V_2+V_3 = 2.4+9.6+48 = 60V$$
Example 2: Find the voltage across and charge on each capacitor of the network of Fig. 4 after each has charged up to its final value.
Fig. 4:Example 2
Solution:
After the capacitor has been charged up to its final value, the network can be redrawn as follows;
Fig. 5:Capacitor acting as open circuit after fully charged.
$$ V_{C1} = {(2Ω)(72 V) \over (2Ω + 7Ω)}= 16 V$$ $$ V_{C2} = {(7Ω)(72 V) \over (2Ω + 7Ω)}= 56 V$$ Now we know voltage and capacitance across each capacitor, we can find $$ Q_{1} = C_1 V_{C1} = (2 \times 10^{-6} F)(16V) = 32 \mu C$$ $$ Q_{2} = C_2 V_{C2} = (3 \times 10^{-6} F)(56V) = 168 \mu C$$