Capacitors in Series and Parallel

Draw circuit diagrams of capacitors in series and parallel.
Explain the effects on electric charge, voltage, and capacitance in series and parallel configurations.
Explain why a series connection acts as a voltage divider and a parallel connection acts as a current (charge) divider.
Calculate the equivalent capacitance of capacitor networks in series and/or parallel (STEM_GP12EM-IIId-24).

Multiple connected capacitors behave as a single equivalent capacitor.
The total capacitance depends on individual capacitors and their connection type.
Common connection types are series and parallel, allowing easy calculation of total capacitance.

Series Connection:
- The charge is the same throughout the circuit because capacitors are on a single path.
- Charges only separate on charged plates, conserving charge.
- $QT = Q1 = Q2 = Q3…$
Parallel Connection:
- Charge splits up according to the number of paths.
- The total charge is the sum of individual charges.
- $QT = Q1 + Q2 + Q3…$

Parallel Connection:
- Electric potential (voltage) stays the same throughout.
- The same work is done on a charge no matter which parallel path is taken.
- $VT = V1 = V2 = V3…$
Series Connection:
- Electric potential (voltage) changes across the electric path.
- The total work done is the sum of the work done on individual capacitors.
- $VT = V1 + V2 + V3…$

To find the total capacitance in series ( $C_S$ ), consider the voltage across individual capacitors.
Using $V = \frac{Q}{C}$ , the voltage across individual capacitors is:
- $V1 = \frac{Q}{C1}$ , $V2 = \frac{Q}{C2}$ , $V3 = \frac{Q}{C3}$ respectively.
The total voltage is the sum of individual voltages: $VT = V1 + V2 + V3…$
Substituting the voltage expressions:
- $VT = \frac{Q}{C1} + \frac{Q}{C2} + \frac{Q}{C3}…$
- $\frac{Q}{CS} = \frac{Q}{C1} + \frac{Q}{C2} + \frac{Q}{C3}…$
Canceling the Q's, the equation for total capacitance in series is:
- $\frac{1}{CS} = \frac{1}{C1} + \frac{1}{C2} + \frac{1}{C3}…$
As the number of capacitors increases in a series connection, the total capacitance becomes smaller.
This is because the series connection increases the effective plate separation, leading to smaller capacitance.
C_S < individual capacitances.

Find the total capacitance for three capacitors connected in series with capacitances 1.000 µF, 5.000 µF, and 8.000 µF.
$\frac{1}{C_S} = \frac{1}{1.000} + \frac{1}{5.000} + \frac{1}{8.000}$
$\frac{1}{C_S} = 1 + 0.2 + 0.125 = 1.325$
Inverting to find $C_S$ :
- $C_S = \frac{1}{1.325} = 0.755 \mu F$

To find the total capacitance in parallel ( $C_P$ ), consider the equation $Q = CV$ .
Solving for Q:
- $QT = CP V$
Individual charges are:
- $Q1 = C1 V$ , $Q2 = C2 V$ , $Q3 = C3 V$
Entering these into the equation $QT = Q1 + Q2 + Q3…$ gives:
- $CP V = C1 V + C2 V + C3 V + …$
Canceling V from the equation:
- $CP = C1 + C2 + C3 + …$
The total capacitance in parallel is simply the sum of individual capacitances.
C_P > individual capacitances.

Find the total capacitance for three capacitors connected in parallel with capacitances 1.000 µF, 5.000 µF, and 8.000 µF.
Using the formula $CP = C1 + C2 + C3 + …$
- $C_P = 1.000 + 5.000 + 8.000 = 14.000 \mu F$

As the number of capacitors in parallel increases, the total capacitance becomes larger.
This is because the equivalent capacitor in parallel has a larger plate area, increasing the capacitance.

More complex connections can be combinations of series and parallel.
To find the total capacitance, identify series and parallel parts, compute their capacitances, and then find the total capacitance of the combination.